1105 chapter sixteen - valenciafd.valenciacollege.edu/file/tsmith143/ch16.pdf16.1 solutions 1105...

16.1 SOLUTIONS 1105

CHAPTER SIXTEEN

Solutions for Section 16.1

Exercises

1. Mark the values of the function on the plane, as shown in Figure 16.1, so that you can guess respectively at the smallestand largest values the function takes on each small rectangle.

Lower sum =∑

f(xi, yi)∆x∆y

= 4∆x∆y + 6∆x∆y + 3∆x∆y + 4∆x∆y

= 17∆x∆y

= 17(0.1)(0.2) = 0.34.

Upper sum =∑

f(xi, yi)∆x∆y

= 7∆x∆y + 10∆x∆y + 6∆x∆y + 8∆x∆y

= 31∆x∆y

= 31(0.1)(0.2) = 0.62.

1.0 1.1 1.22.0

2.2

2.4

← x

↓y

5

4

3

7

6

5

10

8

4

-� ∆x = 0.1

6

?

∆y = 0.2

Figure 16.1

2. In the subrectangle in the top left in Figure 16.4, it appears that f(x, y) has a maximum value of about 9. In the subrect-angle in the top middle, f(x, y) has a maximum value of 10. Continuing in this way, and multiplying by ∆x and ∆y, wehave

Overestimate = (9 + 10 + 12 + 7 + 8 + 10 + 5 + 7 + 8)(10)(5) = 3800.

Similarly, we findUnderestimate = (7 + 7 + 8 + 4 + 5 + 7 + 1 + 3 + 6)(10)(5) = 2400.

Thus, we expect that

2400 ≤∫

R

f(x, y)dA ≤ 3800.

3. (a) If we take the partition of R consisting of just R itself, we get

Lower bound for integral = minRf ·AR = 0 · (4− 0)(4− 0) = 0.

Similarly, we getUpper bound for integral = maxRf ·AR = 4 · (4− 0)(4− 0) = 64.

1106 Chapter Sixteen /SOLUTIONS

(b) The estimates asked for are just the upper and lower sums. We partition R into subrectangles R(a,b) of width 2 andheight 2, where (a, b) is the lower-left corner ofR(a,b). The subrectangles are thenR(0,0),R(2,0),R(0,2), andR(2,2).

(0, 0)

(0, 4)

(4, 0)

(4, 4)

Figure 16.2

Then we find the lower sum

Lower sum =∑

(a,b)

AR(a,b)· minR(a,b)

f =∑

(a,b)

4 · (Min of f on R(a,b))

= 4∑

(a,b)

(Min of f on R(a,b))

= 4(f(0, 0) + f(2, 0) + f(0, 2) + f(2, 2))

= 4(√

0 · 0 +√

2 · 0 +√

0 · 2 +√

2 · 2)

= 8.

Similarly, the upper sum is

Upper sum = 4∑

(a,b)

(Max of f on R(a,b))

= 4(f(2, 2) + f(4, 2) + f(2, 4) + f(4, 4))

= 4(√

2 · 2 +√

4 · 2 +√

2 · 4 +√

4 · 4)

= 24 + 16√

2 ≈ 46.63.

The upper sum is an overestimate and the lower sum is an underestimate, so we can get a better estimate by averagingthem to get 16 + 8

√2 ≈ 27.3.

4. (a) We first find an over- and underestimate of the integral, using four subrectangles. On the first subrectangle (0 ≤ x ≤3, 0 ≤ y ≤ 4), the function f(x, y) appears to have a maximum of 100 and a minimum of 79. Continuing in thisway, and using the fact that ∆x = 3 and ∆y = 4, we have

Overestimate = (100 + 90 + 85 + 79)(3)(4) = 4248,

andUnderestimate = (79 + 68 + 61 + 55)(3)(4) = 3156.

A better estimate of the integral is the average of the overestimate and the underestimate:

Better estimate =4248 + 3156

2= 3702.

(b) The average value of f(x, y) on this region is the value of the integral divided by the area of the region. Since thearea of R is (6)(8) = 48, we approximate

Average value =1

Area

∫

R

f(x, y)dA ≈ 1

48· 3702 = 77.125.

We see in the table that the values of f(x, y) on this region vary between 55 and 100, so an average value of 77.125is reasonable.

5. Partition R into subrectangles with the lines x = 0, x = 0.5, x = 1, x = 1.5, and x = 2 and the lines y = 0, y = 1,y = 2, y = 3, and y = 4. Then we have 16 subrectangles, each of which we denote R(a,b), where (a, b) is the location ofthe lower-left corner of the subrectangle.

16.1 SOLUTIONS 1107

We want to find a lower bound and an upper bound for the volume above each subrectangle. The lower bound for thevolume of R(a,b) is

0.5(Min of f on R(a,b))

because the area of R(a,b) is 0.5 · 1 = 0.5. The function f(x, y) = 2 + xy increases with both x and y over the wholeregion R, as shown in Figure 16.3. Thus,

Min of f on R(a,b) = f(a, b) = 2 + ab,

because the minimum on each subrectangle is at the corner closest to the origin.

xy

z

Figure 16.3

(a, b)

� R(a, b)

y

2

4

x

(a, b)

Figure 16.4

Similarly,Max of f on R(a,b) = f(a+ 0.5, b+ 1) = 2 + (a+ 0.5)(b+ 1).

So we have

Lower sum =∑

(a,b)

0.5(2 + ab) = 0.5∑

(a,b)

(2 + ab)

= 16 + 0.5∑

(a,b)

ab

Since a = 0, 0.5, 1, 1.5 and b = 0, 1, 2, 3, expanding this sum gives

Lower sum = 16 + 0.5 ( 0 · 0 + 0 · 1 + 0 · 2 + 0 · 3+ 0.5 · 0 + 0.5 · 1 + 0.5 · 2 + 0.5 · 3+ 1 · 0 + 1 · 1 + 1 · 2 + 1 · 3+ 1.5 · 0 + 1.5 · 1 + 1.5 · 2 + 1.5 · 3)

= 25.

Similarly, we can compute the upper sum:

Upper sum =∑

(a,b)

0.5(2 + (a+ 0.5)(b+ 1)) = 0.5∑

(a,b)

(2 + (a+ 0.5)(b+ 1))

= 16 + 0.5∑

(a,b)

(a+ 0.5)(b+ 1)

= 41.

6. Since f(x, y) is measured in micrograms per square meter, and we are integrating over an area measured in squaremeters, the units of the integral are micrograms. The integral represents the total quantity of pollution, in micrograms, inthe region R.

7. The exact value of the integral is 40/3.

8. The value of the integral is around −2.4.

Problems

9. The function being integrated is f(x, y) = 1, which is positive everywhere. Thus, its integral over any region is positive.


10. The function being integrated is f(x, y) = 1, which is positive everywhere. Thus, its integral over any region is positive.

11. The function being integrated is f(x, y) = 5x. Since x > 0 in R, f is positive in R and thus the integral is positive.

12. The function being integrated is f(x, y) = 5x, which is an odd function in x. Since B is symmetric with respect to x, thecontributions to the integral cancel out, as f(x, y) = −f(−x, y). Thus, the integral is zero.

13. The region D is symmetric both with respect to x and y axes. The function being integrated is f(x, y) = 5x, which is anodd function in x. Since D is symmetric with respect to x, the contributions to the integral cancel out. Thus, the integralof the function over the region D is zero.

14. The function being integrated, f(x, y) = y3 + y5, is an odd function in y while D is symmetric with respect to y. Then,by symmetry, the positive and negative contributions of f will cancel out and thus its integral is zero.

15. In a region such as B in which y < 0, the quantity y3 + y5 is less than zero. Thus, its integral is negative.

16. The region R is symmetric with respect to y and the integrand is an odd function in y, so the integral over R is zero.

17. The function being integrated, f(x, y) = y − y3 is always negative in the region B since in that region −1 < y < 0 and|y3| < |y|. Thus, the integral is negative.

18. The function being integrated, f(x, y) = y − y3, is an odd function in y while D is symmetric with respect to y. Bysymmetry, the integral is zero.

19. The region D is symmetric both with respect to x and y axes. The function being integrated is odd with respect to y in theregion D. Thus, its integral is zero.

20. Since D is a disk of radius 1, in the region D, we have |y| < 1. Thus, −π/2 < y < π/2. Thus, cos y is always positivein the region D and thus its integral is positive.

21. The function f(x, y) = ex is positive for any value of x. Thus, its integral is always positive for any region, such as D,with nonzero area.

22. The region D is symmetric both with respect to x and y axes. Looking at the contributions to the integral of the functionf(x, y) = xex, we can see that any contribution made by the point (x, y), where x > 0, is greater than the correspondingcontribution made by (−x, y), since ex > 1 > e−x for x > 0. Thus, the integral of f in the region D is positive.

23. The region D is symmetric both with respect to x and y axes. The function f(x, y) = xy2 is odd with respect to x, andthus the contributions to the integral from (x, y) and (−x, y) cancel. Thus the integral is zero in the region D.

24. The function f(x, y) is odd with respect to x, and thus the integral is zero in region B, which is symmetric with respectto x.

25. We use four subrectangles to find an overestimate and underestimate of the integral:

Overestimate = (15 + 9 + 9 + 5)(4)(3) = 456,

Underestimate = (5 + 2 + 3 + 1)(4)(3) = 132.

A better estimate of the integral is the average of the two:∫

R

f(x, y)dA ≈ 456 + 132

2= 294.

The units of the integral are milligrams, and the integral represents the total number of mg of mosquito larvae in this 8meter by 6 meter section of swamp.

26. The question is asking which graph has more volume under it, and from inspection, it appears that it would be the graphfor the mosquitos.

27. Let’s break up the room into 25 sections, each of which is 1 meter by 1 meter and has area ∆A = 1.We shall begin our sum as an upper estimate starting with the lower left corner of the room and continue across the

bottom and moving upward using the highest temperature, Ti, in each case. So the upper Riemann sum becomes∑25

i=1Ti∆A = T1∆A+ T2∆A+ T3∆A+ · · ·+ T25∆A

= ∆A(T1 + T2 + T3 + · · ·+ T25)

= (1)(31 +29 + 28 + 27 + 27+

29 +28 + 27 + 27 + 26+

27 +27 + 26 + 26 + 26+

26 +26 + 25 + 25 + 25+

25 +24 + 24 + 24 + 24)

= (1)(659) = 659.

16.1 SOLUTIONS 1109

In the same way, the lower Riemann sum is formed by taking the lowest temperature, ti, in each case:∑25

i=1ti∆A = t1∆A+ t2∆A+ t3∆A+ · · ·+ t25∆A

= ∆A(t1 + t2 + t3 + · · ·+ t25)

= (1)(27 +27 + 26 + 26 + 25+

26 +26 + 25 + 25 + 25+

25 +24 + 24 + 24 + 24+

24 +23 + 23 + 23 + 23+

23 +21 + 20 + 21 + 22)

= (1)(602) = 602.

So, averaging the upper and lower sums we get: 630.5.To compute the average temperature, we divide by the area of the room, giving

Average temperature =630.5

(5)(5)≈ 25.2◦C.

Alternatively we can use the temperature at the central point of each section ∆A. Then the sum becomes∑25

i=1T ′i∆A = ∆A

∑25

i=1T ′i

= (1)(29 +28 + 27 + 26.5 + 26+

27 +27 + 26 + 26 + 25.5+

26 +25.5 + 25 + 25 + 25+

25 +24 + 24 + 24 + 24+

24 +23 + 22 + 22.5 + 23)

= (1)(630) = 630.

Then we get

Average temperature =

∑25

i=1T ′i∆A

Area=

630

(5)(5)≈ 25.2◦C.

28. The total area of the squareR is (1.5)(1.5) = 2.25. See Figure 16.5. On a disk of radius≈ 0.5 the function has a value of 3or more, giving a total contribution to the integral of at least (3)·(π ·0.52) ≈ 2.3. On less than half of the rest of the squarethe function has a value between−2 and 0, giving a contribution to the integral of between (1/2·2.25)(−2) = −2.25 and0. Since the positive contribution to the integral is therefore greater in magnitude than the negative contribution,

∫Rf dA

is positive.

−1.0 −0.5 0 0.5 1.0 1.5 2.0−1.0

−0.5

0

0.5

1.0

1.5

2.0

x

y

−1

0

1

2

2

3

4

Figure 16.5


29. The total number of tornados, per year, in a certain region, is the integral of the frequency of tornados in that region. Inorder to approximate it, we first subdivide the states into smaller regions of 100 miles2, as shown in the figure in the text.We will find the upper and lower bounds for the frequency of tornados, and then take the average of the two. We do this byfinding the highest frequency of tornados in each subdivision, and then add up all the frequencies, and then do the samefor the low frequencies.

(a) For the high frequency in Texas, going left to right, top to bottom, we get: [(9 + 8) + (9 + 8 + 9 + 9 + 8) + (7 +7 + 9 + 9 + 8) + (0 + 3 + 4 + 6 + 8 + 7 + 7 + 6) + (0 + 1 + 4 + 6 + 7 + 7) + (2 + 4 + 5) + (2 + 3)]. This equals182 tornados. For the low frequency, we get: [(7 + 7) + (7 + 7 + 7 + 7 + 5) + (3 + 5 + 6 + 7 + 5) + (0 + 0 + 0 +1 + 4 + 6 + 7 + 6) + (0 + 0 + 0 + 3 + 5 + 5) + (0 + 1 + 3) + (0 + 0)]. This equals 114 tornados. The average ofthe two equals (182 + 114)/2 = 148 tornados.

(b) For the high frequency in Florida, going left to right, top to bottom, we get: [(5+5+5+9)+(9+9)+(7+7)+(5)].This equals 61 tornados. For the low frequency, we get: [(5 + 5 + 5 + 5) + (7 + 7) + (7 + 5) + (5)]. This equals 51tornados. The average of the two is equal to (61 + 51)/2 = 56 tornados.

(c) For the high frequency in Arizona, going left to right, top to bottom, we get: [(1+1+0)+(1+1+0)+(0+0+0)+(0 + 0)]. This equals 4 tornados. For the low frequency, we get: [(0 + 0 + 0) + (0 + 0 + 0) + (0 + 0 + 0) + (0 + 0)].This equals 0 tornados. The average of the two is equal to (4 + 0)/2 = 2 tornados.

30. Let R be the region 0 ≤ x ≤ 60, 0 ≤ y ≤ 8. Then

Volume =

∫

R

w(x, y) dA

Lower estimate: 10·2(1+4+8+10+10+8+0+3+4+6+6+4+0+1+2+3+3+2+0+0+1+1+1+1) = 1580.Upper estimate:10 ·2(8+13+16+17+17+16+4+8+10+11+11+10+3+4+6+7+7+6+1+2+3+4+4+3) = 3820.The average of the two estimates is 2700 cubic feet.

31. (a) The graph of f looks like the graph of g in the xz-plane slid parallel to itself forward and backward in the y direction,since the value of y does not affect the value of f .

(b) The solid bounded by the graph of f is a cylinder, hence∫

R

f dA = Volume under surface =

(Length iny direction

)(Cross-sectional area

in xz-plane

)

= (d− c)∫ b

a

g(x) dx

32. (a) On the xy-plane, z = 0, so the equation of the edge of the base is x2 + y2 = 15, a circle of radius√

15.(b) The area of the base is π(

√15)2 = 15π meters2.

(c) The cross-section at z = 10 has equation x2 + y2 = 5, a circle of radius√

5.(d) The area of the cross-section is π(

√5)2 = 5π meters2.

(e) At height z, the cross-section is the circlex2 + y2 = 15− z.

This is a circle of radius√

15− z, so

A(z) = π(√

15− z)2 = (15− z)π meters2.

(f) The approximate volume of the slice is A(z)∆z meters3.(g) We have

Volume of pile =

∫ 15

0

(15− z)π dz =

(15z − z2

2

)π

∣∣∣∣15

0

=225π

2meters3.

33. Letf(x, y) =

1

1 + a1x2 + y2

and letg(x, y) =

1

1 + a2x2 + y2

16.2 SOLUTIONS 1111

where 0 < a1 < a2. For all points (x, y) we have f(x, y) ≥ g(x, y) > 0, so in Riemann sum approximations for∫Rf dA and

∫Rg dA using the same subdivision of R for both integrals, we have

∑f(xi, yj)∆x∆y ≥

∑g(xi, yj)∆x∆y.

It follows that∫Rf dA ≥

∫Rg dA, and so increasing the value of a decreases the value of the integral

∫R

11+ax2+y2 dA.

34. Take a Riemann sum approximation to ∫

R

fdA ≈∑

i,j

f(xi, yi)∆A.

Then, using the fact that |a+ b| ≤ |a|+ |b| repeatedly, we have:∣∣∣∣∫

R

fdA

∣∣∣∣ ≈ |∑

i,j

f(xi, yi)∆A| ≤∑|f(xi, yi)∆A|.

Now |f(xi, yj)∆A| = |f(xi, yj |∆A since ∆A is non-negative, so∣∣∣∣∫

R

fdA

∣∣∣∣ ≤∑

i,j

|f(xi, yi)∆A| =∑

i,j

|f(xi, yj)|∆A.

But the last expression on the right is a Riemann sum approximation to the integral∫R|f |dA, so we have

∣∣∣∣∫

R

fdA

∣∣∣∣ ≈∣∣∣∣∣∑

i,j

f(xi, yj)∆A

∣∣∣∣∣ ≤∑

i,j

|f(xi,j )|∆A ≈∫

R

|f |dA.

Thus, ∣∣∣∣∫

R

fdA

∣∣∣∣ ≤∫

R

|f |dA.


Exercises

1. We evaluate the inside integral first:∫ 4

0

(4x+ 3y) dx = (2x2 + 3yx)

∣∣∣4

0= 32 + 12y.

Therefore, we have∫ 3

0

∫ 4

0

(4x+ 3y) dxdy =

∫ 3

0

(32 + 12y) dy = (32y + 6y2)∣∣∣3

0= 150.


0

(x2 + y2) dy =

(x2y +

y3

3

)∣∣∣∣y=3

y=0

= 3x2 + 9.

Therefore, we have ∫ 2

0

∫ 3

0

(x2 + y2) dydx =

∫ 2

0

(3x2 + 9) dx = (x3 + 9x)∣∣∣2

0= 26.

3. We evaluate the inside integral first: ∫ 2

0

(6xy) dy = (3xy2)∣∣∣2

0= 12x.


0

∫ 2

0

(6xy) dydx =

∫ 3

0

(12x) dx = (6x2)

∣∣∣3

0= 54.



0

(x2y) dy =

(x2y2

2

)∣∣∣∣y=2

y=0

= 2x2.


0

∫ 2

0

(x2y) dydx =

∫ 1

0

(2x2) dx =

(2x3

3

)∣∣∣∣1

0

=2

3.

5.∫ 4

1

∫ 2

1

f dy dx or∫ 2

1

∫ 4

1

f dx dy

6. This region lies between x = 0 and x = 4 and between the lines y = 3x and y = 12, and so the iterated integral is∫ 4

0

∫ 12

3x

f(x, y) dydx.

Alternatively, we could have set up the integral as follows:∫ 12

0

∫ y/3

0

f(x, y) dxdy.

7. The line connecting (−1, 1) and (3,−2) is3x+ 4y = 1

ory =

1− 3x

4So the integral becomes ∫ 3

−1

∫ (1−3x)/4

−2

f dy dx or∫ 1

−2

∫ (1−4y)/3

−1

f dx dy

8. The line on the left (through points (0, 0) and (3, 6)) is the line y = 2x; the line on the right (through points (3, 6) and(5, 0)) is the line y = −3x+ 15. See Figure 16.6. One way to set up this iterated integral is:

∫ 6

0

∫ (15−y)/3

y/2

f(x, y) dxdy.

The other option for setting up this integral requires two separate integrals, as follows:∫ 3

0

∫ 2x

0

f(x, y) dydx+

∫ 5

3

∫ −3x+15

0

f(x, y) dydx.

R

y = 2x

y = −3x+ 15

x

y

Figure 16.6

9. Two of the sides of the triangle have equations x =y − 1

2and x =

y − 5

−2. So the integral is

∫ 3

1

∫ − 12

(y−5)

12

(y−1)

f dx dy

16.2 SOLUTIONS 1113

10. The line connecting (1, 0) and (4, 1) is

y =1

3(x− 1)

So the integral is ∫ 4

1

∫ 2

(x−1)/3

f dy dx

11.∫ 3

1

∫ 4

0

ex+y dxdy =

∫ 3

1

exey∣∣∣∣4

0

dx =

∫ 3

1

ex(e4 − 1) dx = (e4 − 1)(e2 − 1)e. See Figure 16.7.

1 3

y = 4

x

y

Figure 16.7

22x

yy = x

Figure 16.8

12.∫ 2

0

∫ x

0

ex2

dydx =

∫ 2

0

ex2

y

∣∣∣∣x

0

dx =

∫ 2

0

xex2

dx =1

2ex

2

∣∣∣∣2

0

=1

2(e4 − 1). See Figure 16.8.

13.∫ 5

1

∫ 2x

x

sinx dy dx =

∫ 5

1

sinx · y∣∣2xxdx

=

∫ 5

1

sinx · x dx

= (sinx− x cosx)∣∣51

= (sin 5− 5 cos 5)− (sin 1− cos 1) ≈ −2.68.

See Figure 16.9.

1 5x

y

y = x

y = 2x

Figure 16.9

1 2 4

1

4

x

y

y = 4

y = xx =√y

Figure 16.10

14.∫ 4

1

∫ y

√y

x2y3 dxdy =

∫ 4

1

y3 x3

3

∣∣∣∣y

√y

dy


=1

3

∫ 4

1

(y6 − y 92 ) dy

=1

3

(y7

7− y11/2

11/2

)∣∣∣∣4

1

=1

3

[(47

7− 411/2 × 2

11

)−(

1

7− 2

11

)]≈ 656.082

See Figure 16.10.

15. The region of integration ranges from x = 0 to x = 3 and from y = 0 to y = 2x, as shown in Figure 16.11. To evaluatethe integral, we evaluate the inside integral first:

∫ 2x

0

(x2 + y2) dy =

(x2y +

y3

3

)∣∣∣∣y=2x

y=0

= x2(2x) +(2x)3

3= 2x3 +

8x3

3=

14

3x3.


0

∫ 2x

0

(x2 + y2) dydx =

∫ 3

0

(14

3x3)dx =

(14

12x4) ∣∣∣

3

0= 94.5.

3

6 y = 2x

x

y

Figure 16.11

−3 −2 3

−3

3

x

y

R

x2 + y2 = 9

Figure 16.12

16. See Figure 16.12.∫ 0

−2

∫ 0

−√

9−x2

2xy dydx =

∫ 0

−2

x y2

∣∣∣∣0

−√

9−x2

dx

= −∫ 0

−2

x(9− x2) dx

=

∫ 0

−2

(x3 − 9x) dx

=

(x4

4− 9

2x2

)∣∣∣∣0

−2

= −4 + 18 = 14

17.∫

R

√x+ y dA =

∫ 2

0

∫ 1

0

√x+ y dx dy

=

∫ 2

0

2

3(x+ y)

32

∣∣∣∣1

0

dy

=2

3

∫ 2

0

((1 + y)32 − y 3

2 ) dy

16.2 SOLUTIONS 1115

=2

3· 2

5[(1 + y)

52 − y 5

2 ]

∣∣∣∣2

0

=4

15((3

52 − 2

52 )− (1− 0))

=4

15(9√

3− 4√

2− 1) = 2.38176

18. In the other order, the integral is ∫ 1

0

∫ 2

0

√x+ y dy dx.

First we keep x fixed and calculate the inside integral with respect to y:

∫ 2

0

√x+ y dy =

2

3(x+ y)3/2

∣∣∣∣y=2

y=0

=2

3

[(x+ 2)3/2 − x3/2

].

Then the outside integral becomes

∫ 1

0

2

3

[(x+ 2)3/2 − x3/2

]dx =

2

3

[2

5(x+ 2)5/2 − 2

5x5/2

] ∣∣∣∣1

0

=2

3· 2

5

[35/2 − 1− 25/2

]= 2.38176

Note that the answer is the same as the one we got in Exercise 17.

19.∫

R

(5x2 + 1) sin 3y dA =

∫ 1

−1

∫ π/3

0

(5x2 + 1) sin 3y dy dx

=

∫ 1

−1

(5x2 + 1)

(−1

3cos 3y

∣∣∣π/3

0

)dx

=2

3

∫ 1

−1

(5x2 + 1) dx

=2

3(5

3x3 + x)

∣∣∣∣1

−1

=2

3(10

3+ 2) =

32

9

20. The region of integration, R, is shown in Figure 16.13.Integrating first over y, as shown in the diagram, we obtain

∫

R

xy dA =

∫ 1

0

(∫ 1−x

0

xy dy

)dx =

∫ 1

0

xy2

2

∣∣∣∣1−x

0

dx =

∫ 1

0

1

2x(1− x)2 dx

Now integrating with respect to x gives∫

R

xy dA =(

1

4x2 − 1

3x3 +

1

8x4)∣∣∣

1

0=

1

24.


1

1

y = 1− x

x

y

Figure 16.13

x

y

−1 1

1

y = −x+ 1y = x+ 1

Figure 16.14

21. It would be easier to integrate first in the x direction from x = y − 1 to x = −y + 1, because integrating first in the ydirection would involve two separate integrals.

∫

R

(2x+ 3y)2 dA =

∫ 1

0

∫ −y+1

y−1

(2x+ 3y)2 dx dy

=

∫ 1

0

∫ −y+1

y−1

(4x2 + 12xy + 9y2) dx dy

=

∫ 1

0

[4

3x3 + 6x2y + 9xy2

]−y+1

y−1

dy

=

∫ 1

0

[8

3(−y + 1)3 + 9y2(−2y + 2)] dy

=

[− 2

3(−y + 1)4 − 9

2y4 + 6y3

]1

0

= −2

3(−1)− 9

2+ 6 =

13

6

See Figure 16.14.

22. The region is bounded by x = 1, x = 4, y = 2, and y = 2x. Thus

Volume =

∫ 4

1

∫ 2x

2

(6x2y) dydx.

To evaluate this integral, we evaluate the inside integral first:∫ 2x

2

(6x2y) dy = (3x2y2)

∣∣∣2x

2= 3x2(2x)2 − 3x2(22) = 12x4 − 12x2.


1

∫ 2x

2

(6x2y) dydx =

∫ 4

1

(12x4 − 12x2) dx =(

12

5x5 − 4x3

) ∣∣∣4

1= 2203.2.

The volume of this object is 2203.2.

23. To find the average value, we evaluate the integral∫ 3

0

∫ 6

0

(x2 + 4y) dydx,

and then divide by the area of the base region.To evaluate this integral, we evaluate the inside integral first:

∫ 6

0

(x2 + 4y) dy = (x2y + 2y2)

∣∣∣y=6

y=0= 6x2 + 72.

16.2 SOLUTIONS 1117


0

∫ 6

0

(x2 + 4y)dydx =

∫ 3

0

(6x2 + 72)dx = (2x3 + 72x)∣∣∣3

0= 270.

The value of the integral is 270. The area of the base region is 3 · 6 = 18. To find the average value of the function, wedivide the value of the integral by the area of the base region:

Average value =1

Area

∫ 3

0

∫ 6

0

(x2 + 4y) dydx =1

18· 270 = 15.

The average value is 15. This is reasonable, since the smallest value of f(x, y) on this region is 0, and the largest value is32 + 4 · 6 = 33.

24. To find the average value, we first find the value of the integral∫ 4

0

∫ 3

0

(xy2) dydx.

We evaluate the inside integral first: ∫ 3

0

(xy2) dy =

(xy3

3

)∣∣∣y=3

y=0= 9x.


0

∫ 3

0

(xy2) dydx =

∫ 4

0

(9x) dx =

(9x2

2

)∣∣∣4

0= 72.

The value of the integral is 72. To find the average value, we divide the value of the integral by the area of the region:

Average value =1

Area

∫ 4

0

∫ 3

0

(xy2) dydx =72

3 · 4 = 6.

The average value of f(x, y) on this rectangle is 6. This is reasonable since the smallest value of xy2 on this region is 0and the largest value is 4 · 32 = 36.

25. (a) See Figure 16.15.

y = x x+ y = 1

(1/2, 1/2)

R

x

y

Figure 16.15

(b) If we integrate with respect to x first, we have∫

R

f(x, y) dA =

∫ 1/2

0

∫ 1−y

y

f(x, y) dx dy.

If we integrate with respect to y first, the integral must be split into two parts, so∫

R

f(x, y) dA =

∫ 1/2

0

∫ x

0

f(x, y) dy dx+

∫ 1

1/2

∫ 1−x

0

f(x, y) dy dx.

(c) If f(x, y) = x,∫

R

x dA =

∫ 1/2

0

∫ 1−y

y

x dx dy =

∫ 1/2

0

x2

2

∣∣∣∣1−y

y

dy

=1

2

∫ 1/2

0

(1− y)2 − y2 dy =1

2

∫ 1/2

0

1− 2y dy

=1

2(y − y2)

∣∣∣∣1/2

0

=1

8.


Alternatively,∫

R

x dA =

∫ 1/2

0

∫ x

0

x dy dx+

∫ 1

1/2

∫ 1−x

0

x dy dx

=

∫ 1/2

0

xy

∣∣∣∣x

0

dx+

∫ 1

1/2

xy

∣∣∣∣1−x

0

dx

=

∫ 1/2

0

x2 dx+

∫ 1

1/2

x(1− x) dx

=x3

3

∣∣∣∣1/2

0

+

(x2

2− x3

3

)∣∣∣∣1

1/2

=1

24+

1

2− 1

3− 1

8+

1

24=

1

8.

26. (a)

2

4

x

y

x = −(y − 4)/2 or y = −2x+ 4

Figure 16.16

(b)∫ 2

0

∫ −2x+4

0g(x, y) dy dx.

Problems

27. As given, the region of integration is as shown in Figure 16.17. Reversing the limits gives∫ 1

0

∫ x

0

ex2

dydx =

∫ 1

0

(yex

2

∣∣∣∣x

0

)dx =

∫ 1

0

xex2

dx

=ex

2

2

∣∣∣∣1

0

=e− 1

2.

1

1

x

y

x = y x = 1

Figure 16.17

x

y

y = x

1

Figure 16.18

16.2 SOLUTIONS 1119

28. The function sin (x2) has no elementary antiderivative, so we try integrating with respect to y first. The region of integra-tion is shown in Figure 16.18. Changing the order of integration, we get

∫ 1

0

∫ 1

y

sin (x2) dx dy =

∫ 1

0

∫ x

0

sin (x2) dy dx

=

∫ 1

0

sin (x2) · y∣∣∣∣x

0

dx

=

∫ 1

0

sin (x2) · x dx

= −cos (x2)

2

∣∣∣∣1

0

= −cos 1

2+

1

2=

1

2(1− cos 1) = 0.23.

29. As given, the region of integration is as shown in Figure 16.19.Reversing the limits gives

∫ 1

0

∫ x2

0

√2 + x3 dydx =

∫ 1

0

(y√

2 + x3

∣∣∣∣x2

0

) dx

=

∫ 1

0

x2√

2 + x3 dx

=2

9(2 + x3)

32

∣∣∣∣1

0

=2

9(3√

3− 2√

2).

x

y

1

1

x = 1

x =√y

Figure 16.19

9x

y

3

x = 9

x = y2

Figure 16.20

30. As given, the region of integration is as shown in Figure 16.20.Reversing the limits gives

∫ 9

0

∫ √x

0

y sin (x2) dydx =

∫ 9

0

(y2 sin (x2)

2

∣∣∣∣

√x

0

)dx

=1

2

∫ 9

0

x sin (x2) dx

= −cos (x2)

4

∣∣∣∣9

0

=1

4− cos (81)

4= 0.056.


31. The region of the integration is shown in Figure 16.21. To make the integration easier, we want to change the order of theintegration and get

∫ 1

0

∫ e

ey

x

lnxdx dy =

∫ e

1

∫ ln x

0

x

lnxdy dx

=

∫ e

1

x

lnx· y∣∣∣∣ln x

0

dx

=

∫ e

1

x dx =x2

2

∣∣∣∣e

1

=1

2(e2 − 1).

x

y

1 e

(e, 1)

Figure 16.21

−4 4

8

x

y

R

y = 2x+ 8

y = −2x+ 8

Figure 16.22

32. Order reversed:∫ 8

0

∫ (8−y)/2

(y−8)/2

f(x, y) dx dy. See Figure 16.22.

33. (a) We divide the base region into four subrectangles as shown in Figure 16.23. The height of the object at each point(x, y) is given by f(x, y) = xy, we label each corner of the subrectangles with the value of the function at that point.(See Figure 16.23.) Since Volume = Height× Length×Width, and ∆x = 2 and ∆y = 3, we have

Overestimate = (12 + 24 + 6 + 12)(2)(3) = 324,

andUnderestimate = (0 + 6 + 0 + 0)(2)(3) = 36.

We average these to obtain

Volume ≈ 324 + 36

2= 180.

2 4

3

6

f = 0

f = 0

f = 0

f = 0

f = 6

f = 12

f = 0

f = 12

f = 24

x

y

Figure 16.23

(b) We have f(x, y) = xy, so

Volume =

∫ 4

0

∫ 6

0

xy dydx =

∫ 4

0

(xy2

2

)∣∣∣∣y=6

y=0

dx =

∫ 4

0

18x dx = 9x2∣∣∣4

0= 144.

The volume of this object is 144. Notice that 144 is between the over- and underestimates, 324 and 36, found inpart (a).

16.2 SOLUTIONS 1121

34. (a) The contour f(x, y) = 1 lies in the xy-plane and has equation

2e−(x−1)2−y2

= 1,

so

−(x− 1)2 − y2 = ln(1/2)

(x− 1)2 + y2 = ln 2 = 0.69.

This is the equation of a circle centered at (1, 0) in the xy-plane.Other contours are of the form

2e−(x−1)2−y2

= c

−(x− 1)2 − y2 = ln(c/2).

Thus, all the contours are circles centered at the point (1, 0).(b) The cross-section has equation z = f(1, y) = e−y

2

. If x = 1, the base region in the xy-plane extends fromy = −

√3 to y =

√3. See Figure 16.24, which shows the circular region below W in the xy-plane. So

Area =

∫ √3

−√

3

e−y2

dy.

(c) Slicing parallel to the y-axis, we get

Volume =

∫ 2

−2

∫ √4−x2

−√

4−x2

e−(x−1)2−y2

dy dx.

−2 2

−2

2(1,√

3)y =√

4− x2

y = −√

4− x2 (1,−√

3)

x

y

Figure 16.24: Region beneath W in thexy-plane

35. The intersection of the graph of f(x, y) = 25− x2− y2 and xy-plane is a circle x2 + y2 = 25. The given solid is shownin Figure 16.25.Thus the volume of the solid is

V =

∫

R

f(x, y) dA

=

∫ 5

−5

∫ √25−y2

−√

25−y2

(25− x2 − y2) dx dy.


x y

z

f(x, y) = 25− x2 − y2

Figure 16.25

x

y

z

f(x, y) = 25− x2 − y2

z = 16

Figure 16.26

36. The intersection of the graph of f(x, y) = 25− x2 − y2 and the plane z = 16 is a circle, x2 + y2 = 32. The given solidis shown in Figure 16.26.

Thus, the volume of the solid is

V =

∫

R

(f(x, y)− 16) dA

=

∫ 3

−3

∫ √9−y2

−√

9−y2

(9− x2 − y2) dx dy.

37. The solid is shown in Figure 16.27, and the base of the integral is the triangle as shown in Figure 16.28.

x

y

z

2

−4

4

� y − x = 4backside

2x+ y + z = 4-y = 0

Figure 16.27

−4 2

4

x

y

y − x = 4 2x+ y = 4

Figure 16.28

Thus, the volume of the solid is

V =

∫

R

z dA

=

∫

R

(4− 2x− y) dA

=

∫ 4

0

∫ (4−y)/2

y−4

(4− 2x− y) dx dy.

16.2 SOLUTIONS 1123

38.

Volume =

∫ 2

0

∫ 2

0

xy dy dx =

∫ 2

0

1

2xy2

∣∣∣∣2

0

dx

=

∫ 2

0

2x dx

= x2

∣∣∣∣2

0

= 4

39. The region of integration is shown in Figure 16.29. Thus

Volume =

∫ 1

0

∫ x

0

(x2 + y2) dy dx =

∫ 1

0

(x2y +

y3

3

)∣∣∣∣y=x

y=0

dx =

∫ 1

0

4

3x3 dx =

x4

3

∣∣∣∣1

0

=1

3.

1

1

x

y

x = y x = 1

Figure 16.29

9x

y

3

x = 9

x = y2

Figure 16.30

40. The region of integration is shown in Figure 16.30. Thus,

Volume =

∫ 9

0

∫ √x

0

(x+ y) dy dx =

∫ 9

0

(xy +

y2

2

)∣∣∣∣y=√x

y=0

dx

=

∫ 9

0

(x3/2 +

x

2

)dx =

(2

5x5/2 +

x2

4

)∣∣∣∣9

0

=2349

20= 117.45.

41. The plane 2x+ y + z = 4 cuts the xy-plane in the line 2x+ y = 4, so the region of integration is the triangle shown inFigure 16.31. We want to find the volume under the graph of z = 4− 2x− y. Thus,

Volume =

∫ 2

0

∫ −2x+4

0

(4− 2x− y) dy dx =

∫ 2

0

(4y − 2xy − y2

2

)∣∣∣∣−2x+4

0

dx

=

∫ 2

0

(4(−2x+ 4)− 2x(−2x+ 4)− (−2x+ 4)2

2

)dx

=

∫ 2

0

(2x2 − 8x+ 8) dx =(

2

3x3 − 4x2 + 8x

)∣∣∣2

0=

16

3.


2

4

x

y

x = −(y − 4)/2 or y = −2x+ 4

Figure 16.31

42. Let R be the triangle with vertices (1, 0), (2, 2) and (0, 1). Note that (3x + 2y + 1) − (x + y) = 2x + y + 1 > 0 forx, y > 0, so z = 3x+ 2y + 1 is above z = x+ y on R. We want to find

Volume =

∫

R

((3x+ 2y + 1)− (x+ y)) dA =

∫

R

(2x+ y + 1) dA.

We need to express this in terms of double integrals.

1 2

1

2

x

y

R1

R2

y = 1 + 0.5x

y = 1− xy = 2x− 2

(2, 2)

O

Figure 16.32

To do this, divide R into two regions with the line x = 1 to make regions R1 for x ≤ 1 and R2 for x ≥ 1. SeeFigure 16.32. We want to find

∫

R

(2x+ y + 1) dA =

∫

R1

(2x+ y + 1) dA+

∫

R2

(2x+ y + 1) dA.

Note that the line connecting (0, 1) and (1, 0) is y = 1− x, and the line connecting (0, 1) and (2, 2) is y = 1 + 0.5x. So∫

R1

(2x+ y + 1) dA =

∫ 1

0

∫ 1+0.5x

1−x(2x+ y + 1) dy dx.

The line between (1, 0) and (2, 2) is y = 2x− 2, so∫

R2

(2x+ y + 1) dA =

∫ 2

1

∫ 1+0.5x

2x−2

(2x+ y + 1) dy dx.

16.2 SOLUTIONS 1125

We can now compute the double integral for R1:

∫ 1

0

∫ 1+0.5x

1−x(2x+ y + 1) dy dx =

∫ 1

0

(2xy +

y2

2+ y

)∣∣∣∣1+0.5x

1−xdx

=

∫ 1

0

(21

8x2 + 3x

)dx

=(

7

8x3 +

3

2x2) ∣∣∣∣

1

0

dx

=19

8,

and the double integral for R2:

∫ 2

1

∫ 1+0.5x

2x−2

(2x+ y + 1) dy dx =

∫ 2

1

(2xy + y2/2 + y)

∣∣∣∣1+0.5x

2x−2

dx

=

∫ 1

0

(−39

8x2 + 9x+

3

2

)dx

=(−13

8x3 +

9

2x2 +

3

2x) ∣∣∣∣

2

1

=29

8.

So, Volume =19

8+

29

8=

48

8= 6.

43. We want to calculate the volume of the tetrahedron shown in Figure 16.33.

x

y

z

1/a

1/c

1/b

ax+ by = 1

Figure 16.33

We first find the region in the xy-plane where the graph of ax + by + cz = 1 is above the xy-plane. When z = 0we have ax + by = 1. So the region over which we want to integrate is bounded by x = 0, y = 0 and ax + by = 1.Integrating with respect to y first, we have

Volume =

∫ 1/a

0

∫ (1−ax)/b

0

z dy dx =

∫ 1/a

0

∫ (1−ax)/b

0

1− by − axc

dy dx

=

∫ 1/a

0

(y

c− by2

2c− axy

c

)∣∣∣∣y=(1−ax)/b

y=0

dx

=

∫ 1/a

0

1

2bc(1− 2ax+ a2x2) dx

=1

6abc.


44. The region bounded by the x-axis and the graph of y = x− x2 is shown in Figure 16.34. The area of this region is

A =

∫ 1

0

(x− x2)dx = (x2

2− x3

3)

∣∣∣∣1

0

=1

2− 1

3=

1

6.

12

10 x

y14

Figure 16.34

So the average distance to the x-axis for points in the region is

Average distance =

∫Ry dA

area(R)

∫

R

y dA =

∫ 1

0

(∫ x−x2

0

y dy

)dx

=

∫ 1

0

(x2

2− x3 +

x4

2

)dx =

1

6− 1

4+

1

10=

1

60.

Therefore the average distance is 1/601/6

= 1/10.

45. Assume the length of the two legs of the right triangle are a and b, respectively. See Figure 16.35. The line through (a, 0)and (0, b) is given by y

b+ x

a= 1. So the area of this triangle is

A =1

2ab.

x

y

b

a

Figure 16.35

Thus the average distance from the points in the triangle to the y-axis (one of the legs) is

Average distance =1

A

∫ a

0

∫ − bax+b

0

x dy dx

=2

ab

∫ a

0

(− bax2 + bx

)dx

=2

ab

(− b

3ax3 +

b

2x2) ∣∣∣∣

a

0

=2

ab

(a2b

6

)=a

3.

16.2 SOLUTIONS 1127

Similarly, the average distance from the points in the triangle to the x-axis (the other leg) is

Average distance =1

A

∫ b

0

∫ − aby+a

0

y dx dy

=2

ab

∫ b

0

(−aby2 + ay

)dy

=2

ab

(ab2

6

)=b

3.

46. (a) We have

Average value of f =1

Area of Square

∫

Square

f dA

=1

4

∫ 2

0

∫ 2

0

(ax2 + bxy + cy2) dydx =1

4

∫ 2

0

(ax2y + bx

y2

2+ c

y3

3

)∣∣∣∣y=2

y=0

dx

=1

4

∫ 2

0

(2ax2 + 2bx+

8

3c)dx =

1

4

(2

3ax3 + bx2 +

8

3cx) ∣∣∣∣

2

0

=1

4

(16

3a+ 4b+

16

3c)

=4

3a+ b+

4

3c

The average value will be 20 if and only if (4/3)a+ b+ (4/3)c = 20.(b) Since (4/3)a+ b+ (4/3)c = 20, we must have b = 20− (4/3)a− (4/3)c. Any function f(x, y) = ax2 + (20−

(4/3)a− (4/3)c)xy+ cy2 where a and c are any real numbers is a correct solution. For example, a = 1, c = 3 leadsto the function f(x, y) = x2 + (44/3)xy + 3y2, and a = −3, c = 0 leads to the function f(x, y) = −3x2 + 24xy,both of which have average value 20 on the given square. See Figures 16.36 and 16.37.

x

y

Figure 16.36: f(x, y) = x2+ 443xy+3y2

x

y

Figure 16.37: f(x, y) = −3x2 + 24xy

47. (a) We have

Average value of f =1

Area of Rectangle

∫

Rectanglef dA

=1

6

∫ 2

x=0

∫ 3

y=0

(ax+ by) dydx =1

6

∫ 2

0

(axy + b

y2

2

)∣∣∣∣y=3

y=0

dx

=1

6

∫ 2

0

(3ax+

9

2b)dx =

1

6

(3

2ax2 +

9

2bx) ∣∣∣∣

2

0

=1

6(6a+ 9b)

= a+3

2b.


The average value will be 20 if and only if a+ (3/2)b = 20.This equation can also be expressed as 2a + 3b = 40, which shows that f(x, y) = ax + by has average value

of 20 on the rectangle 0 ≤ x ≤ 2, 0 ≤ y ≤ 3 if and only if f(2, 3) = 40.(b) Since 2a+ 3b = 40, we must have b = (40/3)− (2/3)a. Any function f(x, y) = ax+ ((40/3)− (2/3)a)y where

a is any real number is a correct solution. For example, a = 1 leads to the function f(x, y) = x + (38/3)y, anda = −3 leads to the function f(x, y) = −3x+(46/3)y, both of which have average value 20 on the given rectangle.See Figure 16.38 and 16.39.

x

y

Figure 16.38: f(x, y) = x+ 383y

x

y

Figure 16.39: f(x, y) = −3x+ 463y

48. (a) One solution would be to arrange that the minimum values of f on the square occur at the corners, so that the cornervalues give an underestimate of the average. See Figure 16.40.

1

1

100

98

96

94

x

y

Figure 16.40

1

1

0

0.1

0.2

0.3

x

y

Figure 16.41

(b) One solution would be to arrange that the maximum values of f on the square occur at the corners, so that the cornervalues give an overestimate of the average. See Figure 16.41.

49. The force, ∆F , acting on ∆A, a small piece of area, is given by

∆F ≈ p∆A,where p is the pressure at that point. Thus, if R is the rectangle, the total force is given by

F =

∫

R

p dA.

We choose coordinates with the origin at one corner of the plate. See Figure 16.42.

a

b(a, b)

x

y

Figure 16.42

16.3 SOLUTIONS 1129

Suppose p is proportional to the square of the distance from the corner represented by the origin. Then we have

p = k(x2 + y2), for some positive constant k.

Thus, we want to compute∫

R

k(x2 + y2)dA. Rewriting as an iterated integral, we have

F =

∫

R

k(x2 + y2) dA =

∫ b

0

∫ a

0

k(x2 + y2) dxdy = k

∫ b

0

(x3

3+ xy2

∣∣∣∣a

0

)dy

= k

∫ b

0

(a3

3+ ay2

)dy = k

(a3y

3+ a

y3

3

∣∣∣∣b

0

)

=k

3(a3b+ ab3).


Exercises

1.∫

W

f dV =

∫ 2

0

∫ 1

−1

∫ 3

2

(x2 + 5y2 − z) dz dy dx

=

∫ 2

0

∫ 1

−1

(x2z + 5y2z − 1

2z2)

∣∣∣∣3

2

dy dx

=

∫ 2

0

∫ 1

−1

(x2 + 5y2 − 5

2) dy dx

=

∫ 2

0

(x2y +5

3y3 − 5

2y)

∣∣∣∣1

−1

dx

=

∫ 2

0

(2x2 +10

3− 5) dx

= (2

3x3 − 5

3x)

∣∣∣∣2

0

=16

3− 10

3= 2

2.∫

W

f dV =

∫ 1

0

∫ 1

0

∫ 2

0

(ax+ by + cz) dz dy dx

=

∫ 1

0

∫ 1

0

(2ax+ 2by + 2c) dy dx

=

∫ 1

0

(2ax+ b+ 2c) dx

= a+ b+ 2c


3.∫

W

f dV =

∫ a

0

∫ b

0

∫ c

0

e−x−y−z dz dy dx

=

∫ a

0

∫ b

0

∫ c

0

e−xe−ye−z dz dy dx

=

∫ a

0

∫ b

0

e−xe−y(−e−z)∣∣∣∣c

0

dy dx

=

∫ a

0

∫ b

0

e−xe−y(−e−c + 1) dy dx

= (1− e−c)∫ a

0

e−x(−e−y)

∣∣∣∣b

0

dx

= (1− e−b)(1− e−c)∫ a

0

e−x dx

= (1− e−a)(1− e−b)(1− e−c)

4. ∫

W

f dV =

∫ π

0

∫ π

0

∫ π

0

sinx cos(y + z) dz dy dx

=

∫ π

0

∫ π

0

sinx sin(y + z)

∣∣∣∣π

0

dy dx

=

∫ π

0

∫ π

0

sinx[sin(y + π)− sin y] dy dx

=

∫ π

0

∫ π

0

sinx(−2 sin y) dy dx

= −2

∫ π

0

sinx(− cos y)

∣∣∣∣π

0

dx

= −2

∫ π

0

2 sinx dx

= −4(− cosx)

∣∣∣∣π

0

= (−4)(2) = −8

5. The region is the half cylinder in Figure 16.43.


7. The region is the quarter sphere in Figure 16.45.

11

1

x

y

z

Figure 16.43

x y

z

1 1

1

Figure 16.44

11

1

x

y

z

Figure 16.45

16.3 SOLUTIONS 1131


9. The region is the cylinder in Figure 16.47.

10. The region is the hemisphere in Figure 16.48.

11

1

x

y

z

Figure 16.46

x

y

z

1

1

1

Figure 16.47

11

1

xy

z

Figure 16.48

11. The region is the hemisphere in Figure 16.49.



11

1

x

y

z

Figure 16.49

1

1

1

x

y

z

Figure 16.50

11

1

xy

z

Figure 16.51

Problems

14. A slice throughW for a fixed value of x is a semi-circle the boundary of which is y2 = r2−x2−z2, so the inner integralis ∫ √r2−x2−z2

0

f(x, y, z) dy.

Lining up these stacks parallel to y-axis gives a slice from z = −√r2 − x2 to z =

√r2 − x2 giving

∫ √r2−x2

−√r2−x2

∫ √r2−x2−z2

0

f(x, y, z) dy dz.

Finally, there is a slice for each x between −r and r, so the integral we want is

∫ r

−r

∫ √r2−x2

−√r2−x2

∫ √r2−x2−z2

0

f(x, y, z) dy dz dx.


15. A slice throughW for a fixed value of x is a semi-circle the boundary of which is y2 = r2−x2−z2, so the inner integralis ∫ √r−x2−z2

−√r−x2−z2

f(x, y, z) dy.

Lining up these stacks parallel to y-axis gives a slice from z = 0 to z =√r2 − x2 giving

∫ √r−x2

0

∫ √r−x2−z2

−√r−x2−z2

f(x, y, z) dy dz.

Finally, there is a slice for each x between 0 and r, so the integral we want is∫ r

0

∫ √r−x2

0

∫ √r−x2−z2

−√r−x2−z2


16. A slice through W for a fixed value of x is a semi-circle the boundary of which is y2 = 4− z2, so the inner integral is∫ √4−z2

0

f(x, y, z) dy.

Lining up these stacks parallel to y-axis gives a slice from z = −2 to z = 2 giving∫ 2

−2

∫ √4−z2

0

f(x, y, z) dy dz.

Finally, there is a slice for each x between 0 and 1, so the integral we want is∫ 1

0

∫ 2

−2

∫ √4−z2

0


17. A slice through W for a fixed value of y is a semi-circle the boundary of which is z2 = r2 − x2, so the inner integral is∫ √r2−x2

0

f(x, y, z) dz.

Lining up these stacks parallel to z-axis gives a slice from x = −r to x = r giving∫ r

−r

∫ √r2−x2

0

f(x, y, z) dz dx.

Finally, there is a slice for each y between 0 and 1, so the integral we want is∫ 1

0

∫ r

−r

∫ √r2−x2

0

f(x, y, z) dz dx dy.

18. The required volume, V , is given by

V =

∫ 10

0

∫ 10−x

0

∫ 10

x+y

dzdydx

=

∫ 10

0

∫ 10−x

0

(10− (x+ y)) dydx

=

∫ 10

0

[10y − xy − 1

2y2]y=10−x

y=0dx

=

∫ 10

0

1

2(10− x)2 dx

=500

3

16.3 SOLUTIONS 1133

19. The pyramid is shown in Figure 16.52. The planes y = 0, and y−x = 4, and 2x+ y+ z = 4 intersect the plane z = −6in the lines y = 0, y − x = 4, 2x+ y = 10 on the z = −6 plane as shown in Figure 16.53.

x

y

z

2

−4

4

� 2x+ y + z = 4-y = 0

� z = −6 bottom

� y − x = 4 backside

(−4, 0,−6)

(5, 0,−6)

(2, 6,−6)

Figure 16.52

(−4, 0,−6) (5, 0,−6)

(2, 6,−6)

z = −6 plane

x

y

y − x = 4 2x+ y − 6 = 4

Figure 16.53

These three lines intersect at the points (−4, 0,−6), (5, 0,−6), and (2, 6,−6). Let R be the triangle in the planesz = −6 with the above three points as vertices. Then, the volume of the solid is

V =

∫ 6

0

∫ (10−y)/2

y−4

∫ 4−2x−y

−6

dz dx dy

=

∫ 6

0

∫ (10−y)/2

y−4

(10− 2x− y) dx dy = 162

=

∫ 6

0

(10x− x2 − xy)

∣∣∣∣(10−y)/2

y−4

dy

=

∫ 6

0

(9y2

4− 27y + 81) dy

= 162

20. Figure 16.54 shows a slice through the region for a fixed x.

10

1

2

3

z = y

z = 3y

y

z

Figure 16.54


The required volume, V , is given by

V =

∫ 2

1

∫ 1

0

∫ 3y

y

dz dy dx

=

∫ 2

1

∫ 1

0

z|3yy dy dx

=

∫ 2

1

∫ 1

0

2y dy dx

=

∫ 2

1

y2∣∣10dx

=

∫ 2

1

dx

= 1.

21. Figure 16.55 shows a slice through the region for a fixed value of y.

10

1

z = x

z = x2

x

z

Figure 16.55

We break the region into small cubes of volume ∆V = ∆x∆y∆z. A stack of cubes vertically above the point (x, z)in the xz-plane gives the strip shown in Figure 16.55 and so the inner integral is

∫ x

x2

dz

The plane and the surface meet when x = x2, giving x(1 − x) = 0, so x = 0 or x = 1. Lining up the stacks parallel tothe z-axis gives a slice from x = 0 to x = 1. Thus, the limits on the middle integral are

∫ 1

0

∫ x

x2

dz dx.

Finally, there is a slice for each y between 0 and 3, so the integral we want is∫ 3

0

∫ 1

0

∫ x

x2

dz dx dy.

The required volume, V , is given by

V =

∫ 3

0

∫ 1

0

∫ x

x2

dz dx dy

16.3 SOLUTIONS 1135

=

∫ 3

0

∫ 1

0

(x− x2

)dx dy

=

∫ 3

0

x2

2− x3

3

∣∣∣∣1

0

dy

=

∫ 3

0

1

6dy

=1

6

∫ 3

0

dy

=1

6· 3

=1

2.


V =

∫ 5

0

∫ 5−x

0

∫ x+y

0

dz dy dx

=

∫ 5

0

∫ 5−x

0

(x+ y) dy dx

=

∫ 5

0

xy +1

2y2∣∣∣y=5−x

y=0dx

=

∫ 5

0

(x(5− x) +

1

2(5− x)2

)dx

=125

3.


V =

∫ 5

0

∫ 3

0

∫ x2

0

dz dy dx

=

∫ 5

0

∫ 3

0

x2 dy dx

=

∫ 5

0

x2y

∣∣∣∣y=3

y=0

dx

=

∫ 5

0

3x2 dx

= 125.

24. (a) The vectors ~u = ~i − ~j and ~v = ~i − ~k lie in the required plane so ~p = ~u × ~v = ~i + ~j + ~k is perpendicularto this plane. Let (x, y, z) be a point in the plane, then (x − 1)~i + y~j + z~k is perpendicular to ~p , so ((x − 1)~i +

y~j + z~k ) · (~i +~j +~j ) = 0 and so(x− 1) + y + z = 0.

Therefore, the equation of the required plane is x+ y + z = 1.(b) The required volume, V , is given by

V =

∫ 1

0

∫ 1−x

0

∫ 1−x−y

0

dz dy dx


=

∫ 1

0

∫ 1−x

0

(1− x− y) dy dx

=

∫ 1

0

y − xy − 1

2y2∣∣∣1−x

0dx

=

∫ 1

0

(1− x− x(1− x)− 1

2(1− x)2

)dx

=

∫ 1

0

1

2(1− x)2dx

=1

6.

25. (a) The equation of the surface of the whole cylinder along the y-axis is x2 + z2 = 1. The part we want is

z =√

1− x2 0 ≤ y ≤ 10.

See Figure 16.56.

x1

y10

z

Figure 16.56

(b) The integral is∫

D

f(x, y, z) dV =

∫ 10

0

∫ 1

−1

∫ √1−x2

0

f(x, y, z) dzdxdy.

26. The region of integration is shown in Figure 16.57, and the mass of the given solid is given by

x

y

z

3

2

6

x3

+ y2

+ z6

= 1

or z = −2x− 3y + 6

�x3

+ y2

= 1

or y = − 23x+ 2

Figure 16.57

mass =

∫

R

δ dV

16.3 SOLUTIONS 1137

=

∫ 3

0

∫ − 23x+2

0

∫ −2x−3y+6

0

(x+ y) dzdydx

=

∫ 3

0

∫ − 23x+2

0

(x+ y)z

∣∣∣∣−2x−3y+6

0

dydx

=

∫ 3

0

∫ − 23x+2

0

(x+ y)(−2x− 3y + 6) dydx

=

∫ 3

0

∫ − 23x+2

0

(−2x2 − 3y2 − 5xy + 6x+ 6y) dydx

=

∫ 3

0

(−2x2y − y3 − 5

2xy2 + 6xy + 3y2

) ∣∣∣∣− 2

3x+2

0

dx

=

∫ 3

0

(14

27x3 − 8

3x2 + 2x+ 4

)dx

=(

7

54x4 − 8

9x3 + x2 + 4x

) ∣∣∣∣3

0

=7

54· 34 − 8

9· 33 + 32 + 12 =

21

2− 3 =

15

2.

27. The pyramid is shown in Figure 16.58. The planes y = 0, and y−x = 4, and 2x+ y+ z = 4 intersect the plane z = −6in the lines y = 0, y − x = 4, 2x+ y = 10 as shown in Figure 16.59.

x

y

z

2

−4

4

� 2x+ y + z = 4-y = 0

� z = −6 bottom

� y − x = 4 backside

(−4, 0,−6)

(5, 0,−6)

(2, 6,−6)

Figure 16.58

(−4, 0,−6) (5, 0,−6)

(2, 6,−6)

z = −6 plane

x

y

y − x = 4 2x+ y − 6 = 4

Figure 16.59

These three lines (the edges of the pyramid) intersect the plane z = −6 at the points (−4, 0,−6), (5, 0,−6), and(2, 6,−6). Let R be the triangle in the plane z = −6 with these three points as vertices. Then, the mass of the solid is

Mass =

∫ 6

0

∫ (10−y)/2

y−4

∫ 4−2x−y

−6

δ(x, y, z) dz dx dy

=

∫ 6

0

∫ (10−y)/2

y−4

∫ 4−2x−y

−6

y dz dx dy

=

∫ 6

0

∫ (10−y)/2

y−4

y(10− 2x− y) dx dy

=

∫ 6

0

y(10x− x2 − xy)

∣∣∣∣x=(10−y)/2

x=y−4

dy

=

∫ 6

0

(9y3

4− 27y2 + 81y) dy

= 243.


28. From the problem, we know that (x, y, z) is in the cube which is bounded by the three coordinate planes, x = 0, y = 0,z = 0 and the planes x = 2, y = 2, z = 2. We can regard the value x2 + y2 + z2 as the density of the cube. The averagevalue of x2 + y2 + z2 is given by

average value =

∫V

(x2 + y2 + z2) dV

volume(V )

=

∫ 2

0

∫ 2

0

∫ 2

0(x2 + y2 + z2) dxdydz

8

=

∫ 2

0

∫ 2

0

(x3

3+ (y2 + z2)x

) ∣∣20dydz

8

=

∫ 2

0

∫ 2

0

(83

+ 2y2 + 2z2)dydz

8

=

∫ 2

0

(83y + 2

3y3 + 2z2y

) ∣∣20dz

8

=

∫ 2

0

(163

+ 163

+ 4z2)dz

8

=

(323z + 4

3z3) ∣∣2

0

8

=

(643

+ 323

)

8= 4.

29. Positive. The function√x2 + y2 is positive, so its integral over the solid W is positive.

30. Positive. If (x, y, z) is any point inside the solid W then√x2 + y2 < z. Thus the integrand z −

√x2 + y2 > 0, and so

its integral over the solid W is positive.

31. Zero. The value of x is positive above the first and fourth quadrants in the xy-plane, and negative (and of equal absolutevalue) above the second and third quadrants. The integral of x over the entire solid cone is zero because the integrals overthe two halves of the cone cancel.

32. Zero. The value of y is positive on the half of the cone above the second and third quadrants and negative (of equalabsolute value) on the half of the cone above the third and fourth quadrants. The integral of y over the entire solid cone iszero because the integrals over the four quadrants cancel.

33. Positive. Since z is positive on W , its integral is positive.

34. Zero. You can see this in several ways. One way is to observe that xy is positive on part of the cone above the first andthird quadrants (where x and y are of the same sign) and negative (of equal absolute value) on the part of the cone abovethe second and fourth quadrants (where x and y have opposite signs). These add up to zero in the integral of xy over allof W .

Another way to see that the integral is zero is to write the triple integral as an iterated integral, say integrating firstwith respect to x. For fixed y and z, the x-integral is over an interval symmetric about 0. The integral of x over such aninterval is zero. If any of the inner integrals in an iterated integral is zero, then the triple integral is zero.

35. Zero. Write the triple integral as an iterated integral, say integrating first with respect to x. For fixed y and z, the x-integralis over an interval symmetric about 0. The integral of x over such an interval is zero. If any of the inner integrals in aniterated integral is zero, then the triple integral is zero.

36. Negative. If (x, y, z) is any point inside the cone then z < 2. Hence the function z − 2 is negative on W and so is itsintegral.

37. Positive. The function e−xyz is a positive function everywhere so its integral over W is positive.

38. Positive. The function√x2 + y2 is positive, so its integral over the solid W is positive.

39. Positive. If (x, y, z) is any point inside the solid W then√x2 + y2 < z. Thus z −

√x2 + y2 > 0, and so its integral

over the solid W is positive.

40. Positive. The value of x is positive on the half-cone, so its integral is positive.

41. Zero. y is positive on the half of the half-cone above the first quadrant in the xy-plane and negative (of equal absolutevalue) on the half of the half-cone above the fourth quadrant. The integral of y over W is zero because the integrals over

16.3 SOLUTIONS 1139

each half add up to zero.

42. Positive. Since z is positive on W , its integral is positive.

43. Zero. You can see this in several ways. One way is to observe that xy is positive on part of the cone above the firstquadrant (where x and y are of the same sign) and negative (of equal absolute value) on the part of the cone above thefourth quadrant (where x and y have opposite signs). These add up to zero in the integral of xy over all of W .

Another way to see that the integral is zero is to write the triple integral as an iterated integral, say integrating firstwith respect to y. For fixed x and z, the y-integral is over an interval symmetric about 0. The integral of y over such aninterval is zero. If any of the inner integrals in an iterated integral is zero, then the triple integral is zero.

44. Zero. Write the triple integral as an iterated integral, say integrating first with respect to y. For fixed x and z, the y-integralis over an interval symmetric about 0. The integral of y over such an interval is zero. If any of the inner integrals in aniterated integral is zero, then the triple integral is zero.

45. Negative. If (x, y, z) is any point inside the cone then z < 2. Hence the function z − 2 is negative on W and so is itsintegral.


47. Orient the region as shown in Figure 16.60 and use Cartesian coordinates with origin at the center of the sphere. Theequation of the sphere is x2 + y2 + z2 = 25, and we want the volume between the planes z = 3 and z = 5. The planez = 3 cuts the sphere in the circle x2 + y2 + 32 = 25, or x2 + y2 = 16.

Volume =

∫ 4

−4

∫ √16−x2

−√

16−x2

∫ √25−x2−y2

3

dzdydx.

x

y

z

5

Sphere isx2 + y2 + z2 = 25

Ix2 + y2 = 16

ICircle isx2 + y2 = 16

6

?

3

6

?

2

Figure 16.60

x

y

z

Figure 16.61

48. The intersection of two cylinders x2 + z2 = 1 and y2 + z2 = 1 is shown in Figure 16.61. This region is bounded by foursurfaces:

z = −√

1− x2, z =√

1− x2, y = −√

1− z2, and y =√

1− z2

So the volume of the given solid is

V =

∫ 1

−1

∫ √1−x2

−√

1−x2

∫ √1−z2

−√

1−z2

dy dz dx

49. Set up axes as in Figure 16.62.


2

2

2

x

y

z

Figure 16.62

The slanting plane has equation z = 2− x− y and the line where it intersects the xy-plane has equation y = 2− x.The mass of the bottom part is δ1 times its volume, VBottom, where

VBottom =

∫ 2

0

∫ 2−x

0

∫ 2−x−y

0

dz dy dx =

∫ 2

0

∫ 2−x

0

z

∣∣∣2−x−y

0dy dx

=

∫ 2

0

∫ 2−x

0

(2− x− y) dy dx =

∫ 2

0

((2− x)y − y2

2

)∣∣∣∣2−x

0

dx

=

∫ 2

0

(2− x)2

2dx =

−(2− x)3

6

∣∣∣∣2

0

=4

3m3.

Since the cube has volume 23 = 8 m3, the upper part has volume VTop = 8− 4/3 = 20/3 m3. Thus

Mass = VBottomδ1 + VTopδ2 =4

3δ1 +

20

3δ2 gm.

50. The mass m is given by

m =

∫

W

1 dV =

∫ 1

0

∫ 1

0

∫ x+y+1

0

1 dz dy dx

=

∫ 1

0

∫ 1

0

(x+ y + 1) dy dx

=

∫ 1

0

(xy + y2/2 + y)

∣∣10dx

=

∫ 1

0

(x+ 3/2) dx = 2 gm.

Then the x-coordinate of the center of mass is given by

x =1

2

∫

W

x dV =1

2

∫ 1

0

∫ 1

0

∫ x+y+1

0

x dz dy dx

=1

2

∫ 1

0

∫ 1

0

x(x+ y + 1) dy dx

=1

2

∫ 1

0

(x2y + xy2/2 + xy)

∣∣10dx

=1

2

∫ 1

0

(x2 + 3/2x) dx = 13/24 cm.

An essentially identical calculation (since the region is symmetric in x and y) gives y = 13/24 cm.

16.3 SOLUTIONS 1141

Finally, we compute z:

z =1

2

∫

W

z dV =1

2

∫ 1

0

∫ 1

0

∫ x+y+1

0

z dz dy dx

=1

2

∫ 1

0

∫ 1

0

(x+ y + 1)2/2 dy dx

=1

2

∫ 1

0

(x+ y + 1)3/6∣∣10dx

=1

12

∫ 1

0

((x+ 2)3 − (x+ 1)3) dx = 25/24 cm.

So (x, y, z) = (13/24, 13/24, 25/24).

51. The mass m is given by

m =

∫

W

1 dV =

∫ 1

0

∫ (1−x)/2

0

∫ (1−x−2y)/3

0

1 dz dy dx

=

∫ 1

0

∫ (1−x)/2

0

1− x− 2y

3dy dx

=1

3

∫ 1

0

(y − xy − y2)

∣∣∣∣(1−x)/2

0

dx

=1

3

(∫ 1

0

1− x2− x1− x

2−(

1− x2

)2)dx

=1

3

∫ 1

0

((1− x)2

2− (1− x)2

4

)dx

=1

3

((1− x)3

12

)∣∣∣∣1

0

= 1/36 gm.

Then the coordinates of the center of mass are given by

x = 36

∫

W

x dV = 36

∫ 1

0

∫ (1−x)/2

0

∫ (1−x−2y)/3

0

x dz dy dx = 1/4 cm.

and

y = 36

∫

W

y dV = 36

∫ 1

0

∫ (1−x)/2

0

∫ (1−x−2y)/3

0

y dz dy dx = 1/8 cm.

and

z = 36

∫

W

z dV = 36

∫ 1

0

∫ (1−x)/2

0

∫ (1−x−2y)/3

0

z dz dy dx = 1/12 cm.

52. The volume V of the solid is 1 · 2 · 3 = 6. We need to compute

m

6

∫

W

x2 + y2 dV =m

6

∫ 1

0

∫ 2

0

∫ 3

0

x2 + y2 dz dy dx

=m

6

∫ 1

0

∫ 2

0

3(x2 + y2) dy dx

=m

2

∫ 1

0

(x2y + y3/3)∣∣20dx

=m

2

∫ 1

0

(2x2 + 8/3) dx = 5m/3


53. The volume of the solid is 8abc, so we need to evaluate

m

8abc

∫

W

(y2 + z2) dV =m

8abc

∫ c

−c

∫ b

−b

∫ a

−a(y2 + z2) dx dy dz

=m

8abc

∫ c

−c

∫ b

−b2a(y2 + z2) dy dz

=m

4bc

∫ c

−c(y3/3 + yz2)

∣∣b−b dz

=m

2c

∫ c

−c(b2/3 + z2) dz

= m(b2 + c2)/3

54. By the definition, we have that

a+ b =m

V

∫

W

(y2 + z2) dV +m

V

∫

W

(x2 + z2) dV

=m

V

∫

W

(x2 + y2 + 2z2) dV

=m

V

∫

W

(x2 + y2) dV +m

V

∫

W

(2z2) dV

= c+m

V

∫

W

(2z2) dV

Since z2 is always positive, the integral∫W

(2z2) dV will be positive, thus a+ b > c.


Exercises

1.∫ 2π

0

∫ √2

0

f rdr dθ

2.∫ π/2

0

∫ 1/2

0

f rdr dθ

3.∫ 3π/2

π/2

∫ 2

1

f rdr dθ

4.∫ 3π/4

π/4

∫ 2

0

f rdr dθ

5. See Figure 16.63.

x

y

θ = −π/2

θ = π/2

r = 4

Figure 16.63

−1

1

x

y

R

r = 1

Figure 16.64


16.4 SOLUTIONS 1143



1 2x

y

r = 2

�

r = 1

Figure 16.65

x

y

θ = π/6

θ = π/3

r = 1

Figure 16.66

x

y

θ = 3π/4

θ = 3π/2�

r = 4

�

r = 3

Figure 16.67



x

y

θ = π/4

r = 1/ cos θor r cos θ = 1or x = 1

Figure 16.68

x

y

r = 2/ sin θor r sin θ = 2or y = 2

θ = π/4

Figure 16.69


12. By using polar coordinates, we get∫

R

sin(x2 + y2)dA =

∫ 2π

0

∫ 2

0

sin(r2)r dr dθ

=

∫ 2π

0

−1

2cos(r2)

∣∣∣∣2

0

dθ

= −1

2

∫ 2π

0

(cos 4− cos 0) dθ

= −1

2(cos 4− 1) · 2π = π(1− cos 4)

13. The region is pictured in Figure 16.70.

1 2

1

2

x

y

Figure 16.70


Using polar coordinates, we get

∫

R

(x2 − y2)dA =

∫ π/2

0

∫ 2

1

r2(cos2 θ − sin2 θ)rdr dθ =

∫ π/2

0

(cos2 θ − sin2 θ) · 1

4r4

∣∣∣∣2

1

dθ

=15

4

∫ π/2

0

(cos2 θ − sin2 θ) dθ

=15

4

∫ π/2

0

cos 2θ dθ

=15

4· 1

2sin 2θ

∣∣∣∣π/2

0

= 0.

14. The presence of the term x2+y2 suggests that we should convert the integral into polar coordinates. Since√x2 + y2 = r,

the integral becomes

∫

R

√x2 + y2 dxdy =

∫ 2π

0

∫ 3

2

r2 drdθ =

∫ 2π

0

r3

3

∣∣∣∣3

2

dθ =

∫ 2π

0

19

3dθ =

38π

3.

15. (a)

x

y

1

3

y = x/3

Figure 16.71

(b)∫ 1

0

∫ 3y

0

f(x, y) dx dy.

(c) For polar coordinates, on the line y = x/3, tan θ = y/x = 1/3, so θ = tan−1(1/3). On the y-axis, θ = π/2. Thequantity r goes from 0 to the line y = 1, or r sin θ = 1, giving r = 1/ sin θ and f(x, y) = f(r cos θ, r sin θ). Thusthe integral is ∫ π/2

tan−1(1/3)

∫ 1/ sin θ

0

f(r cos θ, r sin θ)r dr dθ.

Problems

16. By the given limits 0 ≤ x ≤ −1, and −√

1− x2 ≤ y ≤√

1− x2, the region of integration is in Figure 16.72.In polar coordinates, we have

∫ 3π/2

π/2

∫ 1

0

r cos θ r dr dθ =

∫ 3π/2

π/2

cos θ(

1

3r3) ∣∣∣∣

1

0

dθ

=1

3

∫ 3π/2

π/2

cos θ dθ

=1

3sin θ

∣∣∣∣3π/2

π/2

=1

3(−1− 1) = −2

3

16.4 SOLUTIONS 1145

x

y

−1

Figure 16.72

−√

6

√6

x

y

y = x

√6

y = −x

Figure 16.73

17. From the given limits, the region of integration is in Figure 16.73.In polar coordinates,−π/4 ≤ θ ≤ π/4. Also,

√6 = x = r cos θ. Hence, 0 ≤ r ≤

√6/ cos θ. The integral becomes

∫ √6

0

∫ x

−xdy dx =

∫ π/4

−π/4

∫ √6/cos θ

0

r dr dθ

=

∫ π/4

−π/4

(r2

2

∣∣∣∣

√6/cos θ

0

)dθ =

∫ π/4

−π/4

6

2 cos2 θdθ

= 3 tan θ

∣∣∣∣π/4

−π/4= 3 · (1− (−1)) = 6.

Notice that we can check this answer because the integral gives the area of the shaded triangular region which is 12·√

6 ·(2√

6) = 6.

18. From the given limits, the region of integration is in Figure 16.74.

2x

y

√2

2

√2

x

y

π/4

x = y

Figure 16.74

So, in polar coordinates, we have,

∫ π/4

0

∫ 2

0

(r2 cos θ sin θ)r dr dθ =

∫ π/4

0

cos θ sin θ(

1

4r4) ∣∣∣∣

2

0

dθ

= 4

∫ π/4

0

sin(2θ)

2dθ

= − cos(2θ)

∣∣∣∣π/4

0

= 0− (−1) = 1.


19. The graph of f(x, y) = 25−x2−y2 is an upside down bowl, and the region whose volume we want is contained betweenthe bowl (above) and the xy-plane (below). We must first find the region in the xy-plane where f(x, y) is positive. To dothat, we set f(x, y) ≥ 0 and get x2 + y2 ≤ 25. The disk x2 + y2 ≤ 25 is the region R over which we integrate.

Volume =

∫

R

(25− x2 − y2) dA =

∫ 2π

0

∫ 5

0

(25− r2) rdr dθ

=

∫ 2π

0

(25

2r2 − 1

4r4) ∣∣∣∣

5

0

dθ

=625

4

∫ 2π

0

dθ

=625π

2

20. First, let’s find where the two surfaces intersect.√

8− x2 − y2 =√x2 + y2

8− x2 − y2 = x2 + y2

x2 + y2 = 4

So z = 2 at the intersection. See Figure 16.75.

x

y

z

�

R2

�

�

2

x2 + y2 = 4�

Figure 16.75

The volume of the ice cream cone has two parts. The first part (which is the volume of the cone) is the volume of the

solid bounded by the plane z = 2 and the cone z =√x2 + y2. Hence, this volume is given by

∫

R

(2−√x2 + y2) dA,

where R is the disk of radius 2 centered at the origin, in the xy-plane. Using polar coordinates, we have:∫

R

(2−

√x2 + y2

)dA =

∫ 2π

0

∫ 2

0

(2− r) · r dr dθ

=

∫ 2π

0

[(r2 − r3

3

)∣∣∣∣2

0

]dθ

=4

3

∫ 2π

0

dθ

= 8π/3

16.4 SOLUTIONS 1147

The second part is the volume of the region above the plane z = 2 but inside the sphere x2 + y2 + z2 = 8, which is given

by∫

R

(√

8− x2 − y2 − 2) dA where R is the same disk as before. Now

∫

R

(√

8− x2 − y2 − 2) dA =

∫ 2π

0

∫ 2

0

(√

8− r2 − 2)rdr dθ

=

∫ 2π

0

∫ 2

0

r√

8− r2 dr dθ −∫ 2π

0

∫ 2

0

2r dr dθ

=

∫ 2π

0

(−1

3(8− r2)3/2

∣∣∣∣2

0

)dθ −

∫ 2π

0

r2

∣∣∣∣2

0

dθ

= −1

3

∫ 2π

0

(43/2 − 83/2) dθ −∫ 2π

0

4 dθ

= −1

3· 2π(8− 16

√2)− 8π

=2π

3(16√

2− 8)− 8π

=8π(4√

2− 5)

3

Thus, the total volume is the sum of the two volumes, which is 32π(√

2− 1)/3.

21. (a)

Total Population =

∫ 3π/2

π/2

∫ 4

1

δ(r, θ) rdr dθ.

(b) We know that δ(r, θ) decreases as r increases, so that eliminates (iii). We also know that δ(r, θ) decreases as thex-coordinate decreases, but x = r cos θ. With a fixed r, x is proportional to cos θ. So as the x-coordinate decreases,cos θ decreases and (i) δ(r, θ) = (4− r)(2 + cos θ) best describes this situation.

(c)∫ 3π/2

π/2

∫ 4

1

(4− r)(2 + cos θ) rdr dθ =

∫ 3π/2

π/2

(2 + cos θ)(2r2 − 1

3r3)

∣∣∣∣4

1

dθ

= 9

∫ 3π/2

π/2

(2 + cos θ) dθ

= 9

[2θ + sin θ

]3π/2

π/2

= 18(π − 1) ≈ 39

Thus, the population is around 39,000.

22. (a) The volume, V , is given by

V =

∫

x2+y2≤a2

e−(x2+y2) dA.

Converting to polar coordinates gives

V =

∫ 2π

0

∫ a

0

e−r2

r dr dθ = θ

∣∣∣∣2π

0

(−1

2e−r

2) ∣∣∣∣

a

0

= 2π(

1

2− 1

2e−a

2)

= π(1− e−a2

).

(b) As a→∞, the value of e−a2 → 0, so the volume tends to π.

23. The density function is given byρ(r) = 10− 2r

where r is the distance from the center of the disk. So the mass of the disk in grams is∫

R

ρ(r) dA =

∫ 2π

0

∫ 5

0

(10− 2r)rdr dθ


=

∫ 2π

0

[5r2 − 2

3r3

]5

0

dθ

=

∫ 2π

0

125

3dθ =

250π

3(grams)

24. A rough graph of the base of the spring is in Figure 16.76, where the coil is roughly of width 0.01 inches. The volume isequal to the product of the base area and the height. To calculate the area we use polar coordinates, taking the followingintegral:

Area =

∫ 4π

0

∫ 0.26+0.04θ

0.25+0.04θ

rdrdθ

=1

2

∫ 4π

0

(0.26− 0.04θ)2 − (0.25− 0.04θ)2dθ

=1

2

∫ 4π

0

0.01 · (0.51 + 0.08θ)dθ

= 0.0051 · 2π +1

4(0.0008θ2)

∣∣∣∣4π

0

= 0.0636

Therefore, the volume= 0.0636 · 0.2 = 0.0127 in3.

Figure 16.76

25. The charge density is δ = k/r, where k is a constant.

Total charge =

∫

Disk

δ dA =

∫ R

0

∫ 2π

0

k

rr dθ dr = k

∫ R

0

∫ 2π

0

dθ dr = k

∫ R

0

2π dr = 2kπR.

Thus the total charge is proportional to R with constant of proportionality 2kπ.

26. (a) We must first decide where to put the origin. We locate the origin at the center of one disk and locate the center of thesecond disk at the point (1, 0). See Figure 16.77. (Other choices of origin are possible.)

12

1

−√

3/2

√3/2

x2 + y2 = 1

(x− 1)2 + y2 = 1

x

y

Figure 16.77

π/3

r = 1

r = 2 cos θ

x

y

Figure 16.78

By symmetry, the points of intersection of the circles are half-way between the centers, at x = 1/2. The y-valuesat these points are given by

y = ±√

1− x2 = ±√

1−(

1

2

)2

= ±√

3

2.

16.5 SOLUTIONS 1149

We integrate in the x-direction first, so that it is not necessary to set up two integrals. The right-side of the circlex2 + y2 = 1 is given by

x =√

1− y2.

The left side of the circle (x− 1)2 + y2 = 1 is given by

x = 1−√

1− y2.

Thus the area of overlap is given by

Area =

∫ √3/2

−√

3/2

∫ √1−y2

1−√

1−y2

dxdy.

(b) In polar coordinates, the circle centered at the origin has equation r = 1. See Figure 16.78. The other circle, (x −1)2 + y2 = 1, can be written as

x2 − 2x+ 1 + y2 = 1

x2 + y2 = 2x,

so its equation in polar coordinates isr2 = 2r cos θ,

and, since r 6= 0,r = 2 cos θ.

At the top point of intersection of the two circles, x = 1/2, y =√

3/2, so tan θ =√

3, giving θ = π/3.Figure 16.78 shows that if we integrate with respect to r first, we have to write the integral as the sum of two

integrals. Thus, we integrate with respect to θ first. To do this, we rewrite

r = 2 cos θ as θ = arccos(r

2

).

This gives the top half of the circle; the bottom half is given by

θ = − arccos(r

2

).

Thus the area is given by

Area =

∫ 1

0

∫ arccos(r/2)

− arccos(r/2)

r dθdr.


Exercises

1. (a) A vertical plane perpendicular to the x-axis: x = 2.(b) A cylinder: r = 3.(c) A sphere: ρ =

√3.

(d) A cone: φ = π/4.(e) A horizontal plane: z = −5.(f) A vertical half-plane: θ = π/4.

2. ∫

W

f dV =

∫ 1

−1

∫ 3π/4

π/4

∫ 4

0

(r2 + z2) rdr dθ dz

=

∫ 1

−1

∫ 3π/4

π/4

(64 + 8z2) dθ dz

=

∫ 1

−1

π

2(64 + 8z2) dz

= 64π +8

3π =

200

3π


3. ∫

W

f dV =

∫ 3

−1

∫ 2π

0

∫ 1

0

(sin (r2)) rdr dθ dz

=

∫ 3

−1

∫ 2π

0

(−1

2cos r2)

∣∣∣∣1

0

dθ dz

= −1

2

∫ 3

−1

∫ 2π

0

(cos 1− cos 0) dθ dz

= −π∫ 3

−1

(cos 1− 1) dz = −4π(cos 1− 1) = 4π(1− cos 1)

4. ∫

W

f dV =

∫ 5

0

∫ 2π

0

∫ π

π/2

1

ρ· ρ2 sinφ dφ dθ dρ

=

∫ 5

0

∫ 2π

0

∫ π

π/2

ρ sinφ dφ dθ dρ

=

∫ 5

0

∫ 2π

0

ρ dθ dρ

= 2π

∫ 5

0

ρ dρ = 25π

5. ∫

W

f dV =

∫ 2π

0

∫ π/4

0

∫ 2

1

(sinφ)ρ2 sinφ dρ dφ dθ

=

∫ 2π

0

∫ π/4

0

∫ 2

1

ρ2 sin2 φ dρ dφ dθ

=7

3

∫ 2π

0

∫ π4

0

sin2 φ dφ dθ

=7

3

∫ 2π

0

∫ π/4

0

1− cos 2φ

2dφ dθ

=7

6

∫ 2π

0

(φ− 1

2sin 2φ)

∣∣∣∣π/4

0

dθ

=7

6

∫ 2π

0

(π

4− 1

2) dθ

=7

6· 2π(

π

4− 1

2) =

7π(π − 2)

12

6. Using Cartesian coordinates, we get:∫ 3

0

∫ 1

0

∫ 5

0

f dz dy dx

7. Using cylindrical coordinates, we get:∫ 1

0

∫ 2π

0

∫ 4

0

f · rdr dθ dz

8. Using cylindrical coordinates, we get:

∫ 4

0

∫ π/2

0

∫ 2

0

f · rdr dθ dz

16.5 SOLUTIONS 1151

9. Using spherical coordinates, we get:∫ π

0

∫ π

0

∫ 3

2

f · ρ2 sinφ dρ dφ dθ

10. Using spherical coordinates, we get:

∫ 2π

0

∫ π/6

0

∫ 3

0

f · ρ2 sinφ dρ dφ dθ

11. We use Cartesian coordinates, oriented as shown in Figure 16.79. The slanted top has equation z = mx, where m is theslope in the x-direction, so m = 1/5. Then if f is an arbitrary funtion, the triple integral is

∫ 5

0

∫ 2

0

∫ x/5

0

f dzdydx.

Other answers are possible.

x

y

z

12

5

(5, 2, 0)

(5, 2, 1)?

(5, 0, 1)

Figure 16.79

12. We choose cylindrical coordinates oriented as in Figure 16.80. The cone has equation z = r. Since we have a half conescooped out of a half cylinder, θ varies between 0 and π. Thus, if f is an arbitrary function, the integral is

∫ π

0

∫ 2

0

∫ r

0

fr dzdrdθ.


2

2z = r

π/4

x

z

Figure 16.80

13.

x

y

z

Figure 16.81

R is one eighth of a sphere of radius 1, below the xy-plane and under the first quadrant.


14. (a) The region of integration is the region between the cone z = r, the xy-plane and the cylinder r = 3. In sphericalcoordinates, r = 3 becomes ρ sinφ = 3, so ρ = 3/ sinφ. The cone is φ = π/4 and the xy-plane is φ = π/2. SeeFigure 16.82. Thus, the integral becomes

∫ 2π

0

∫ π/2

π/4

∫ 3/ sinφ

0

ρ2 sinφ dρdφdθ.

x

z

3

r = 3

z = r

π4

Figure 16.82: Region of integration isbetween the cone and the xy-plane

(b) The original integral is easier to evaluate, so∫ 2π

0

∫ 3

0

∫ r

0

r dzdrdθ =

∫ 2π

0

∫ 3

0

zr∣∣∣z=r

z=0drdθ =

∫ 2π

0

∫ 3

0

r2 drdθ = 2π · r3

3

∣∣∣∣3

0

= 18π.

Problems

15. In spherical coordinates, the spherical cap is part of the surface ρ =√

2. If α is the angle at the vertex of the cone, wehave tan(α/2) = 2/2 = 1, so α/2 = π/4. Since the cone is below the xy-plane, the angle φ ranges from 3π/4 to π.Thus, the integral is given by ∫ 2π

0

∫ π

3π/4

∫ √2

0

f(ρ, φ, θ)ρ2 sinφ dρ dφ dθ.

16. In cylindrical coordinates, the spherical cap has equation z = −√

2− r2. If α is the angle at the vertex of the cone, wehave tan(α/2) = 2/2 = 1, so α/2 = π/4. The cone has equation z = −r. Thus, the integral is

∫ 2π

0

∫ 1

0

∫ −√2−r2

r

g(r, θ, z)r dz dr dθ.

17. In rectangular coordinates, the spherical cap has equation z = −√

2− x2 − y2. If α is the angle at the vertex of thecone, we have tan(α/2) = 2/2 = 1, so α/2 = π/4. The cone has equation z = −

√x2 + y2. Thus, the integral is

∫ √2

−√

2

∫ √2−x2

−√

2−x2

∫ −√x2+y2

−√

2−x2−y2

h(x, y, z) dz dy dx.

18. Orient the cone as shown in Figure 16.83 and use cylindrical coordinates with the origin at the vertex of the cone. Since theangle at the vertex of the cone is a right angle, the anglesAOB andCOB are both π/4. Thus,OB = 5 cosπ/4 = 5/

√2.

The curved surface of the cone has equation z = r, so

Volume =

∫ 2π

0

∫ 5/√

2

0

∫ 5/√

2

r

r dz dr dθ

=

∫ 2π

0

∫ 5/√

2

0

rz

∣∣∣∣∣

z=5/√

2

z=r

dr dθ =

∫ 2π

0

∫ 5/√

2

0

r

(5√2− r)dr dθ

16.5 SOLUTIONS 1153

= θ

∣∣∣∣∣

2π

0

(5√2

r2

2− r3

3

)∣∣∣∣∣

5/√

2

0

= 2π

(5√2· 52

22− 53

2 · 3 ·√

2

)

= 2π · 53

2√

2

(1

2− 1

3

)=

53π

6√

2= 46.28 cm3.

5/√

2

ABC

O

�π/4

^π/4 5

r

z

Figure 16.83

19. (a) In Cartesian coordinates, the bottom half of the sphere x2 + y2 + z2 = 1 is given by z = −√

1− x2 − y2. Thus

∫

W

dV =

∫ 1

0

∫ √1−x2

0

∫ 0

−√

1−x2−y2

dz dy dx.

(b) In cylindrical coordinates, the sphere is r2 + z2 = 1 and the bottom half is given by z = −√

1− r2. Thus∫

W

dV =

∫ π/2

0

∫ 1

0

∫ 0

−√

1−r2r dz dr dθ.

(c) In spherical coordinates, the sphere is ρ = 1. Thus,∫

W

dV =

∫ π/2

0

∫ π

π/2

∫ 1

0

ρ2 sinφ dρ dφ dθ.

20. (a) Since the cone has a right angle at its vertex, it has equation

z =√x2 + y2.

The sphere has equation x2 + y2 + z2 = 1, so the top half is given by

z =√

1− x2 − y2.

The cone and the sphere intersect in the circle

x2 + y2 =1

2, z =

1√2.

See Figure 16.84. Thus ∫

W

dV =

∫ 1/√

2

−1/√

2

∫ √(1/2)−x2

−√

(1/2)−x2

∫ √1−x2−y2

√x2+y2

dz dy dx.

x

y

z

1

Ix2 + y2 = 1

2

ICircle isx2 + y2 = 1

2

6

?

1√2 π

4

Figure 16.84


(b) In cylindrical coordinates, the cone has equation z = r and the sphere has equation z =√

1− r2. Thus

∫

W

dV =

∫ 2π

0

∫ 1/√

2

0

∫ √1−r2

r

r dz dr dθ.

(c) In spherical coordinates, the cone has equation φ = π/4 and the sphere is ρ = 1. Thus

∫

W

dV =

∫ 2π

0

∫ π/4

0

∫ 1

0


21. (a) Since the cone has a right angle at its vertex, it has equation

z =√x2 + y2.

Figure 16.85 shows the plane with equation z = 1/√

2. The plane and the cone intersect in the circle x2 +y2 = 1/2.Thus, ∫

W

dV =

∫ 1/√

2

−1/√

2

∫ √(1/2)−x2

−√

(1/2)−x2

∫ 1

√x2+y2

dz dy dx.

x

y

z1√2

1√2 π

4 1

Figure 16.85

(b) In cylindrical coordinates the cone has equation z = r, so

∫

W

dV =

∫ 2π

0

∫ 1/√

2

0

∫ 1

r

r dz dr dθ.

(c) In spherical coordinates, the cone has equation φ = π/4 and the plane z = 1/√

2 has equation ρ cosφ = 1/√

2.Thus ∫

W

dV =

∫ 2π

0

∫ π/4

0

∫ 1/(√

2 cosφ)

0


22. The region is a solid cylinder of height 1, radius 1 with base on the xy-plane and axis on the z-axis. Use cylindricalcoordinates:

∫ 1

0

∫ 1

−1

∫ √1−x2

−√

1−x2

1

(x2 + y2)1/2dy dx dz =

∫ 1

0

∫ 2π

0

∫ 1

0

1

rr dr dθ dz

=

∫ 1

0

∫ 2π

0

r

∣∣∣∣1

0

dθ dz

=

∫ 1

0

∫ 2π

0

dθ dz = 2π.

16.5 SOLUTIONS 1155

23. The region of integration is half of a ball centered at the origin, radius 1, on the x ≥ 0 side. Since the integral is symmetric,we can integrate over the quarter unit ball (x ≥ 0, y ≥ 0) and multiply the result by 2. Use spherical coordinates:

∫ 1

0

∫ √1−x2

−√

1−x2

∫ √1−x2−z2

−√

1−x2−z2

1

(x2 + y2 + z2)1/2dy dz dx

= 2

∫ π/2

0

∫ π

0

∫ 1

0

1

ρρ2 sinφ dρ dφ dθ

= 2

∫ π/2

0

∫ π

0

ρ2

2

∣∣∣∣1

0

sinφ dφ dθ

=

∫ π/2

0

(− cosφ)

∣∣∣∣π

0

dθ

= (−(−1)− 0) · π = π.

24. Use cylindrical coordinates: when r2 = x2 + y2 = 1, then x2 + y2 + z2 = 1 + z2 = 2 so z = ±1. The region W isshown in Figure 16.86.

∫

W

(x2 + y2) dV =

∫ 1

−1

∫ 2π

0

∫ √2−z2

1

r2 · r drdθdz

=

∫ 1

−1

∫ 2π

0

r4

4

∣∣∣∣

√2−z2

1

dθdz =1

4

∫ 1

−1

∫ 2π

0

((2− z2)2 − 1

)dθdz

=2π

4

∫ 1

−1

(3− 4z2 + z4

)dz

=π

2

(3z − 4

3z3 +

z5

5

)∣∣∣∣1

−1

=28π

15.

x

y

z

Cylinderx2 + y2 = 1

Spherex2 + y2 + z2 = 2

−1

1

Figure 16.86

25. (a) The angle φ takes on values in the range 0 ≤ φ ≤ π. Thus, sinφ is nonnegative everywhere inW1, and so its integralis positive.

(b) The function φ is symmetric across the xy plane, such that for any point (x, y, z) in W1, with z 6= 0, the point(x, y,−z) has a cosφ value with the same magnitude but opposite sign of the cosφ value for (x, y, z). Furthermore,if z = 0, then (x, y, z) has a cosφ value of 0. Thus, with cosφ positive on the top half of the sphere and negative onthe bottom half, the integral will cancel out and be equal to zero.

26. (a) The integral is negative. In W2, we have 0 < z < 1. Thus, z2 − z is negative throughout W2 and thus its integral isnegative.

(b) On the top half of the sphere, z is nonnegative, but x can be both positive and negative. Thus, since W2 is symmetricwith respect to the yz plane, the contribution of a point (x, y, z) will be canceled out by its reflection (−x, y, z).Thus, the integral is zero.


27. The region whose volume we want is shown in Figure 16.87:

x

y

z

θ = π6

θ = π3

56

?

2

Figure 16.87

Using cylindrical coordinates, the volume is given by the integral:

V =

∫ 2

0

∫ π/3

π/6

∫ 5

0

r dr dθ dz

=

∫ 2

0

∫ π/3

π/6

r2

2

∣∣∣∣5

0

dθ dz

=25

2

∫ 2

0

∫ π/3

π/6

dθ dz

=25

2

∫ 2

0

(π

3− π

6

)dz

=25

2· π

6· 2 =

25π

6.

28. (a) In cylindrical coordinates, the cone is z = r and the sphere is r2 + z2 = 4. The surfaces intersect where z2 + z2 =2z2 = 4. So z =

√2 and r =

√2.

Volume =

∫ 2π

0

∫ √2

0

∫ √4−r2

r

r dzdrdθ.

(b) In spherical coordinates, the cone is φ = π/4 and the sphere is ρ = 2.

Volume =

∫ 2π

0

∫ π/4

0

∫ 2

0


29. (a)∫ 2π

0

∫ π

0

∫ 2

1


(b)∫ 2π

0

∫ 2

0

∫ √4−r2

−√

4−r2r dzdrdθ −

∫ 2π

0

∫ 1

0

∫ √1−r2

−√

1−r2r dzdrdθ.

30. Orient the region as shown in Figure 16.88 and use cylindrical coordinates with the origin at the center of the sphere. Theequation of the sphere is x2 + y2 + z2 = 25, or r2 + z2 = 25. If z = 3, then r2 + 32 = 25, so r2 = 16 and r = 4.

Volume =

∫ 2π

0

∫ 4

0

∫ √25−r2

3

r dzdrdθ =

∫ 2π

0

∫ 4

0

rz

∣∣∣∣∣

z=√

25−r2

z=3

drdθ

16.5 SOLUTIONS 1157

=

∫ 2π

0

∫ 4

0

(r√

25− r2 − 3r) drdθ = θ

∣∣∣∣∣

2π

0

(− (25− r2)3/2

3− 3r2

2

)∣∣∣∣∣

4

0

= 2π((

125

3− 27

3

)− 24

)=

52π

3= 54.45 cm3.

5

56

?

3

6

?2

Spherer2 + z2 = 25

-� 4

r

z

Figure 16.88

31. Orient the region as shown in Figure 16.89 and use spherical coordinates with the origin at the center of the sphere. Theequation of the sphere is x2 + y2 + z2 = 25, or ρ = 5. The plane z = 3 is the plane ρ cosφ = 3, so ρ = 3/ cosφ. InFigure 16.89, angle AOB is given by

cosφ =3

5, so φ = arccos(3/5).

The volume is given by

V =

∫ 2π

0

∫ arccos(3/5)

0

∫ 5

3/ cosφ

ρ2 sinφ dρdφdθ = θ

∣∣∣∣∣

2π

0

∫ arccos(3/5)

0

sinφρ3

3

∣∣∣∣∣

p=5

p=3/ cosφ

dφ

= 2π

∫ arccos(3/5)

0

(125

3− 9

cos3 φ

)sinφ dφ

= 2π

(∫ arccos(3/5)

0

125

3sinφ dφ−

∫ arccos(3/5)

0

9

cos3 φsinφ dφ

)

= 2π

((−125

3cosφ

) ∣∣∣∣arccos(3/5)

0

− 9(

1

2 cos2φ) ∣∣∣∣

arccos(3/5)

0

)

= 2π

(−125

3

(3

5− 1)− 9

2

(1

(3/5)2− 1

))

= 2π(−125

3

(−2

5

)− 9

2

(16

9

))

= 2π(

50

3− 8)

=52π

3≈ 54.45 cm3.

O

φ

Sphereρ = 5

BA

6

?

3

6

?2

�

�

5

x

z

Figure 16.89


32. The density function can be rewritten as δ(ρ, φ, θ) = ρ. So the mass is

∫

W

δ(P ) dV =

∫ 2π

0

∫ π/4

0

∫ 3

0

ρ · ρ2 sinφ dρ dφ dθ

=

∫ 2π

0

∫ π/4

0

81

4sinφ dφ dθ

=81

4

∫ 2π

0

(−√

2

2+ 1) dθ

=81

4· 2π · (−

√2

2+ 1) =

81

4π(−√

2 + 2)

33. Using spherical coordinates:

M =

∫ π

0

∫ 2π

0

∫ 3

0

(3− ρ)ρ2 sinφ dρ dθ dφ

=

∫ π

0

∫ 2π

0

[ρ3 − ρ4

4

]3

0

sinφ dθ dφ

=27

4

∫ π

0

∫ 2π

0

sinφ dθ dφ

=27

4· 2π · (− cosφ)

∣∣∣∣π

0

=27

2π · [−(−1) + 1] = 27π.

34. (a) We use spherical coordinates. Since δ = 9 where ρ = 6 and δ = 11 where ρ = 7, the density increases at 2 gm/cm3

for each cm increase in radius. Thus, since density is a linear function of radius, the slope of the linear function is 2.Its equation is

δ − 11 = 2(ρ− 7) so δ = 2ρ− 3.

(b) Thus,

Mass =

∫ 2π

0

∫ π

0

∫ 7

6

(2ρ− 3)ρ2 sinφ dρ dφ dθ.

(c) Evaluating the integral, we have

Mass = 2π

(− cosφ

∣∣∣∣π

0

)(2ρ4

4− 3ρ3

3

∣∣∣∣7

6

)= 2π · 2(425.5) = 1702π gm = 5346.991 gm.

35. The distance from a point (x, y, z) to the origin is given by√x2 + y2 + z2. Thus we want to evaluate

∫R

√x2 + y2 + z2 dV

Vol(R)

where R is the region bounded by the hemisphere z =√

8− x2 − y2 and the cone z =√x2 + y2. See Figure 16.90.

We will use spherical coordinates.

16.5 SOLUTIONS 1159

x

y

z

�

�

2

x2 + y2 = 4�

Figure 16.90

In spherical coordinates, the quantity ρ goes from 0 to√

8, and θ goes from 0 to 2π, and φ goes from 0 to π/4(because the angle of the cone is π/4). Thus we have

∫

R

√x2 + y2 + z2 dV =

∫ 2π

0

∫ π/4

0

∫ √8

0

ρ(ρ2 sinφ) dρdφdθ

=

∫ 2π

0

∫ π/4

0

sinφ · ρ4

4

∣∣∣∣

√8

0

dφdθ

=

∫ 2π

0

∫ π/4

0

16 sinφ dφdθ

=

∫ 2π

0

16(− cosφ)

∣∣∣∣π/4

0

dθ

=

∫ 2π

0

16

(1−√

2

2

)dθ

= 32

(1−√

2

2

)π

From Problem 20 of Section 16.4 we know that Vol(R) = 32π(√

2− 1)/3, therefore

Average distance =

∫R

√x2 + y2 + z2 dV

Vol(R)

=32(

1−√

22

)π

[32(√

2− 1)π/3]=

3√2.

36. (a) First we must choose a coordinate system, since none is given. We pick the xy-plane to be the fixed plane and thez-axis to be the line perpendicular to the plane. Then the distance from a point to the plane is |z|, so the density at apoint is given by

Density = ρ = k|z|.Using cylindrical coordinates for the integral, we find

Mass =

∫ 2π

0

∫ a

0

∫ √a2−r2

−√a2−r2

k|z|r dzdrdθ.


(b) By symmetry, we can evaluate this integral over the top half of the sphere, where |z| = z. Then

Mass = 2

∫ 2π

0

∫ a

0

∫ √a2−r2

0

kzr dzdrdθ = 2k

∫ 2π

0

∫ a

0

z2

2r

∣∣∣∣z=√a2−r2

z=0

drdθ

= k

∫ 2π

0

∫ a

0

r(a2 − r2) drdθ = k2π

(r2

2a2 − r4

4

)∣∣∣∣a

0

= 2πk

(a4

2− a4

4

)=πka4

2.

37. (a) We use the axes shown in Figure 16.91. Then the sphere is given by r2 + z2 = 25, so

Volume =

∫ 2π

0

∫ 5

1

∫ √25−r2

−√

25−r2r dzdrdθ.

(b) Evaluating gives

Volume = 2π

∫ 5

1

rz

∣∣∣∣z=√

25−r2

z=−√

25−r2dr = 2π

∫ 5

1

2r√

25− r2 dr

= 2π(−2

3

)(25− r2)3/2

∣∣∣5

1

=4π

3(24)3/2 = 64

√6π = 492.5 mm3.

1 5

5

r2 + z2 = 25x2 + y2 + z2 = 25

x

y

Figure 16.91

x

y

z

a b

Figure 16.92

38. We must first decide on coordinates. We pick spherical coordinates with the common center of the two spheres as theorigin. We imagine the half-melon with the flat side horizontal and the positive z-axis going through the curved surface.See Figure 16.92. The volume is given by the integral

Volume =

∫ 2π

0

∫ π/2

0

∫ b

a


Evaluating gives

Volume =

∫ 2π

0

∫ π/2

0

sinφρ3

3

∣∣∣∣p=b

p=a

dφdθ = 2π(− cosφ)∣∣∣π/2

0

(b3

3− a3

3

)=

2π

3(b3 − a3).

To check our answer, notice that the volume is the difference between the volumes of two half spheres of radius a and b.These half spheres have volumes 2πb3/3 and 2πa3/3, respectively.

16.5 SOLUTIONS 1161

39. The total volume of the cone is 13πr2h = 1

3π · 12 · 1 = 1

3π, so the total mass is 1

3π (since the density is always 1). The

center of mass z-coordinate is given by

z =3

π

∫

C

z dV

Using cylindrical coordinates to evaluate this integral gives

z =3

π

∫ 2π

0

∫ 1

0

∫ z

0

zr dr dz dθ

=3

π

∫ 2π

0

∫ 1

0

z3

2dz dθ

=3

π

∫ 2π

0

1

8dθ =

3

4

40. (a) The mass m of the cone is given by∫Cδ dV . In cylindrical coordinates this is

m =

∫ 2π

0

∫ 1

0

∫ z

0

z2r dr dz dθ

=

∫ 2π

0

∫ 1

0

z4

2dz dθ

=

∫ 2π

0

1

10dθ =

π

5

(b) The center of mass z-coordinate is given by

z =5

π

∫

C

z · z2 dV

Using cylindrical coordinates to evaluate this integral gives

z =5

π

∫ 2π

0

∫ 1

0

∫ z

0

z3r dr dz dθ

=5

π

∫ 2π

0

∫ 1

0

z5

2dz dθ

=5

π

∫ 2π

0

1

12dθ =

5

6

Comparing this answer with the center of mass in Problem 39, where the density was constant, it makes sensethat the center of mass would be higher in this problem, since more mass is concentrated near the top of the cone.

41. We first need to find the mass of the solid, using cylindrical coordinates:

m =

∫ 2π

0

∫ 1

0

∫ √z/a

0

r dr dz dθ

=

∫ 2π

0

∫ 1

0

z

2adz dθ

=

∫ 2π

0

1

4adθ =

π

2a

It makes sense that the mass would vary inversely with a, since increasing a makes the paraboloid skinnier. Now forthe z-coordinate of the center of mass, again using cylindrical coordinates:

z =2a

π

∫ 2π

0

∫ 1

0

∫ √z/a

0

zr dr dz dθ

=2a

π

∫ 2π

0

∫ 1

0

z2

2adz dθ

=2a

π

∫ 2π

0

1

6adθ =

2

3


42. The volume of the hemisphere is 23πa3 so its mass is 2

3πa3b. To find the location of the center of mass, we place the

base of the hemisphere on the xy-plane with the origin at its center, so we can describe it in spherical coordinates by0 ≤ ρ ≤ a, 0 ≤ φ ≤ π

2and 0 ≤ θ ≤ 2π. Then the x-coordinate of the center of mass is, integrating using spherical

coordinates:

x =3

2πa3b

∫ a

0

∫ π2

0

∫ 2π

0

ρ sin(φ) cos(θ) · ρ2 sin(φ) dθ dφ dρ = 0

since the first integral∫ 2π

0cos(θ) dθ is zero. A similar computation shows that y = 0. Now for the z-coordinate:

z =3

2πa3b

∫ a

0

∫ π2

0

∫ 2π

0

ρ cos(φ) · ρ2 sin(φ) dθ dφ dρ

=3

2πa3b· 2π

∫ a

0

∫ π2

0

ρ3 cos(φ) sin(φ) dφ dρ

=3

a3b

∫ a

0

ρ3 sin2(φ)

2

∣∣∣∣π2

0

dρ

=3

2a3b

∫ a

0

ρ3 dρ =3a

8b

So the x and y-coordinates are located at the center of the base, while the z-coordinate is located 3a8b

above the center ofthe base.

43. The sum of the three moments of inertia I for the ball B will be

3I =3

4πa3

∫

B

(y2 + z2) dV +3

4πa3

∫

B

(x2 + z2) dV +3

4πa3

∫

B

(x2 + y2) dV

=3

4πa3

∫

B

(2x2 + 2y2 + 2z2) dV,

which, in spherical coordinates is

3

2πa3

∫

B

(x2 + y2 + z2) dV =3

2πa3

∫ a

0

∫ π

0

∫ 2π

0

ρ2 · ρ2 sin(φ) dθ dφ dρ

=3

a3

∫ a

0

∫ π

0

ρ4 sin(φ) dφ dρ

=6

a3

∫ a

0

ρ4 dρ =6

5a2.

Thus 3I = 65a2, so I = 2

5a2.

44. First we need to find the volume of the cone. In spherical coordinates we find:

V =

∫ a

0

∫ π3

0

∫ 2π

0

ρ2 sin(φ) dθ dφ dρ =πa3

3

Now, to find the moment of inertia about the z-axis we need to compute the integral 3πa3

∫Wx2 + y2 dV . We can do

this in spherical coordinates as

3

πa3

∫

W

x2 + y2 dV =3

πa3

∫ a

0

∫ π3

0

∫ 2π

0

(ρ2 sin2(φ) cos2(θ) + ρ2 sin2(φ) sin2(θ)) · ρ2 sin(φ) dθ dφ dρ

=3

πa3

∫ a

0

∫ π3

0

∫ 2π

0

ρ4 sin3(φ) dθ dφ dρ

=6

a3

∫ a

0

∫ π3

0

ρ4 sin3(φ) dφ dρ

=6

a3

5

24

∫ a

0

ρ4 dρ =a2

4.

16.5 SOLUTIONS 1163

45. Assume the base of the cylinder sits on the xy-plane with center at the origin. Because the cylinder is symmetric about thez-axis, the force in the horizontal x or y direction is 0. Thus we need only compute the vertical z component of the force.We are going to use cylindrical coordinates; since the force is G ·mass/(distance)2, a piece of the cylinder of volumedV located at (r, θ, z) exerts on the unit mass a force with magnitude G(δ dV )/(r2 + z2). See Figure 16.93.

Vertical componentof force

=G(δ dV )

r2 + z2· cosφ =

Gδ dV

r2 + z2· z√

r2 + z2=

Gδz dV

(r2 + z2)3/2.

Adding up all the contributions of all the dV ’s, we obtain

Vertical force =

∫ H

0

∫ 2π

0

∫ R

0

Gδzr

(r2 + z2)3/2drdθdz

=

∫ H

0

∫ 2π

0

(Gδz)

(− 1√

r2 + z2

)∣∣∣∣R

0

dθdz

=

∫ H

0

∫ 2π

0

(Gδz) ·(− 1√

R2 + z2+

1

z

)dθdz

=

∫ H

0

2πGδ

(1− z√

R2 + z2

)dz

= 2πGδ(z −√R2 + z2)

∣∣∣∣H

0

= 2πGδ(H −√R2 +H2 +R) = 2πGδ(H +R−

√R2 +H2)

√r2 + z2

z

φ

Figure 16.93

46. In the system used in this book the volume element is dV = ρ2 sinφ dρ dφ dθ. In the system shown in the problem, φand θ have been interchanged and ρ changed to r. So the volume element is dV = r2 sin θ dr dθ dφ.

47. The charge density is δ = kz, where k is a constant. In cylindrical coordinates,

Total charge =

∫

Cylinder

δ dV =

∫ h

0

∫ R

0

∫ 2π

0

kzr dθ dr dz = k

∫ h

0

∫ R

0

2πzr dr dz

= kπ

∫ h

0

R2z dz = k(πR2)h2

2=kπ

2R2h2.

Thus, the total charge is proportional to R2h2 with constant of proportionality kπ/2.

48. The charge density is δ = k/ρ. Integrating in spherical coordinates,

Total charge =

∫ 2π

0

∫ π

0

∫ R

0

k

ρρ2 sinφ dρ dφ dθ = k

∫ 2π

0

∫ π

0

R2

2sinφ dφ dθ

= 4πkR2

2= 2πkR2.

Thus, the total charge is proportional to R2 with constant of proportionality 2πk.


49. Using spherical coordinates,

Stored energy =1

2

∫ b

a

∫ π

0

∫ 2π

0

εE2 ρ2 sinφ dθ dφ dρ =q2

32π2ε

∫ b

a

∫ π

0

∫ 2π

0

1

ρ2sinφ dθ dφ dρ

=q2

8πε

∫ b

a

1

ρ2dρ =

q2

8πε

(1

a− 1

b

).

50. Use cylindrical coordinates, with the z-axis being the axis of the cable. Consider a piece of cable of length 1. Then

Stored energy =1

2

∫ b

a

∫ 1

0

∫ 2π

0

εE2 r dθ dz dr =q2

8π2ε

∫ b

a

∫ 1

0

∫ 2π

0

1

rdθ dz dr

=q2

4πε

∫ b

a

1

rdr =

q2

4πε(ln b− ln a) =

q2

4πεlnb

a.

So the stored energy is proportional to ln(b/a) with constant of proportionality q2/4πε.


Exercises

1. We have p(x, y) = 0 for all points (x, y) satisfying x ≥ 3, since all such points lie outside the region R. Therefore thefraction of the population satisfying x ≥ 3 is 0.

2. The fraction is 0, since∫ 1

1xy dx = 0, so

∫∞−∞∫ 1

1p(x, y) dx dy =

∫ 1

0

∫ 1

1xy dx dy=0.

3. Since x+ y ≤ 3 for all points (x, y) in the region R, the fraction of the population satisfying x+ y ≤ 3 is 1.

4. Since p(x, y) = 0 for any (x, y) with x < 0 and also p(x, y) = 0 for any (x, y) with y > 1 or y < 0, the fraction of thepopulation is given by the double integral:

∫ 1

0

∫ 1

0

xy dx dy =

∫ 1

0

x2y

2

∣∣∣∣1

0

dy =

∫ 1

0

y

2dy =

y2

4

∣∣∣∣1

0

=1

4.

5. Since p(x, y) = 0 for all (x, y) outside the rectangle R, the population is given by the volume under the graph of p overthe region inside the rectangle R and to the right of the line x = y. Therefore the fraction of the population is given bythe double integral:

∫ 1

0

∫ 2

y

xy dx dy =

∫ 1

0

x2y

2

∣∣∣∣2

y

dy =

∫ 1

0

(2y − y3

2

)dy =

(y2 − y4

8

)∣∣∣∣1

0

=7

8.

6. Since p(x, y) = 0 for all (x, y) outside the rectangle R, the population is given by the volume under the graph of p overthe region inside the rectangle R and below the line x+ y = 1. This is the same as the region bounded by the x-axis, they-axis, and the line x+ y = 1. Therefore the fraction of the population is given by the double integral:

∫ 1

0

∫ 1−y

0

xy dx dy =

∫ 1

0

x2y

2

∣∣∣∣1−y

0

dy =

∫ 1

0

(1− y)2y

2dy =

(y2

4− y3

3+y4

8

)∣∣∣∣1

0

=1

24.

7. The fraction of the population is given by the double integral:

∫ 1/2

0

∫ 1

0

xy dx dy =

∫ 1/2

0

x2y

2

∣∣∣∣1

0

dy =

∫ 1/2

0

y

2dy =

y2

4

∣∣∣∣1/2

0

=1

16.

16.6 SOLUTIONS 1165

8. We are looking for points inside the circle x2 + y2 = 1 and inside the rectangle R. In the first quadrant, all of the circleand its interior lies inside the rectangle R. Thus the fraction of the population we want is given by the volume under thegraph of p over the region inside the circle x2 + y2 = 1 in the first quadrant. We evaluate this double integral using polarcoordinates:

∫ π/2

0

∫ 1

0

(r cos θ)(r sin θ) r dr dθ =

∫ π/2

0

r4

4cos θ sin θ

∣∣∣∣1

0

dθ =1

4

∫ π/2

0

cos θ sin θ dθ.

Making the substitution w = sin θ, we get:∫ π/2

0

cos θ sin θ dθ =

∫ 1

0

w dw =1

2.

Thus the fraction is (1/4)(1/2) = 1/8.

9. (a) The entries in this table are non-negative and their sum is 1.(b) Add up the values along the row x = 2: 0.2 + 0.1 + 0 = 0.3.(c) Add up the columns with y = 1 and y = 2: 0.3 + 0.2 + 0.1 + 0 + 0.2 + 0.1 + 0 + 0 = 0.9.(d) Add up the values in the grid corresponding x = 1, 2, 3 and y = 1, 2: 0.3 + 0.2 + 0.2 + 0.1 + 0.1 + 0 + 0 + 0 = 0.9.

10. No, p is not a joint density function. Since p(x, y) = 0 outside the region R, the volume under the graph of p is the sameas the volume under the graph of p over the region R, which is 2 not 1.

11. Yes, p is a joint density function. The values of p(x, y) are nonnegative, since p(x, y) = 1/2 for all points inside R andp(x, y) = 0 for all other points. The volume under the graph of p over the region R is (1/2)(5− 4)(0− (−2)) = 1.

12. No, p is a not joint density function, because p(x, y) < 0 for some points (x, y) in the region R. For example,p(−0.7, 0.1) = −0.6.

13. Yes, p is a joint density function. Since x ≤ y everywhere in the region R, we have p(x, y) = 6(y − x) ≥ 0 for all xand y in R, and p(x, y) = 0 for all other (x, y). To check that p is a joint density function, we check that the total volumeunder the graph of p over the region R is 1:

∫

R

p(x, y) dA =

∫ 1

0

∫ y

0

6(y − x) dx dy =

∫ 1

0

6

(yx− x2

2

)∣∣∣∣y

0

dy =

∫ 1

0

3y2 dx = y3

∣∣∣∣1

0

= 1.

14. Yes, p is a joint density function. In the region R we have 1 ≥ x2 + y2, so p(x, y) = (2/π)(1− x2 − y2) ≥ 0 for all xand y in R, and p(x, y) = 0 for all other (x, y). To check that p is a joint density function, we check that the total volumeunder the graph of p over the region R is 1. Using polar coordinates, we get:

∫

R

p(x, y)dA =2

π

∫ 2π

0

∫ 1

0

(1− r2)r dr dθ =2

π

∫ 2π

0

(r2

2− r4

4

)∣∣∣∣1

0

dθ =2

π

∫ 2π

0

1

4dθ = 1.

15. Yes, p is a joint density function. Since e−x−y is always positive, p(x, y) = xye−x−y ≥ 0 for all x and y in R, andhence for all x and y. To check that p is a joint density function, we check that the total volume under the graph of p overthe region R is 1. Since e−x−y = e−xe−y , we have

∫

R

xye−x−ydA =

∫ ∞

0

∫ ∞

0

xye−x−ydx dy =

∫ ∞

0

ye−y(∫ ∞

0

xe−xdx

)dy.

Using integration by parts:

∫ ∞

0

xe−xdx = limb→∞

(−xe−x − e−x)

∣∣∣∣b

0

= (0− 0)− (0− 1) = 1.

Thus ∫

R

xye−x−ydA =

∫ ∞

0

ye−y(∫ ∞

0

xe−xdx

)dy =

∫ ∞

0

ye−ydy = 1.


16. (a)∫ 1

0

∫ 1

1/3

2

3(x+ 2y) dx dy =

∫ 1

0

2

3(1

2x2 + 2xy)

∣∣11/3

dy

=

∫ 1

0

2

3

[(1

2+ 2y)− (

1

18+

2

3y)]dy

=2

3

∫ 1

0

(4

9+

4

3y) dy

=2

3

(4

9y +

2

3y2)∣∣1

0

=2

3

(10

9

)=

20

27.

(b) It is easier to calculate the probability that x < (1/3)+y does not happen, that is, the probability that x ≥ (1/3)+y,and subtract it from 1. The probability that x ≥ (1/3) + y is

∫ 1

1/3

∫ x−(1/3)

0

2

3(x+ 2y) dy dx =

∫ 1

1/3

2

3(xy + y2)

∣∣x−(1/3)

0dx

=2

3

∫ 1

1/3

(x(x− 1

3) + (x− 1

3)2) dx

=2

3

∫ 1

1/3

(2x2 − x+1

9) dx

=2

3(2

3x3 − 1

2x2 +

1

9x)∣∣11/3

=2

3

[(2

3− 1

2+

1

9)− (

2

81− 1

18+

1

27)]

= 44/243.

Thus, the probability that x < (1/3) + y is 1− (44/243) = 199/243.

13

x < 13

+ y

Figure 16.94

Problems

17. (a) We know that∫∞−∞∫∞−∞ f(x, y)dydx = 1 for a joint density function. So,

1 =

∫ ∞

−∞

∫ ∞

−∞f(x, y)dydx =

∫ 1

0

∫ 1

x

kxydydx

=1

8k

hence k = 8.

16.6 SOLUTIONS 1167

(b) The region where x < y <√x is sketched in Figure 16.95

10

1

x

yy = x

y =√x

Figure 16.95

So the probability that (x, y) satisfies x < y <√x is given by:

∫ 1

0

∫ √x

x

8xy dy dx =

∫ 1

0

4x(y2)∣∣√x

xdx

=

∫ 1

0

4x(x− x2)dx

= 4(

1

3x3 − 1

4x4) ∣∣∣∣

1

0

= 4(

1

3− 1

4

)

=1

3

This tells us that in choosing points from the region defined by 0 ≤ x ≤ y ≤ 1, that 1/3 of the time we wouldpick a point from the region defined by x < y <

√x. These regions are shown in Figure 16.95.

18. (a) For a density function,

1 =

∫ ∞

−∞

∫ ∞

−∞f(x, y) dy dx =

∫ 2

0

∫ 1

0

kx2 dy dx

=

∫ 2

0

kx2 dx

=kx3

3

∣∣20

=8k

3.

So k = 3/8.(b) ∫ 1

0

∫ 2−y

0

3

8x2 dx dy =

∫ 1

0

1

8(2− y)3 dy =

−1

32(2− y)4

∣∣10

=15

32

(c) ∫ 1/2

0

∫ 1

0

3

8x2 dx dy =

∫ 1/2

0

1

8x3∣∣10dy =

∫ 1/2

0

1

8dy =

1

16.

19. Since ∑

x

∑

y

f(x, y) ∆x∆y ≈∫

R

f(x, y) dx dy

and since x never exceeds 1, and we can assume that no one lives to be over 100, so y does not exceed 100, we have

Fraction ofpolicies

=

∫

R

f(x, y) dx dy =

∫ 100

65

∫ 1

0.8

f(x, y) dx dy,

where R is the rectangle: 0.8 ≤ x ≤ 1, 65 ≤ y ≤ 100.


20. (a) The area of S is (2)(4) = 8. Because the density function p(x, y) is constant on S and the total volume under adensity function above the xy-plane is 1, p(x, y) = 1/8 for (x, y) in S, and p(x, y) = 0 for (x, y) outside S.

(b) The probability that (x, y) is in T is∫

T

f(x, y) dy dx =1

8

∫

T

dy dx =area(T )

8=α

8.

21. (a) Since the exponential function is always positive and λ is positive, p(t) ≥ 0 for all t, and

∫ ∞

0

p(t)dt = limb→∞

−e−λt∣∣∣∣b

0

= limb→∞

−e−bt + 1 = 1.

(b) The density function for the probability that the first substance decays at time t and the second decays at time s is

p(t, s) = λe−λtµe−µs = λµe−λt−µs,

for s ≥ 0 and t ≥ 0, and is zero otherwise.(c) We want the probability that the decay time t of the first substance is less than or equal to the decay time s of the

second, so we want to integrate the density function over the region 0 ≤ t ≤ s. Thus, we compute∫ ∞

0

∫ ∞

t

λµe−λte−µs ds dt =

∫ ∞

0

λe−λt(−e−µs)∣∣∞tdt

=

∫ ∞

0

λe−λte−µt dt

=

∫ ∞

0

λe(−λ+µ)t dt

=−λλ+ µ

e−(λ+µ)t∣∣∞0

=λ

λ+ µ.

So for example, if λ = 1 and µ = 4, then the probability that the first substance decays first is 1/5 .

22. (a) ∫ π6

θ=0

∫ 4

r= 1cos θ

p(r, θ)r dr dθ

(b) ∫ π6

+ π12

θ=π6

∫ 4

r= 1cos θ

p(r, θ)r dr dθ +∫ 2π

6

θ=π6

+ π12

∫ 4

r= 1sin θ

p(r, θ)r dr dθ

23. (a) If t ≤ 0, then F (t) = 0 because the average of two positive numbers can not be negative. If 1 < t then F (t) = 1because the average of two numbers each at most 1 is certain to be less than or equal to 1. For any t, we haveF (t) =

∫Rp(x, y)dA where R is the region of the plane defined by (x + y)/2 ≤ t. Since p(x, y) = 0 outside the

unit square, we need integrate only over the part of R that lies inside the square, and since p(x, y) = 1 inside thesquare, the integral equals the area of that part of the square. Thus, we can calculate the area using area formulas.For 0 ≤ t ≤ 1, we draw the line (x + y)/2 = t, which has x- and y-intercepts of 2t. Figure 16.96 shows that for0 < t ≤ 1/2,

F (t) = Area of triangle =1

2· 2t · 2t = 2t2.

In Figure 16.97, when x = 1, we have y = 2t− 1. Thus, the vertical side of the unshaded triangle is 1− (2t− 1) =2− 2t. The horizontal side is the same length, so for 1/2 < t ≤ 1,

F (t) = Area of Square − Area of triangle = 12 − 1

2(2− 2t)2 − 1− 2(1− t)2.

The final result is:

F (t) =

0 if t ≤ 02t2 if 0 < t ≤ 1/21− 2(1− t)2 if 1/2 < t ≤ 11 if 1 < t

.

16.7 SOLUTIONS 1169

1

2t

1

2t

x+y2

= t

x

y

Figure 16.96: For 0 < t ≤ 12

1 2t

2t

1

x+y2

= t

-� 2− 2t

-�2t− 1

6

?

2− 2t

6?

2t− 1

x

y

Figure 16.97: For1

2< t ≤ 1

(b) The probability density function p(t) of z is the derivative of its cumulative distribution function. We have

p(t) =

0 if t ≤ 04t if 0 < t ≤ 1/24− 4t if 1/2 < t ≤ 10 if 1 < t

.

See Figure 16.98.(c) The values of x and y and equally likely to be near 0, 1/2, and 1. Notice from the graph of the density function in

Figure 16.98 that even though x and y separately are equally likely to be anywhere between 0 and 1, their averagez = (x+ y)/2 is more likely to be near 1/2 than to be near 0 or 1.

0.5 1

2

0x

p(t)

Figure 16.98


Exercises

1. We have∂(x, y)

∂(s, t)=

∣∣∣∣xs xt

ys yt

∣∣∣∣ =

∣∣∣∣5 2

3 1

∣∣∣∣ = −1.

Therefore, ∣∣∣∣∂(x, y)

∂(s, t)

∣∣∣∣ = 1.

2. We have∂(x, y)

∂(s, t)=

∣∣∣∣xs xt

ys yt

∣∣∣∣ =

∣∣∣∣2s −2t

2t 2s

∣∣∣∣ = 4s2 + 4t2.


∂(s, t)

∣∣∣∣ = 4s2 + 4t2.

Notice we can drop the absolute value signs because in this case the Jacobian is nonnegative for all s and t.


3. We have∂(x, y)

∂(s, t)=

∣∣∣∣xs xt

ys yt

∣∣∣∣ =

∣∣∣∣es cos t −es sin t

es sin t es cos t

∣∣∣∣ = (es cos t)2 + (es sin t)2.


∂(s, t)

∣∣∣∣ =∣∣e2s(cos2 t+ sin2 t)

∣∣ = e2s.

Notice we can drop the absolute value signs because the Jacobian in this case is positive for all s and t.

4. We have∂(x, y)

∂(s, t)=

∣∣∣∣xs xt

ys yt

∣∣∣∣ =

∣∣∣∣3s2 − 3t2 −6st

6st 3s2 − 3t2

∣∣∣∣ = 9(s2 − t2)2 + 36s2t2.

Therefore, multiplying out and simplifying∣∣∣∣∂(x, y)

∂(s, t)

∣∣∣∣ = 9∣∣s4 − 2s2t2 + t4 + 4s2t2

∣∣ = 9(s2 + t2)2.

Notice we can drop the absolute value sign since the Jacobian in this case is nonnegative for all s and t.

5. We have

∂(x, y, z)

∂(s, t, u)=

∣∣∣∣∣∣

xs xt xu

ys yt yu

zs zt zu

∣∣∣∣∣∣=

∣∣∣∣∣∣

3 1 2

1 5 −1

2 −1 1

∣∣∣∣∣∣.

This 3× 3 determinant is computed the same way as for the cross product, with the entries 3, 1, 2 in the first row playingthe same role as~i ,~j ,~k . We get

∂(x, y, z)

∂(s, t, u)= ((5)(1)− (−1)(−1))3 + ((−1)(2)− (1)(1))1 + ((1)(−1)− (2)(5))2 = −13.

6. We have

∂(x, y, z)

∂(r, θ, z)=

∣∣∣∣∣∣

xr xθ xz

yr yθ yz

zr zθ zz

∣∣∣∣∣∣=

∣∣∣∣∣∣

cos θ −r sin θ 0

sin θ r cos θ 0

0 0 1

∣∣∣∣∣∣.

This 3 × 3 determinant is computed the way it is for the cross product, with the entries cos θ,−r sin θ, 0 in the first rowplaying the same role as~i ,~j ,~k . We get

∂(x, y, z)

∂(s, t, u)= (r cos θ − 0) cos θ + (0− sin θ)(−r sin θ) + 0

= r cos2 θ + r sin2 θ = r.

7. The square T is defined by the inequalities

0 ≤ s = ax ≤ 1 0 ≤ t = by ≤ 1

that correspond to the inequalities0 ≤ x ≤ 1/a = 10 0 ≤ y ≤ 1/b = 1

that define R. Thus a = 1/10 and b = 1.


0 ≤ s = ax ≤ 1 0 ≤ t = by ≤ 1

that correspond to the inequalities

0 ≤ x ≤ 1/a = 1 0 ≤ y ≤ 1/b = 1/4

that define R. Thus a = 1 and b = 4.

16.7 SOLUTIONS 1171


0 ≤ s = ax ≤ 1 0 ≤ t = by ≤ 1

that correspond to the inequalities

0 ≤ x ≤ 1/a = 50 0 ≤ y ≤ 1/b = 10

that define R. Thus a = 1/50 and b = 1/10.

10. The disc S is defined by the inequalitys2 + t2 = a2x2 + b2y2 ≤ 1

that corresponds to the inequality x2 + y2 ≤ 152 or equivalently

1

152x2 +

1

152y2 ≤ 1

that defines R. Thus a2 = 1/152 and b2 = 1/152. We have a = 1/15, b = 1/15.

11. The disc S is defined by the inequalitys2 + t2 = a2x2 + b2y2 ≤ 1

that corresponds to the inequality x2/4 + y2/9 ≤ 1 that defines R. Thus a2 = 1/4 and b2 = 1/9. We have a = 1/2,b = 1/3.

12. Inverting the change of variables gives x = s− at, y = t.The four edges of R are

y = 0, y = 3, y =1

4x, y =

1

4(x− 10).

The change of variables transforms the edges to

t = 0, t = 3, t =1

4s− 1

4at, t =

1

4s− 1

4at− 10

4.

These are equations for the edges of a rectangle in the st-plane if the last two are of the form: s = (Constant). Thishappens when the t terms drop out, or a = −4. With a = −4 the change of variables gives

∫ ∫

T

∣∣∣∣∂(x, y)

∂(s, t)

∣∣∣∣ ds dt

over the rectangleT : 0 ≤ t ≤ 3, 0 ≤ s ≤ 10.


y = 0, y = 5, y = −1

3x, y = −1

3(x− 10).


t = 0, t = 5, t = −1

3s+

1

3at, t = −1

3s+

1

3at+

10

3.

These are equations for the edges of a rectangle in the st-plane if the last two are of the form: s = (Constant). Thishappens when the t terms drop out, or a = 3. With a = 3 the change of variables gives

∫ ∫

T

∣∣∣∣∂(x, y)

∂(s, t)

∣∣∣∣ ds dt

over the rectangleT : 0 ≤ t ≤ 5, 0 ≤ s ≤ 10.



y = 15, y = 35, y = 2(x− 10) + 15, y = 2(x− 30) + 15.


t = 15, t = 35, t = 2s− 2at− 5, t = 2s− 2at− 45.

These are equations for the edges of a rectangle in the st-plane if the last two are of the form: s = (Constant). Thishappens when the t terms drop out, or a = −1/2. With a = −1/2 the change of variables gives

∫ ∫

T

∣∣∣∣∂(x, y)

∂(s, t)

∣∣∣∣ ds dt

over the rectangle

T : 15 ≤ t ≤ 35,5

2≤ s ≤ 45

2.

Problems

15. Given T = {(s, t) | 0 ≤ s ≤ 3, 0 ≤ t ≤ 2} and{x = 2s− 3t

y = s− 2t

The shaded area in Figure 16.99 is the corresponding region R in the xy-plane.

x

y

(0, 0)

(0,−1)

(−6,−4)

(6, 3)

R1

R2

y = 12x− 1

y = 23x− 1

y = 23x

y = 12x

Figure 16.99

Since∂(x, y)

∂(s, t)=

∣∣∣∣∂x∂s

∂x∂t

∂y∂s

∂y∂t

∣∣∣∣ =

∣∣∣∣2 −3

1 −2

∣∣∣∣ = −1,

∣∣∣∣∂(x, y)

∂(s, t)

∣∣∣∣ = 1.

Thus we get ∫

T

∣∣∣∣∂(x, y)

∂(s, t)

∣∣∣∣ ds dt =

∫ 3

0

ds

∫ 2

0

dt = 6.

Since∫

R

dx dy =

∫

R1

dx dy +

∫

R2

dx dy =

∫ 0

−6

dx

∫ 23x

12x−1

dy +

∫ 6

0

dx

∫ 12x

23x−1

dy

=

∫ 0

−6

(1

6x+ 1

)dx+

∫ 6

0

(−1

6x+ 1

)dx = 3 + 3 = 6,

thus ∫

R

dx dy =

∫

T

∣∣∣∣∂(x, y)

∂(s, t)

∣∣∣∣ ds dt.

16.7 SOLUTIONS 1173

16.

R

2

4

x

y

Figure 16.100

Given T = {(s, t) | 0 ≤ s ≤ 2, s ≤ t ≤ 2} and{x = s2

y = t,

R = {(x, y) | 0 ≤ x ≤ 4,√x ≤ y ≤ 2}.

∂(x, y)

∂(s, t)=

∣∣∣∣∂x∂s

∂x∂t

∂y∂s

∂y∂t

∣∣∣∣ =

∣∣∣∣2s 0

0 1

∣∣∣∣ = 2s,

∣∣∣∣∂(x, y)

∂(s, t)

∣∣∣∣ = 2s since 0 ≤ s ≤ 2.

∫

T

∣∣∣∣∂(x, y)

∂(s, t)

∣∣∣∣ ds dt = 2

∫ 2

0

s ds

∫ 2

s

dt = 2

∫ 2

0

s(2− s) ds = 2

[s2 − s3

3

]2

0

=8

3.

So,∫

R

dx dy =

∫ 4

0

dx

∫ 2

√x

dy =

∫ 4

0

(2−√x) dx

=[2x− 2

3x3/2

]4

0= 8− 16

3=

8

3.

Thus ∫

R

dx dy =

∫

T

∣∣∣∣∂(x, y)

∂(s, t)

∣∣∣∣ ds dt.

17. Given x = ρ sinφ cos θ

y = ρ sinφ sin θ

z = ρ cosφ,

∂(x, y, z)

∂(ρ, φ, θ)=

∣∣∣∣∣∣

∂x∂ρ

∂x∂φ

∂x∂θ

∂y∂ρ

∂y∂φ

∂y∂θ

∂z∂ρ

∂z∂φ

∂z∂θ

∣∣∣∣∣∣=

∣∣∣∣∣∣

sinφ cos θ ρ cosφ cos θ −ρ sinφ sin θ

sinφ sin θ ρ cosφ sin θ ρ sinφ cos θ

cosφ −ρ sinφ 0

∣∣∣∣∣∣

= cosφ

∣∣∣∣ρ cosφ cos θ −ρ sinφ sin θ

ρ cosφ sin θ ρ sinφ cos θ

∣∣∣∣+ ρ sinφ

∣∣∣∣sinφ cos θ −ρ sinφ sin θ

sinφ sin θ ρ sinφ cos θ

∣∣∣∣= cosφ(ρ2 cos2 θ cosφ sinφ+ ρ2 sin2 θ cosφ sinφ)

+ρ sinφ(ρ sin2 φ cos2 θ + ρ sin2 φ sin2 θ)

= ρ2 cos2 φ sinφ+ ρ2 sin3 φ

= ρ2 sinφ.


18. Given {x = 3s− 4t

y = 5s+ 2t,

we have {s = 1

13(x+ 2y)

t = 126

(3y − 5x).

Since∂(x, y)

∂(s, t)=

∣∣∣∣∂x∂s

∂x∂t

∂y∂s

∂y∂t

∣∣∣∣ =

∣∣∣∣3 −4

5 2

∣∣∣∣ = 26,

∂(s, t)

∂(x, y)=

∣∣∣∣∂s∂x

∂s∂y

∂t∂x

∂t∂y

∣∣∣∣ =

∣∣∣∣113

213

− 526

326

∣∣∣∣ =(

3

26

)(1

13

)+(

5

26

)(2

13

)=

1

26.

So∂(x, y)

∂(s, t)· ∂(s, t)

∂(x, y)= 26 · 1

26= 1.

19. Given {x = 2s+ t

y = s− t,we have

∂(x, y)

∂(s, t)=

∣∣∣∣∂x∂s

∂x∂t

∂y∂s

∂y∂t

∣∣∣∣ =

∣∣∣∣2 1

1 −1

∣∣∣∣ = −3,

hence ∣∣∣∣∂(x, y)

∂(s, t)

∣∣∣∣ = 3.

We get ∫

R

(x+ y) dA =

∫

T

3s

∣∣∣∣∂(x, y)

∂(s, t)

∣∣∣∣ dsdt =

∫

T

(3s)(3) ds dt = 9

∫

T

s ds dt,

where T is the region in the st-plane corresponding to R.

x

y

(0, 0)

(3,−3)

(5,−2)

(2, 1)

R

Figure 16.101

s

t

(0, 0)

(0, 3)

(1, 0)

(1, 3)

T

Figure 16.102

Now, we need to find T .As {

x = 2s+ t

y = s− t or

{s = 1

3(x+ y)

t = 13(x− 2y),

so from the above transformation and Figure 16.101, T is the shaded area in Figure 16.102. Therefore∫

R

(x+ y) dA = 9

∫ 1

0

s ds

∫ 3

0

dt = (27)(1

2) = 13.5.

16.7 SOLUTIONS 1175

20. Given {x = s

2

y = t3,

we have∂(x, y)

∂(s, t)=

∣∣∣∣∂x∂s

∂x∂t

∂y∂s

∂y∂t

∣∣∣∣ =

∣∣∣∣12

0

0 13

∣∣∣∣ =1

6

hence ∣∣∣∣∂(x, y)

∂(s, t)

∣∣∣∣ =1

6.

So ∫

R

(x2 + y2) dA =

∫

T

(s2

4+t2

9

)(1

6

)ds dt,

where R is the region bounded by the curve 4x2 + 9y2 = 36 and T is the corresponding region in the st-plane. Since thecurve 4x2 + 9y2 = 36 corresponds to the curve s2 + t2 = 36 under the change of coordinates, T is the region boundedby the curve s2 + t2 = 36. So

∫

R

(x2 + y2) dA =1

24

∫

T

s2 ds dt+1

54

∫

T

t2 ds dt.

Now polar coordinates (r, θ) are used to evaluate the above integral. As{s = r cos θ

t = r sin θand

{s2 = r2 cos2 θ

t2 = r2 sin2 θ,

∣∣∣∣∂(s, t)

∂(r, θ)

∣∣∣∣ =

∣∣∣∣cos θ −r sin θ

sin θ r cos θ

∣∣∣∣ = r.

So

1

24

∫

T

s2 ds dt =1

24

∫ 2π

0

cos2 θ dθ

∫ 6

0

r3 dr =27

2

∫ 2π

0

cos2 θ dθ

=(

27

2

)(1

2

)∫ 2π

0

(1 + cos 2θ) dθ =(

27

4

)(2π) = 13.5π

and

1

54

∫

T

t2 ds dt =1

54

∫ 2π

0

sin2 θ dθ

∫ 6

0

r3 dr = 6

∫ 2π

0

sin2 θ dθ = 3

∫ 2π

0

(1− cos 2θ) dθ = (3)(2π) = 6π.

Thus, ∫

R

(x2 + y2) dA = 13.5π + 6π = 19.5π.

21. The area of the ellipse is∫ ∫

Rdx dy where R is the region x2 + 2xy + 2y2 ≤ 1. We must change variables in both the

area element dA = dx dy and the region R.Inverting the variable change gives x = s− t, y = t. Thus

∂(x, y)

∂(s, t)=

∣∣∣∣∂x∂s

∂x∂t

∂y∂s

∂y∂t

∣∣∣∣ =

∣∣∣∣1 −1

0 1

∣∣∣∣ = 1.

Therefore

dx dy =

∣∣∣∣∂(x, y)

∂(s, t)

∣∣∣∣ ds dt = ds dt.

The region of integration is

x2 + 2xy + 2y2 = (s− t)2 + 2(s− t)t+ 2t2 = s2 + t2 ≤ 1.

Let T be the unit disc s2 + t2 ≤ 1. We have∫ ∫

R

dx dy =

∫ ∫

T

ds dt = Area of T = π.


22. We must change variables in the area element dA = dx dy, the integrand sin(x+ y) and the region R.Inverting the variable change gives x = (s+ t)/2, y = (t− s)/2. Thus

∂(x, y)

∂(s, t)=

∣∣∣∣∂x∂s

∂x∂t

∂y∂s

∂y∂t

∣∣∣∣ =

∣∣∣∣1/2 1/2

−1/2 1/2

∣∣∣∣ =1

2.

Therefore

dx dy =

∣∣∣∣∂(x, y)

∂(s, t)

∣∣∣∣ ds dt =1

2ds dt.

The integrand is sin(x+ y) = sin t.The region of integration is

x2 + y2 =(s+ t

2

)2

+(t− s

2

)2

=s2 + t2

2≤ 1.

Let T be the disc s2 + t2 ≤ 2 of radius√

2. We have∫ ∫

R

sin(x+ y) dx dy =

∫ ∫

T

1

2(sin t) ds dt = 0.

The final integral is zero by symmetry, the integral over the part of the disc where t < 0 cancelling the integral over thepart where t > 0.

23. We must change variables in the area element dA = dx dy, the integrand x and the region R.Inverting the variable change gives x =

√s− t, y = s where we use the positive square root because the region R

is in the first quadrant where x ≥ 0. Thus

∂(x, y)

∂(s, t)=

∣∣∣∣∂x∂s

∂x∂t

∂y∂s

∂y∂t

∣∣∣∣ =

∣∣∣∣1/(2√s− t) −1/(2

√s− t)

1 0

∣∣∣∣ =1

2√s− t .

Therefore

dx dy =

∣∣∣∣∂(x, y)

∂(s, t)

∣∣∣∣ ds dt =1

2√s− t ds dt.

The integrand is x =√s− t.

The region of integration can be transformed by examination of its boundaries. The left and right boundaries of Rare given by y − x2 = t = 0 and y − x2 = t = −9. The bottom and top boundaries of R are given by y = s = 0 andy = s = 16.

Let T be the rectangle 0 ≤ s ≤ 16, −9 ≤ t ≤ 0 of area (9)(16) = 144. We have∫ ∫

R

x dx dy =

∫ ∫

T

√s− t 1

2√s− t ds dt =

1

2(Area of T ) = 72.

24. We must change variables in the differential dr dθ, the integrand 1/r2 and the region R.Inverting the variable change gives r = 1/s, θ = t. Thus

∂(r, θ)

∂(s, t)=

∣∣∣∣∂r∂s

∂r∂t

∂θ∂s

∂θ∂t

∣∣∣∣ =

∣∣∣∣−1/s2 0

0 1

∣∣∣∣ = − 1

s2.

Therefore

dr dθ =

∣∣∣∣∂(r, θ)

∂(s, t)

∣∣∣∣ ds dt =1

s2ds dt.

The integrand is 1/r2 = s2.The region R : 1 ≤ r < ∞, 0 ≤ θ ≤ 2π transforms to the rectangle T : 0 ≤ s ≤ 1, 0 ≤ t ≤ 2π of area 2π in the

st-plane.We have ∫ ∫

R

1

r3r dr dθ =

∫ ∫

T

s2 1

s2ds dt =

∫ 2π

0

∫ 1

0

ds dt = 2π.

SOLUTIONS to Review Problems for Chapter Sixteen 1177

25. Given {s = xy

t = xy2,

we have∂(s, t)

∂(x, y)=

∣∣∣∣∂s∂x

∂s∂y

∂t∂x

∂t∂y

∣∣∣∣ =

∣∣∣∣y x

y2 2xy

∣∣∣∣ = xy2 = t.

Since∂(s, t)

∂(x, y)· ∂(x, y)

∂(s, t)= 1,

∂(x, y)

∂(s, t)= t so

∣∣∣∣∂(x, y)

∂(s, t)

∣∣∣∣ =1

t

So ∫

R

xy2 dA =

∫

T

t

∣∣∣∣∂(x, y)

∂(s, t)

∣∣∣∣ ds dt =

∫

T

t(1

t) ds dt =

∫

T

ds dt,

where T is the region bounded by s = 1, s = 4, t = 1, t = 4.Then ∫

R

xy2 dA =

∫ 4

1

ds

∫ 4

1

dt = 9.

26. Let {s = x− yt = x+ y,

that is

{x = 1

2(s+ t)

y = 12(t− s),

we get∂(x, y)

∂(s, t)=

∣∣∣∣12

12

− 12

12

∣∣∣∣ =1

2.

Hence

I =

∫

R

cos

(x− yx+ y

)dx dy =

∫

T

cos(s

t

) ∣∣∣∣∂(x, y)

∂(s, t)

∣∣∣∣ ds dt =1

2

∫

T

cos(s

t

)ds dt,

where R is the triangle bounded by x+ y = 1, x = 0, y = 0 and T is its image which is the triangle bounded by t = 1,s = −t, s = t.Then

I =1

2

∫ 1

0

∫ t

−tcos(s

t

)ds dt =

1

2

∫ 1

0

t[sin(1)− sin(−1)] dt

=1

2

∫ 1

0

t · 2 sin 1 dt = sin 1

∫ 1

0

t dt =sin 1

2= 0.42.

Solutions for Chapter 16 Review

Exercises


−1 1

−1

1

x

y

x2 + y2 = 1

Figure 16.103

−2

2

x

y

-x2 + y2 = 4

Figure 16.104




y = 1

y = 4

x =√yx = −√y

x

y

Figure 16.105

1

π2

x

y

y = sinx orx = sin−1 y

Figure 16.106



6. We use Cartesian coordinates, oriented so that the cube is in the first quadrant. See Figure 16.108. Then, if f is an arbitraryfunction, the integral is ∫ 2

0

∫ 3

0

∫ 5

0

f dxdydz.

Other answers are possible. In particular, the order of integration can be changed.

7. If we imagine the disk lying horizontally, as in Figure 16.109, we can use cylindrical coordinates with the origin at thecenter of the flat base. Then, if f is an arbitrary function, the triple integral is

∫ 2π

0

∫ 2

0

∫ 3

0

f r drdzdθ.

1 1

1

x

y

z

Figure 16.107

x

y

z

5(5, 3, 0)

(5, 3, 2)2

Figure 16.108

x

y

z

6

?2 cm

� �3 cm

Figure 16.109

8. We use spherical coordinates as in Figure 16.110. Then if f is an arbitrary function, the triple integral is∫ 2π

0

∫ π

π/2

∫ 5

2

fρ2 sinφ dρdφdθ.


9. We use spherical coordinates, as in Figure 16.111. Then if f is an arbitrary function, the triple integral is∫ π/2

0

∫ π

0

∫ 5

0

fρ2 sinφ dρdφdθ.



10. Integrating with respect to x first we get ∫ 4

0

∫ −y+4

y2−2

f(x, y) dx dy

Integrating with respect to y first we get∫ 0

−2

∫ 2x+4

0

f(x, y) dy dx+

∫ 4

0

∫ −x+4

0

f(x, y) dy dx.

See Figure 16.112

x

y

z

25

Figure 16.110

x

y

z

5

−5

5

0

Figure 16.111

−2 4

4

x

y

R

y = 2x+ 4 orx = y

2− 2

y = −x+ 4 orx = −y + 4

Figure 16.112

11. Compute in polar coordinates:∫

R

√x2 + y2 dA =

∫ π

0

∫ 2

1

r · r dr dθ

=

∫ π

0

[r3

3

]2

1

dθ

=

∫ π

0

(8

3− 1

3

)dθ =

7π

3.

12.∫ 10

0

∫ 0.1

0

xexy dy dx =

∫ 10

0

exy|0.10 dx

=

∫ 10

0

(e0.1x − e0) dx

=

((e0.1x

0.1

)− x)∣∣∣∣

10

0

= (10e1 − 10− 10e0)

= 10e− 20 = 10(e− 2).

13.∫ 1

0

∫ 4

3

(sin(2− y)) cos (3x− 7) dx dy =

∫ 1

0

(sin (2− y))

[sin (3x− 7)

3

]∣∣∣∣4

3

dy

=1

3(sin 5− sin 2)

∫ 1

0

sin (2− y) dy

=1

3(sin 5− sin 2) [cos (2− y)]10

=1

3(sin 5− sin 2)(cos 1− cos 2).


14.∫ 1

0

∫ y

0

(sin3 x)(cosx)(cos y) dx dy =

∫ 1

0

(cos y)

[sin4 x

4

] ∣∣∣∣y

0

dy

=1

4

∫ 1

0

(sin4 y)(cos y) dy

=sin5 y

20

∣∣∣∣1

0

=sin5 1

20.

15. First use integration by parts, with y as the variable, u = x2y, u′ = x2, v = sin (xy)x

, v′ = cos (xy). Then,

∫ 4

3

∫ 1

0

x2y cos (xy) dy dx =

∫ 4

3

([xy sin (xy)]10 −

∫ 1

0

x sin (xy) dy

)dx

=

∫ 4

3

(x sinx+ [cos (xy)]10

)dx

=

∫ 4

3

(x sinx+ cosx− 1) dx.

Now use integration by parts again, with u = x, u′ = 1, v = − cosx, v′ = sinx. Then,∫ 4

3

(x sinx+ cosx− 1) dx = [−x cosx]43 +

∫ 4

3

cosx dx+

∫ 4

3

(cosx− 1) dx

= (−x cosx+ 2 sinx− x)|43= −4 cos 4 + 2 sin 4 + 3 cos 3− 2 sin 3− 1.

Thus, ∫ 4

3

∫ 1

0

x2y cos (xy) dy dx = −4 cos 4 + 2 sin 4 + 3 cos 3− 2 sin 3− 1.

16.∫ 1

0

∫ z

0

∫ 2

0

(y + z)7 dx dy dz =

∫ 1

0

∫ z

0

2(y + z)7 dy dz

=

∫ 1

0

2(y + z)8

8

∣∣∣∣z

0

dz

=

∫ 1

0

(2z)8 − z8

4dz

=255

4

∫ 1

0

z8 dz

=255

4

[z9

9

]1

0

=255

4· 1

9

=85

12.


17. The region is shown in Figure 16.113.

1

−1

0

1

x

y

√1− x2

−√

1− x2

Figure 16.113

The integral has the same values in the upper and lower quarter circles, so we integrate over just the upper circle andmultiply by 2. We convert the integral to polar coordinates.

∫ 1

0

∫ √1−x2

−√

1−x2

e−(x2+y2) dy dx = 2

∫ π/2

0

∫ 1

0

e−r2

r dr dθ =

∫ π/2

0

(−e−r2)

∣∣∣∣1

0

dθ

=

∫ π/2

0

1− e−1 dθ

=π

2(1− e−1).

18. We use cylindrical coordinates. The cone has radius r = 2 when z = 4, so its equation is z = 2r. Thus, the integral is∫ 2π

0

∫ 2

0

∫ 4

2r

f(r, θ, z)r dz dr dθ.

19. We use spherical coordinates. The cone has radius 2 when z = 4, so ρ sinφ = 2 when ρ cosφ = 4. Thus tanφ = 1/2,so φ = arctan(1/2). The top of the cone, z = 4, is given by ρ cosφ = 4. Thus, the integral is

∫ 2π

0

∫ arctan(1/2)

0

∫ 4/ cosφ

0

g(ρ, φ, θ)ρ2 sin θ dρ dφ dθ.

20. In rectangular coordinates, a cone has equation z = k√x2 + y2 for some constant k. Since z = 4 when

√x2 + y2 =√

22 = 2, we have k = 2. Thus, the integral is∫ 2

−2

∫ √4−x2

−√

4−x2

∫ 4

2√x2+y2

h(x, y, z) dz dy dx.

21. From Figure 16.114, we have the following iterated integrals:

x

y

z

x2 + y2 + z2 = 1

Figure 16.114


(a)∫

R

f dV =

∫ 1

−1

∫ √1−x2

−√

1−x2

∫ √1−x2−y2

0

f(x, y, z) dzdydx

(b)∫

R

f dV =

∫ 1

−1

∫ √1−y2

−√

1−y2

∫ √1−x2−y2

0

f(x, y, z) dzdxdy

(c)∫

R

f dV =

∫ 1

−1

∫ √1−y2

0

∫ √1−y2−z2

−√

1−y2−z2

f(x, y, z) dxdzdy

(d)∫

R

f dV =

∫ 1

−1

∫ √1−x2

0

∫ √1−x2−z2

−√

1−x2−z2

f(x, y, z) dydzdx

(e)∫

R

f dV =

∫ 1

0

∫ √1−z2

−√

1−z2

∫ √1−x2−z2

−√

1−x2−z2

f(x, y, z) dydxdz

(f)∫

R

f dV =

∫ 1

0

∫ √1−z2

−√

1−z2

∫ √1−y2−z2

−√

1−y2−z2

f(x, y, z) dxdydz

22. The integral is over the region 0 ≤ x2 + y2 ≤ 3, 1 ≤ z ≤ 4− x2 − y2. Using cylindrical coordinates, we get

∫ 2π

0

∫ √3

0

∫ 4−r2

1

1

z2rdz dr dθ =

∫ 2π

0

∫ √3

0

(− rz

)

∣∣∣∣4−r2

1

dr dθ

=

∫ 2π

0

∫ √3

0

(− r

4− r2+r

1)dr dθ

=

∫ 2π

0

[1

2ln(4− r2) +

1

2r2

]√3

0

dθ

=

∫ 2π

0

(1

2ln 1 +

3

2− 1

2ln 4− 0)dθ

= 2π(3

2− ln 2) = π(3− 2 ln 2)

23. The region is a hemisphere 0 ≤ x2 +y2 +z2 ≤ 32, z ≥ 0, so spherical coordinates are appropriate. Recall the conversionformula x = ρ sinφ cos θ. Then the integral in spherical coordinates becomes

∫ 2π

0

∫ π/2

0

∫ 3

0

(ρ sinφ cos θ)2ρ2 sinφ dρ dφ dθ

=

∫ 2π

0

∫ π/2

0

∫ 3

0

ρ4 sin3 φ cos2 θ dρ dφ dθ

=

∫ 2π

0

∫ π/2

0

243

5sin3 φ cos2 θ dφ dθ

=243

5

∫ 2π

0

∫ π/2

0

cos2 θ · sinφ(1− cos2 φ) dφ dθ

=243

5

∫ 2π

0

cos2 θ

[− cosφ+

1

3cos3 φ

]π2

0

dθ

=243

5

∫ 2π

0

cos2 θ[−(−1) +1

3(−1)] dθ

=243

5· 2

3

∫ 2π

0

1 + cos 2θ

2dθ

=81

5(θ +

1

2sin 2θ)

∣∣∣∣2π

0

=81

5(2π + 0) =

162π

5


24. The integral is over the region x, y ≥ 0, x2 + y2 ≤ 1, 0 ≤ z ≤√x2 + y2. Using cylindrical coordinates, we get

∫ π/2

0

∫ 1

0

∫ r

0

(z + r) rdz dr dθ =

∫ π/2

0

∫ 1

0

∫ r

0

(rz + r2) dz dr dθ

=

∫ π/2

0

∫ 1

0

(1

2r3 + r3) dr dθ

=

∫ π/2

0

3

8r4

∣∣∣∣1

0

dθ

=3

8· π

2=

3π

16

25. W is a cylindrical shell, so cylindrical coordinates should be used. See Figure 16.115.

6

?

4

-�1 -�1

x

y

z

Figure 16.115

∫

W

z

(x2 + y2)3/2dV =

∫ 4

0

∫ 2π

0

∫ 2

1

z

r3rdr dθ dz

=

∫ 4

0

∫ 2π

0

∫ 2

1

z

r2dr dθ dz

=

∫ 4

0

∫ 2π

0

(−zr

)

∣∣∣∣2

1

dθ dz

=

∫ 4

0

∫ 2π

0

z

2dθ dz

=

∫ 4

0

z

2· 2π dz =

1

2π · z2

∣∣∣∣4

0

= 8π

Problems

26. Positive, since e−x is always positive.

27. Negative, since y3 is negative on B, where y < 0.

28. Positive, since (x+ y2) is positive on R, where x > 0.

29. Can’t tell, since y3 is both positive and negative for x < 0.

30. Can’t tell, since x < 0 and y2 ≥ 0 on L, where x < 0.

31. Zero. The solid sphere is symmetric and z is positive on the top half and negative (of equal absolute value) on the bottomhalf. The integral of z over the entire solid is zero because the integrals over each half add up to zero.

32. Zero. x is positive on the hemisphere x2 +y2 +z2 ≤ 1, x > 0 and negative (of equal absolute value) on x2 +y2 +z2 ≤ 1,x < 0. The integral of x over the entire solid is zero because the integrals over each half add up to zero.


33. Zero. You can see this in several ways. One way is to observe that xy is positive on the part of the sphere above and belowthe first and third quadrants (where x and y are of the same sign) and negative (of equal absolute value) on the part of thesphere above and below the second and fourth quadrants (where x and y have opposite signs). These add up to zero in theintegral of xy over all of W .

Another way to see that the integral is zero is to write the triple integral as an iterated integral, say integrating firstwith respect to x. For fixed y and z, the x-integral is over an interval symmetric about 0. The integral of x over such aninterval is zero. If any of the inner integrals in an iterated integral is zero, then the triple integral is zero.

34. Zero. Write the triple integral as an iterated integral, say integrating first with respect to x. For fixed y and z, the x-integralis over an interval symmetric about 0. The integral of x over such an interval is zero. If any of the inner integrals in aniterated integral is zero, then the triple integral is zero.

35. Zero. Write the triple integral as an iterated integral, say integrating first with respect to x. Then sin( π2xy) is integrated

over an integral symmetric about the origin, and this integral is zero because sin( π2xy) is an odd function. Since the

innermost integral is zero so is the triple integral.

36. Positive. Since x2 + y2 + z2 ≤ 1 on the solid, we know x2 + y2 ≤ 1. This means that |xy| ≤ 1, since if |x| ≤ |y| then|xy| ≤ |y2| = y2, and if |y| ≤ |x|, then |xy| ≤ |x2| = x2. So |π

2xy| ≤ π

2on the solid, and hence cos( π

2xy) is positive

on the solid W and so is its integral.

37. Negative. Since z2 − 1 ≤ 0 in the sphere, its integral is negative.


39. Positive.√x2 + y2 + z2 is positive on W , so its integral is positive.

40. (a) The equation of the curved surface of this half cylinder along the x-axis is (y − 1)2 + z2 = 1. The part we want is

z =√

1− (y − 1)2 0 ≤ y ≤ 2 0 ≤ x ≤ 10.

(b) The integral ∫

D

f(x, y, z) dV =

∫ 10

0

∫ 2

0

∫ √1−(y−1)2

0

f(x, y, z) dz dy dx.

41. The region of integration is shown in Figure 16.116, and the mass of the given solid is given by

x

y

z

3

4

12

z = 12− 4x− 3y

� 4x+ 3y = 12

or y = 13

(12− 4x)

Figure 16.116

Mass =

∫

R

δ dV

=

∫ 3

0

∫ 13

(12−4x)

0

∫ 12−4x−3y

0

x2 dzdydx


=

∫ 3

0

∫ 13

(12−4x)

0

x2z

∣∣∣∣∣

z=12−4x−3y

z=0

dydx

=

∫ 3

0

∫ 13

(12−4x)

0

x2(12− 4x− 3y) dydx

=

∫ 3

0

x2(12y − 4xy − 3

2y2)

∣∣∣∣∣

y= 13

(12−4x)

0

dx

=(

8x3 − 4x4 +8

15x5) ∣∣∣∣∣

3

0

=108

5.

42. We use spherical coordinates because we are integrating over a sphere and the density has spherical symmetry. D = 2ρ.

M =

∫ 2π

0

∫ π

0

∫ 3

0

(2ρ)ρ2 sinφ dρ dφ dθ.

43. Let the lower left part of the forest be at (0, 0). Then the other corners have coordinates as shown. The population densityfunction is then given by

ρ(x, y) = 10− 2y

The equations of the two diagonal lines are x = −2y/5 and x = 6 + 2y/5. So the total rabbit population in the forest is

∫ 5

0

∫ 6+ 25y

− 25y

(10− 2y) dx dy =

∫ 5

0

(10− 2y)(6 +4

5y) dy

=

∫ 5

0

(60− 4y − 8

5y2) dy

= (60y − 2y2 − 8

15y3)

∣∣∣∣5

0

= 300− 50− 8

15· 125

=2750

15=

550

3≈ 183

44. We must first decide on coordinates. We pick cylindrical coordinates with the z-axis along the axis of the cylinders. Theinsulation stretches from z = 0 to z = l. See Figure 16.117. The volume is given by the integral

Volume =

∫ 2π

0

∫ l

0

∫ a+h

a

r drdzdθ.

Evaluating the integral gives

Volume =

∫ 2π

0

∫ l

0

r2

2

∣∣∣a+h

adzdθ = 2πz

∣∣∣l

0

((a+ h)2

2− a2

2

)= πl((a+ h)2 − a2).

To check our answer, notice that the volume is the difference between the volume of two cylinders of radius a and a+ h.These cylinders have volumes πl(a+ h)2 and πla2, respectively.


x

y

z

a a+ h

-� h -� a

6

?

l

Figure 16.117

45. The plane (x/p) + (y/q) + (z/r) = 1 cuts the axes at the points (p, 0, 0); (0, q, 0); (0, 0, r). Since p, q, r are positive,the region between this plane and the coordinate planes is a pyramid in the first octant. Solving for z gives

z = r

(1− x

p− y

q

)= r − rx

p− ry

q.

The volume, V , is given by the double integral

V =

∫

R

(r − rx

p− ry

q

)dA,

where R is the region shown in Figure 16.118. Thus

V =

∫ p

0

∫ q−qx/p

0

(r − rx

p− ry

q

)dydx

=

∫ p

0

((ry − rxy

p− ry2

2q

)∣∣∣∣y=q−qx/p

y=0

)dx

=

∫ p

0

(r

(q − qx

p

)− r

px

(q − qx

p

)− r

2q

(q − qx

p

)2)dx

=

∫ p

0

rq − 2rqx

p+rqx2

p2− rq2

2q

(1− 2x

p+x2

p2

)dx

=

(rqx− rqx2

p+rqx3

p23− rqx

2+rqx2

p2− rqx3

2p23

)∣∣∣∣p

0

= pqr − pqr +pqr

3− pqr

2+pqr

2− pqr

6=pqr

6.

p

q

R

{ x

p+y

q= 1

y = q − qx

p

x

y

Figure 16.118


46. We must first decide on coordinates. We imagine the vertex of the cone downward, at the origin, with the flat base in theplane z = h, as in Figure 16.119. Then, using cylindrical coordinates as in Figure 16.120, we see that the curved surfaceof the cone has equation z = hr/a. Thus the volume is given by

Volume =

∫ 2π

0

∫ a

0

∫ h

hr/a

r dzdrdθ.

Evaluating gives

Volume =

∫ 2π

0

∫ a

0

rz

∣∣∣∣z=h

z=hr/a

drdθ =

∫ 2π

0

∫ a

0

(hr − hr2

a

)drdθ

= 2π

(hr2

2− hr3

3a

)∣∣∣∣∣

a

0

= 2πh

(a2

2− a2

3

)=πha2

3.

xy

z

-� a

6

?

h

Figure 16.119

a

h

z = hr/a

r

z

Figure 16.120

47. We use spherical coordinates. Since the density, δ, is equal to the distance from the point to the origin, we have

δ = ρ gm/cm3.

Therefore the mass of the hemisphere is given by

Mass =

∫ 2π

0

∫ π/2

0

∫ 2

0

ρ · ρ2 sinφ dρdφdθ =

∫ 2π

0

∫ π/2

0

ρ4

4

∣∣∣∣2

0

sinφ dφdθ

= 4

∫ 2π

0

∫ π/2

0

sinφ dφdθ = 4

∫ 2π

0

(− cosφ)

∣∣∣∣π/2

0

dθ = 4 · 2π · 1 = 8π gm.

48. Since the hole resembles a cylinder, we will use cylindrical coordinates. Let the center of the sphere be at the origin, andlet the center of the hole be the z-axis (see Figure 16.121).

x

y

z

a

R

R

√a2 −R2

Figure 16.121


Then we will integrate from z = −√a2 −R2 to z =

√a2 −R2, and each cross-section will be an annulus. So the

volume is

∫ √a2−R2

−√a2−R2

∫ 2π

0

∫ √a2−z2

R

r dr dθ dz =

∫ √a2−R2

−√a2−R2

∫ 2π

0

1

2(a2 − z2 −R2) dθ dz

= π

∫ √a2−R2

−√a2−R2

(a2 − z2 −R2) dz

= π

[(a2 −R2)(2

√a2 −R2)− 1

3(2(a2 −R2)

32 )

]

=4π

3(a2 −R2)

32

49. We must first decide on coordinates. We pick Cartesian coordinates with the smaller sphere centered at the origin, thelarger one centered at (0, 0,−1). A vertical cross-section of the region in the xz-plane is shown in Figure 16.122. Thesmaller sphere has equation x2 + y2 + z2 = 1. The larger sphere has equation x2 + y2 + (z + 1)2 = 2.

−1

x2 + y2 + z2 = 1

x2 + y2 + (z + 1)2 = 2

x

z

Figure 16.122

Let R represent the region in the xy-plane which lies directly underneath (or above) the region whose volume wewant. The curve bounding this region is a circle, and we find its equation by solving the system:

x2 + y2 + z2 = 1

x2 + y2 + (z + 1)2 = 2

Subtracting the equations gives

(z + 1)2 − z2 = 1

2z + 1 = 1

z = 0.

Since z = 0, the two surfaces intersect in the xy-plane in the circle x2 + y2 = 1. Thus R is x2 + y2 ≤ 1.The top half of the small sphere is represented by z =

√1− x2 − y2; the top half of the large sphere is represented

by z = −1 +√

2− x2 − y2. Thus the volume is given by

Volume =

∫ 1

−1

∫ √1−x2

−√

1−x2

∫ √1−x2−y2

−1+√

2−x2−y2

dzdydx.

Starting to evaluate the integral, we get

Volume =

∫ 1

−1

∫ √1−x2

−√

1−x2

(√

1− x2 − y2 + 1−√

2− x2 − y2) dydx.


We simplify the integral by converting to polar coordinates

Volume =

∫ 2π

0

∫ 1

0

(√1− r2 + 1−

√2− r2

)r drdθ

=

∫ 2π

0

(− (1− r2)3/2

3+r2

2+

(2− r2)3/2

3

)∣∣∣∣1

0

dθ

= 2π

(1

2+

1

3−(−1

3+

23/2

3

))= 2π

(7

6− 2√

2

3

)= 1.41.

50.

f(x, y) =10√2πe−50(x−5)2 6√

2πe−18(y−15)2

=30

πe−50(x−5)2−18(y−15)2

51. Suppose the brick is set up as shown in Figure 16.123.

x

y

z

6

?1

�

��

�

5

3

Figure 16.123

The brick has m/v = density = 1. The moment of inertia about the z-axis is

Iz =

∫ 5/2

−5/2

∫ 3/2

−3/2

∫ 1/2

−1/2

1(x2 + y2) dz dy dx

=

∫ 5/2

−5/2

∫ 3/2

−3/2

(x2 + y2) dy dx

=

∫ 5/2

−5/2

(3x2 +9

4) dx

=125

4+

45

4=

85

2

The moment of inertia about the y-axis is

Iy =

∫ 5/2

−5/2

∫ 3/2

−3/2

∫ 1/2

−1/2

1(x2 + z2) dz dy dx

=

∫ 5/2

−5/2

∫ 3/2

−3/2

(x2 +1

12) dy dx

=

∫ 5/2

−5/2

(3x2 +1

4) dx

=125

4+

5

4=

65

2


The moment of inertia about the x-axis is

Ix =

∫ 5/2

−5/2

∫ 3/2

−3/2

∫ 1/2

−1/2

1(y2 + z2) dz dy dx

=

∫ 5/2

−5/2

∫ 3/2

−3/2

(y2 +1

12) dy dx

=

∫ 5/2

−5/2

(9

4+

1

4) dx

= 5 · 10

4=

25

2

52. Let the ball be centered at the origin. Since a ball looks the same from all directions, we can choose the axis of rotation;in this case, let it be the z-axis. It is best to use spherical coordinates, so then

x2 + y2 = (ρ sinφ cos θ)2 + (ρ sinφ sin θ)2

= ρ2 sin2 φ

Then m/v = Density = 1, so the moment of inertia is

Iz =

∫ R

0

∫ 2π

0

∫ π

0

1(ρ2 sin2 φ)ρ2 sinφ dφ dθ dρ

=

∫ R

0

∫ 2π

0

∫ π

0

ρ4(sinφ)(1− cos2 φ) dφ dθ dρ

=

∫ R

0

∫ 2π

0

ρ4(− cosφ+1

3cos3 φ)

∣∣∣∣π

0

dθ dρ

=

∫ R

0

∫ 2π

0

4

3ρ4 dθ dρ

=

∫ R

0

8π

3ρ4 dρ =

8

15πR5

53. Set up the cylinder with the base centered at the origin on the xy plane, facing up. (See Figure 16.124.) Newton’s Law ofGravitation states that the force exerted between two particles is

F = Gm1m2

ρ2

where G is the gravitational constant, m1 and m2 are the masses, and ρ is the distance between the particles. We take asmall volume element, so m1 = m, and m2 = δdV . In cylindrical coordinates, if m is at (0,0,0) and δdV is at (r, θ, z),(see Figure 16.124), then the distance from m to δdV is given by ρ =

√r2 + z2 for r1 ≤ r ≤ r2 and 0 ≤ z ≤ h.

z

6?z

rm

φ

� δdV

-� r1-� r2

Figure 16.124


Due to the symmetry of the cylinder the sum of all the horizontal forces is zero; the net force on m is vertical. The forceacting on the particle as a result of the small piece dV makes an angle φ with the vertical and therefore has verticalcomponent

Vertical force onparticle from smallpiece of cylinder

=GmδdV

(√r2 + z2)2

· cosφ =GmδdV

r2 + z2· z√

r2 + z2=

Gmδz

(r2 + z2)32

dV.

Thus, since dV = rdzdrdθ,

Total force =

∫ 2π

0

∫ r2

r1

∫ h

0

Gmδz

(r2 + z2)3/2r dzdrdθ

= 2πGmδ

∫ r2

r1

∫ h

0

zr

(r2 + z2)3/2drdz

= 2πGmδ

∫ r2

r1

(1− r

(r2 + h2)12

)dr

= 2πGmδ(r − (r2 + h2)12 )

∣∣∣∣r2

r1

= 2πGmδ(r2 − r1 −√r22 + h2 +

√r21 + h2).

54. The outer circle is a semicircle of radius 4. This is shown in Figure 16.125, with center atD. Thus,CE = 2 andDC = 2,while AD = 4. Notice that angle ADO is a right angle.

O

r − 2

D

2

C

2

E

A

4r

B

Figure 16.125

Suppose the large circle has center O and radius r. Then OA = r and OD = OC − DC = r − 2. ApplyingPythagoras’ Theorem to triangle OAD gives

r2 = 42 + (r − 2)2

r2 = 16 + r2 − 4r + 4

r = 5.

If we put the origin at O, the equation of the large circle is x2 + y2 = 25. In the same coordinates, the equation of thesmall circle, which has center at D = (3, 0), is (x− 3)2 + y2 = 16. The right hand side of the two circles are given by

x =√

25− y2 and x = 3 +√

16− y2.

Since the y-coordinate of A is 4 and the y-coordinate of B is −4, we have

Area =

∫ 4

−4

∫ 3+√

16−y2

√25−y2

1 dxdy

=

∫ 4

−4

(3 +√

16− y2 −√

25− y2) dy

= 13.95.


55. Let’s denote the (x, y) coordinates of the points in the lagoon by L. Since x and y are measured in kilometers and d ismeasured in meters, and 1 km = 1000 m, the volume of a small piece of the lagoon is given by

∆V ≈ d(x, y)(1000∆x)(1000∆y)m3.

Thus, the total volume of the lagoon is given by

V = 10002

∫

L

d(x, y) dxdy.

Changing coordinates using u = x/2 and v = y − f(x) converts the depth function to:

d(x(u, v), y(u, v)) = 40− 160v2 − 160u2 = 160(1

4− u2 − v2) meters.

Thus, the points in the lagoon have (u, v) coordinates in the disk, D, given by u2 + v2 ≤ 1/4.The Jacobian of the transformation is:

∣∣∣∣∂x∂u

∂x∂v

∂y∂u

∂y∂v

∣∣∣∣ =

∣∣∣∣2 0

2f ′(2u) 1

∣∣∣∣ = 2.

Thus, the integral in u, v coordinates is

V = 10002

∫

L

d(x, y) dxdy = 106

∫

D

160(1

4− u2 − v2)2 dudv = 320 · 106

∫

D

(1

4− u2 − v2) dudv.

Converting to polar coordinates, we have

V = 320 · 106

∫ 2π

0

∫ 1/2

0

(1

4− r2)r drdθ = 320 · 1062π(

1

4

r2

2− r4

4)

∣∣∣∣1/2

0

= 107π m3.

CAS Challenge Problems

56. The region is the triangle to the right of the y-axis, below the line y = 1, and above the line y = x. Thus the integral canbe written as

∫ 1

0

∫ 1

xey

2

dydx or as∫ 1

0

∫ y0ey

2

dxdy. The second of these integrals can be evaluated easily by hand:

∫ 1

0

∫ y

0

ey2

dxdy =

∫ 1

0

(ey

2

x

∣∣∣∣x=y

x=0

)dy =

∫ 1

0

yey2

dy

=1

2ey

2

∣∣∣∣1

0

=1

2(e− 1)

The other integral cannot be done by hand with the methods you have learned, but some computer algebra systems willcompute it and give the same answer.

57. In Cartesian coordinates the integral is

∫

D

3√x2 + y2 dA =

∫ 1

−1

∫ √1−x2

−√

1−x2

3√x2 + y2 dydx.

In polar coordinates it is∫

D

3√x2 + y2 dA =

∫ 2π

0

∫ 1

0

3√r2 rdrdθ =

∫ 2π

0

∫ 1

0

r5/3 drdθ

=

∫ 2π

0

3

8dθ =

6π

7

The Cartesian coordinate version requires the use of a computer algebra system. Some CASs may be able to handle itand may give the answer in terms of functions called hypergeometric functions. To compare the answers are the same youmay need to ask the CAS to give a numerical value for the answer. It’s possible your CAS will not be able to handle theintegral at all.

CHECK YOUR UNDERSTANDING 1193

58.∫ 1

0

∫ 0

−1

x+ y

(x− y)3dydx = 1/2 and

∫ 0

−1

∫ 1

0

x+y(x−y)3

dxdy = −1/2. This does not contradict the theorem because the

function is not continuous everywhere inside the region of integration. In fact, it is not even defined at the origin.

59.

Average value for F =1

Area

∫ h

−h

∫ h

−h(a+ bx4 + cy4 + dx2y2 + ex3y3) dx dy

=1

4h2

(4ah2 +

4bh6

5+

4ch6

5+

4dh6

9

)

= a+1

45(9b+ 9c+ 5d)h4

The limit islimh→0

(a+1

45(9b+ 9c+ 5d)h4) = a.

Notice that F (0, 0) = a.

Average value for G =1

4h2

∫ h

−h

∫ h

−h(a sin(kx) + b cos(ky) + c) dx dy

=1

4h2

(4(ch2k + bh sin(hk)

)

k

)

= c+b sin(hk)

hk.

The limit is

limh→0

(c+

b sin(hk)

hk

)= c+ b lim

h→0

sin(hk)

hk= b+ c.

Notice that G(0, 0) = b+ c.Finally,

Average value for H =(a+ b)

(−1 + e2h

) (−2− 2h− h2 + e2h

(2− 2h+ h2

))

4e2hh2.

You may need to simplify the answer given by your CAS to get this form. The limit of this as h → 0 (calculated with aCAS) is 0. This is equal to H(0, 0).

In each case the limit of the average values over smaller and smaller squares centered at the origin is equal to the valueof the function at the origin. We conjecture that this is true in general for a continuous function. This makes sense becausewhen the square is small, the function is approximately constant on the square with value equal to its value at the origin.Therefore the integral is approximately the area times the value of the function, so the average value is approximately thevalue of the function. This approximation gets better and better as h→ 0.

CHECK YOUR UNDERSTANDING

1. False. For example, if f(x, y) < 0 for all (x, y) in the region R, then∫Rf dA is negative.

2. True. The double integral is the limit of the sum∑

f(x, y)∆A =∑

k∆A = k∑

∆A

over rectangles that lie inside the region R. As the area ∆A→ 0, this sum approaches k · Area(R).

3. False. The function f(x, y) = exy is largest at the (1, 1) corner of R, so for any (x, y) in R we have exy ≤ e1·1 = e.Then ∫

R

exy dA = lim∆A→0

∑exy∆A ≤ lim

∆A→0

∑e∆A = e lim

∆A→0

∑∆A = e · Area(R) = e.

So∫Rexy dA ≤ e ≈ 2.7.


4. False. For example, if f = 1, then∫R

1 dA = Area(R) = 6 and∫S

1 dA = Area(S) = 6.

5. True. The double integral is the limit of the sum∑

∆A→0

ρ(x, y)∆A. Each of the terms ρ(x, y)∆A is an approximation of

the total population inside a small rectangle of area ∆A. Thus the limit of the sum of all of these numbers as ∆A → ∞gives the total population of the region R.

6. False. If the graph of f has equal volumes above and below the xy-plane over the region R, the double integral is zerowithout having f(x, y) = 0 everywhere.

7. True. Writing the definition of the integral of g, we have∫

R

g dA = lim∆A→0

∑g(x, y)∆A = lim

∆A→0

∑kf(x, y)∆A = k lim

∆A→0

∑f(x, y)∆A = k

∫

R

f dA.

8. False. As a counterexample, let R be a rectangle with area 2 and take f(x, y) = g(x, y) = 1. Then∫Rf · g dA =∫

R1 dA = Area(R) = 2, but

∫Rf dA =

∫Rg dA = Area(R) · Area(R) = 4.

9. False. There is no reason to expect this to be true, since the behavior of f on one half of R can be completely unrelatedto the behavior of f on the other half. As a counterexample, suppose that f is defined so that f(x, y) = 0 for points(x, y) lying in S, and f(x, y) = 1 for points (x, y) lying in the part of R that is not in S. Then

∫Sf dA = 0, since

f = 0 on all of S. To evaluate∫Rf dA, note that f = 1 on the square S1 which is 0 ≤ x ≤ 1, 1 ≤ y ≤ 2. Then∫

Rf dA =

∫S1f dA = Area(S1) = 1, since f = 0 on S.

10. True. Since all points in the regionR satisfy x < y, it is true that at every point inR, f(x, y) = x+x < x+y = g(x, y).Since all of the values of f in R are less than those of g, the average of the values of f is less than the average of thevalues of g.

11. False. Since the inside integral is performed with respect to x and the outside integral with respect to y, the region ofintegration is the rectangle 5 ≤ x ≤ 12, 0 ≤ y ≤ 1.

12. False. The iterated integral∫ 2

0

∫ 1

0f dxdy is over a rectangle. The correct limits are

∫ 1

0

∫ 2−xx

f dydx.

13. True. For any point in the region of integration we have 1 ≤ x ≤ 2, and so y is between the positive numbers 1 and 8.

14. False. The sign of∫ ba

∫ dcfdydx depends on the behavior of the function f on the region of integration. For example,∫ 2

1

∫ 2

1(−x)dydx = − 3

2.

15. False. The integrals are over different regions, so there is no reason to expect their values to be equal. The region ofintegration of

∫ 1

0

∫ x0f dydx is the triangle with vertices (0, 0), (1, 1) and (1, 0), while the region of

∫ 1

0

∫ y0f dxdy is the

triangle with vertices (0, 0), (1, 1) and (0, 1).

16. True. Since f does not depend on x, the inside integral (which is with respect to x) evaluates to∫ 1

0fdx = xf

∣∣∣x=1

x=0=

(f − 0) = f . Thus ∫ b

a

∫ 1

0

f dxdy =

∫ b

a

f dy.

17. False. The given limits describe only the upper half disk where y ≥ 0. The correct limits are∫ a−a∫√a2−x2

−√a2−x2

fdydx.

18. True. As an example, take the region inside the triangle with vertices (0, 0), (1, 1) and (2, 3). Iterated integration over thisregion using either the dxdy or the dydx orders requires breaking the region into two pieces:

∫ 1

0

∫ 3x/2

x

f dydx+

∫ 2

1

∫ 3x/2

2x−1

f dydx or∫ 1

0

∫ y

2y/3

f dxdy +

∫ 3

1

∫ 12

(y+1)

2y/3

f dxdy

19. True. In the inner integral with respect to y, the function g(x) can be treated as a constant, so∫ b

a

∫ d

c

g(x)h(y) dxdy =

∫ b

a

g(x)

(∫ d

c

h(y) dy

)dx.

The result of the integral∫ dch(y) dy is a constant, so may be factored out of the integral with respect to x. Thus we have

∫ b

a

g(x)

(∫ d

c

h(y) dy

)dx =

(∫ d

c

h(y) dy

)·(∫ b

a

g(x) dx

).

PROJECTS FOR CHAPTER SIXTEEN 1195

20. False. As a counterexample, consider∫ 2

0

∫ 2

0(x+ y) dxdy. We have

∫ 2

0

∫ 2

0

(x+ y) dxdy =

∫ 2

0

(x2

2+ yx

)∣∣∣∣x=2

x=0

dy =

∫ 2

0

(2 + 2y) dy = 8

and ∫ 2

0

x dx+

∫ 2

0

y dy =x2

2

∣∣∣∣x=2

x=0

+y2

2

∣∣∣∣y=2

y=0

= 2 + 2 = 4.

21. False. The integral gives the total mass of the material contained in W.

22. True. The region lies above the square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 and below the plane z = x.

23. False. The given limits only cover the part of the unit ball in the first octant where x ≥ 0, y ≥ 0, and z ≥ 0. To cover theentire unit ball the limits are ∫ 1

−1

∫ √1−x2

−√

1−x2

∫ √1−x2−y2

−√

1−x2−y2

f dzdydx.

24. True. Both sets of limits describe the solid region lying above the triangle x+ y ≤ 1, x ≥ 0, y ≥ 0, z = 0 and below theplane x+ y + z = 1.

25. True. Both sets of limits describe the solid region lying above the rectangle −1 ≤ x ≤ 1, 0 ≤ y ≤ 1, z = 0 and belowthe parabolic cylinder z = 1− x2.

26. False. The iterated integral is of the form like∫ ba

∫ dc

∫ kefdz dy dx only if the rectangular region has faces parallel to the

coordinate axes. More general rectangular regions, such as a cube with one corner at the origin and the opposite corner at(0, 0, 1) will need to be written as the sum of iterated integrals where the limits are not constant.

27. False. As a counterexample, consider f(x, y, z) = 12− x. Then f is positive on half the cube and negative on the other

half. Symmetry can be used to show that∫ 1

0

∫ 1

0

∫ 1

0( 1

2− x)dz dy dx = 0.

28. True. Since∫Wf dV = lim

∑

i,j,k

f(xi, yj , zk)∆V, where (xi, yj , zk) is a point inside the ijk-th subbox of volume ∆V,

and since f > g, we have

lim∑

i,j,k

f(xi, yj , zk)∆V > lim∑

i,j,k

g(xi, yj , zk)∆V =

∫

W

g dV.

29. False. As a counterexample, let W1 be the solid cube 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, 0 ≤ z ≤ 1, and let W2 be the solid cube− 1

2≤ x ≤ 0,− 1

2≤ y ≤ 0,− 1

2≤ z ≤ 0. Then volume(W1) = 1 and volume(W2) = 1

8. Now if f(x, y, z) = −1, then∫

W1f dV = 1 · −1 which is less than

∫W2

f dV = 18· −1.

30. True. If W is the solid region lying under the graph of f and above the region R in the xy-plane, we can compute thevolume of W either using the double integral

∫Rf dA, or using the triple integral

∫W

1 dV.

PROJECTS FOR CHAPTER SIXTEEN

1. (a) We are integrating over the whole plane, so converting to polar coordinates gives∫ ∞

−∞

∫ ∞

−∞e−(x2+y2)dxdy =

∫ 2π

0

∫ ∞

0

e−r2

rdrdθ =

∫ 2π

0

−1

2e−r

2

∣∣∣∣∞

0

dθ =

∫ 2π

0

1

2dθ = π.

(b) Rewriting the integrand as a product gives∫ ∞

−∞

∫ ∞


∫ ∞

−∞

∫ ∞

−∞e−x

2

e−y2

dxdy.

Now e−y2

is a constant as far as the integral with respect to x is concerned, so∫ ∞

−∞

∫ ∞

−∞e−x

2

e−y2

dxdy =

∫ ∞

−∞e−y

2

(∫ ∞

−∞e−x

2

dx

)dy.


We assume that the integral with respect to x converges, and so is a constant as far as the integral withrespect to y is concerned. Thus, we have

∫ ∞

−∞e−y

2

(∫ ∞

−∞e−x

2

dx

)dy =

(∫ ∞

−∞e−x

2

dx

)(∫ ∞

−∞e−y

2

dy

).

But∫∞−∞ e−x

2

dx and∫∞−∞ e−y

2

dy are the same number, so we can write

∫ ∞

−∞

∫ ∞


(∫ ∞

−∞e−x

2

dx

)(∫ ∞

−∞e−y

2

dy

)=

(∫ ∞

−∞e−x

2

dx

)2

.

(c) Using the results of parts (a) and (b), we have(∫ ∞

−∞e−x

2

dx

)2

=

∫ ∞

−∞

∫ ∞

−∞e−(x2+y2)dxdy = π.

Taking square roots and observing that the integral we are looking for is positive, we have∫ ∞

−∞e−x

2

dx =√π.

2. (a) We want to find the average value of |x− y| over the square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1:

Average distance between gates =

∫ 1

0

∫ 1

0

|x− y| dy dx.

Let’s fix x, with 0 ≤ x ≤ 1. Then |x− y| ={y − x for y ≥ xx− y for y ≤ x . Therefore

∫ 1

0

|x− y| dy =

∫ x

0

(x− y) dy +

∫ 1

x

(y − x) dy

=

(xy − y2

2

) ∣∣∣∣x

0

+

(y2

2− xy

) ∣∣∣∣1

x

= x2 − x2

2+

1

2− x− x2

2+ x2

= x2 − x+1

2.

So,

Average distance between gates =

∫ 1

0

∫ 1

0

|x− y| dy dx

=

∫ 1

0

(∫ 1

0

|x− y| dy)dx =

∫ 1

0

(x2 − x+1

2) dx

=x3

3− x2

2+

1

2x

∣∣∣∣1

0

=1

3.

(b) There are (n + 1)2 possible pairs (i, j) of gates, i = 0, . . . , n, j = 0, . . . , n, so the sum given representsthe average distances apart of all such gates. The Riemann sum with ∆x = ∆y = 1/n, if we choose theleast x and y-values in each subdivision is

n−1∑

i=0

n−1∑

j=0

∣∣∣∣i

n− j

n

∣∣∣∣1

n2,

which for large n is just about the same as the other sum. For n = 5 the sum is about 0.389; for n = 10the sum is about 0.364.

1105 chapter sixteen - valenciafd.valenciacollege.edu/file/tsmith143/ch16.pdf16.1 solutions 1105...

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