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7/31/2019 Double Int Gen

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210 CHAPTER 4. MULTIPLE INTEGRALS

Figure 4.7: Bounded region in R2

4.3 Double Integrals Over General Regions

4.3.1 Introduction

When dealing with integrals of functions of one variable, we are always inte-grating over an interval. The only diculty in evaluating the denite integral

Rba f(x) dx came from the function f and the diculty to nd an antiderivativefor it. With functions of two or more variables, not only the function can causethe integral to be dicult, but also the region over which we integrate. In theprevious section, we restricted ourselves to rectangular regions. However, notevery region is rectangular. A region ofR2 can have any shape. Even if thefunction is easy to integrate, if the region is complex enough, the integral willstill be dicult to evaluate. We rst dene the integral of a function over twovariables over a general region. We will then show how this integral can beevaluated over some simple regions which we will call regions of types I and II.

Let us assume we are given a function f(x; y) dened on a closed andbounded region D as shown in gure 4.7. Suppose we want to integrate fover D. Because D is bounded, it can be enclosed in a rectangle R as shown ingure 4.8.

To dene the integral of f over D, we rst dene the function F (x; y) asfollows:

F (x; y) =

f(x; y) if (x; y) 2 D0 if (x; y) 2 R D (4.3)


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4.3. DOUBLE INTEGRALS OVER GENERAL REGIONS 211

Figure 4.8: Bounded region in R2, enclosed in a rectangle.

Denition 330 If

ZZR

F (x; y) dA exists, then we dene

ZZD

f(x; y) dA =

ZZR

F (x; y) dA

This denition makes sense since, as we can see in gures 4.9 and 4.10, theportion of the graph of F which is not 0 is identical to the graph of f: Theportion which is 0 will not contribute to the integral. In particular, this meansthat for our denition, it does not matter which rectangle R we select. In the

case, f(x; y) 0,ZZD

f(x; y) dA corresponds to the volume of the solid which

lies above D and below the graph of z = f(x; y).

We still must be able to compute

ZZR

F (x; y) dA. This is not always a simple

task. But it is for certain regions, which we consider next.

4.3.2 Regions of Type I

When describing a region, one has to give the condition x and y must satisfy

so that a point (x; y) lies in the region. A region is said to be of type I if x isbetween two constants, and y is between two continuous functions of x. Moreprecisely, we have the following denition:

Denition 331 A plane region D is said to be of type I if it is of the form

D =

(x; y) 2 R2 j a x b and g1 (x) y g2 (x)


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Figure 4.9: Graph of f

Figure 4.10: Graph of F


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Figure 4.11: Region of type I

where g1 and g2 are two continuous functions of x.

Possible regions of type I are shown in gures 4.11, 4.12 and 4.13. To evaluateZZD

f(x; y) dA, we enclose D in the rectangle [a; b] [c; d] as shown in gure

4.3.2.We dene F (x; y) as explained in equation 4.3. So, we have

ZZD

f (x; y) dA = ZZR

F (x; y) dA

=

Zba

Zdc

F (x; y) dydx by Fubinis theorem

To evaluate this iterated integral, we rst evaluate the inside integral. Usingthe properties of the denite integral, we have:Zd

c

F (x; y) dy =

Zg1(x)c

F (x; y) dy +

Zg2(x)g1(x)

F (x; y) dy +

Zdg2(x)

F (x; y) dy

=

Zg2(x)g1(x)

F (x; y) dy

since F (x; y) = 0 for y < g1 (x) and y > g2 (x) (see gure 4.3.2). Also when y

is between g1 (x) and g2 (x), F (x; y) = f(x; y), so we haveZdc

F (x; y) dy =

Zg2(x)g1(x)

f(x; y) dy

Therefore, we have the following proposition:


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Proposition 332 If f is continuous on a type I region D such that

D =

(x; y) 2 R2 j a x b and g1 (x) y g2 (x)

then ZZD

f (x; y) dA =

Zba

Zg2(x)g1(x)

f(x; y) dydx (4.4)

Remark 333 Notice that the functionF (x; y) does not appear in the formula.It was simply a device used to derive this formula. When computing an integral,it does not come into play, as the next example will show.

Remark 334 When writingZZD

f(x; y) dA as an iterated integral, it is impor-

tant to remember that the outer integral must be the one with constant limit ofintegrations.

Remark 335 To evaluate a double integral over a general region, the rst stepis always to write it as an iterated integral. How this is done depends on theregion. The region will determine what the limits of integration of the iteratedintegrals are. Hence, it is extremely important to really understand what theregion looks like. Graphing it will help.

Example 336 Write

ZZD

f(x; y) dA as an iterated integral whereD =

(x; y) 2 R2 j 0 x 1 and x2

Here, the region is given explicitly, we know the limits for both x and y, so wecan write the iterated integral.ZZ

D

f(x; y) dA =

Z10

Zx2x2

f(x; y) dydx


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Example 337 Write ZZD

f(x; y) dA as an iterated integral where D is the re-

gion bounded by the x-axis, y = sin x, the lines x = 0 and x = 1.The region is shown in gure 337.

-1.0 -0.8 -0.6 -0.4 -0.2 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0

-1

1

2

x

y

Region D

We see that we can writeD as: D = f(x; y) j 0 x 1 and 0 y sin xg andhence

ZZD

f(x; y) dA =

Z10

Zsinx0

f(x; y) dydx

Example 338 Evaluate

ZZD

x4 2y dA whereD = (x; y) 2 R2 j 1 x 1 and x2 y


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D is clearly a type I region.

ZZD

x4 2y dA = Z1

1

Zx2

x2

x4 2y dydx

=

Z11

hx4y y2

x2x2

idx

=

Z11

x6 x4 + x6 + x4 dx

=

Z11

2x6dx

=2

7x71

1

= 47

4.3.3 Regions of Type II

The condition a point (x; y) must satisfy to be in a type II region is as follows.A region is said to be of type II if y is between two constants, and x is betweentwo continuous functions of y. More precisely, we have the following denition:

Denition 339 A plane region D is said to be of type II if it is of the form

D =

(x; y) 2 R2 j c y d and h1 (y) x h2 (y)

Using the same method as for type I regions, we can show that

Proposition 340 If f is continuous on a type II region D such that

D =

(x; y) 2 R2 j c y d and h1 (y) x h2 (y)

then ZZD

f(x; y) dA =

Zdc

Zh2(y)h1(y)

f(x; y) dxdy (4.5)

Example 341 EvaluateZZD

xy y3

dA if D is the region bounded by the x-axis; the lines x = 1, y = 1 and y = x.It often helps to sketch the region so we can see what type of region it is. Thisregion is shown in gure 4.14. This is a region of type II, it can be describedby D =

(x; y) 2 R2 j 0 y 1 and 1 x y. In type II regions, x is the

variable between two functions of y. Make sure these are functions ofy, not x.


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Figure 4.14: Region bouded by the x axis, and the lines y = 1, x = 1, andy = x.

In this case, the right side of the region is delimited by the line y = x, whichwhen we write as a function of y becomes x = y. Therefore, we haveZZ

D

xy y3 dA = Z1

0

Zy1

xy y3 dxdy

=

Z10

"x2

2y xy3

y

1

#dy

=

Z1

0 y3

2 y4

y

2+ y3

dy

=

Z1

0

y4 y

3

2 y

2

dy

=

y

5

5 y

4

8 y

2

4

1

0

= 2340

4.3.4 Properties

We list without proof several properties the double integral satises. Theseproperties are very similar to the properties of the denite integral of functionsof one variable.

Proposition 342 Assuming that the following integrals exist and that C is aconstant, we have:

1.

ZZD

(f(x; y) g (x; y)) dA =ZZD

f(x; y) dA ZZD

g (x; y) dA


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2. ZZD

Cf (x; y) dA = CZZD

f(x; y) dA

3. Iff(x; y) g (x; y) for all(x; y) inD, thenZZD

f(x; y) dA ZZD

g (x; y) dA

Proposition 343 If D, D1 and D2 are three regions in R2 such that D =

D1 [ D2 and D1 and D2 do not overlap thenZZD

f(x; y) dA =

ZZD1

f(x; y) dA +

ZZD2

f(x; y) dA

This property corresponds to the property for functions of one variable which

says:Rba

f(x) dx =Rca

f (x) dx +Rbc

f(x) dx. It can sometimes be used to evalu-ate integrals over regions which are neither of type I, nor of type II. If we havesuch a region and we can divide it into two (or more regions) of type I or II ,then we can break the integral into integrals over the subregions. Each integralis then an integral of type I or II, which we know how to evaluate.

Proposition 344

ZZD

1dA = A (D) where A (D) denotes the area of D.

This property says that if we integrate the constant function f (x; y) = 1over a region D, we get the area of D. This makes sense. The volume of a solida section D between two parallel planes is (area of the section) height. In thiscase, the height is simply 1. This is also very important as it gives us a way ofcomputing the area of general regions. We illustrate it with an example.

Example 345 Find the area of the region D in the rst quadrant bounded byy = x and y = x2.The region is shown in gure 4.15. We can treat this as a type 1 or a type 2region. We show the solution both ways.

Type 1 region. The area of the region is ZZD

dA where D is the region

described in the problem, in green on gure 4.15. To evaluate this integral,we must rewrite it as an iterated integral treatingD as a type 1 region. Weimagine covering the region with vertical strips as shown in gure 4.16.We would use a strip for every x such that 0 x 1 and for each x, the


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Figure 4.15: Region between y = x and y = x2


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Figure 4.16: Type 1 region

range of y values along the strip would be: x2 y x. Therefore

ZZD

dA =Z10

Zxx2

dydx

=

Z10

yjxx2 dx

=

Z10

x x2 dx

=

x2

2 x

3

3

1

0

=1

2 1

3

=1

6

Type 2 region. This is similar to above, but we imagine covering the regionwith horizontal strips as shown in gure 4.17. we would use such strips

for 0 y 1. For each y, the range of x values along the strip would be


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Figure 4.17: Type 2 region

y x py. Therefore

ZZD

dA =Z10

Zpyy

dxdy

=

Z10

xjpy

y dy

=

Z10

(p

y y) dy

=

0@2

3y

3

2 y2

2

1A1

0

=2

3 1

2

=1

6

Combining the above properties, we have:


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Figure 4.18: Region bounded by y = x2 and y = 4p

x

Proposition 346 If m f(x; y) M for all (x; y) in D, then

mA (D) ZZD

f(x; y) dA M A (D)

Denition 347 (Average Value) We dene the average value of a functionf(x; y) over a region D to be

Average value of f over D =

1

area of DZZD

f(x; y) dA

4.3.5 More Practice Problems

In many cases, a region can be considered of type I or II. One then has to decidewhether to treat it as a type I region or a type II. It may be that either way willwork. However, in some cases, one way may lead to an integral which is verydicult, while the other way leads to an easier integral. In the example thatfollows, we look at cases to illustrate these various options. The more problemyou practice on, the better you will become at identifying the type of a region,and how to perform the integration.

Example 348 EvaluateZZ

px y2 dA where is the region bounded byy = x2 and y = x

1

4 .Once again, it helps to draw the region. It is shown in gure 4.18


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Figure 4.19: Region bounded by y = x2 and y = 4p

x

To completely dene the region, one need to nd the points where the two regionsmeet. For this, we solve

y = x2

y = 4p

x

Combining the two gives

x2 = x1

4

If we raise both sides to the fourth power, we get

x8 = x

x8 x = 0x

x7 1 = 0x = 0 or x = 1

When x = 0, we get y = 0. When x = 1, we get y = 1. We see that this region

can be seen as a type I region: = n(x; y) j 0 x 1 and x2 y x 14o. Note

that y is between two functions of x. can also be seen as a type II region.(seegure 4.19). In this case, =

(x; y) j 0 y 1 and y4 x py. Note that

x is between two functions ofy. We obtained them as follows. From gure 4.19,we see that the region is bounded on the left by y = x

1

4 which can be written asx = y4. It is bounded on the right by y = x2 which can be written as x =

py.

We compute the integral both ways.


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Method 1 Treating as a type I region.

ZZ

px y2 dA = Z1

0

Zx1

4

x2

px y2 dydx

=

Z10

24ypx y3

3

x1

4

x2

35 dx

=

Z10

x

3

4 x3

4

3

!

x5

2 x6

3

!dx

=

Z10

2

3x

3

4 x 52 + x6

3

dx

= 8

21 x

7

4 2

7 x

7

2

+

x7

21

1

0

=8

21 2

7+

1

21

=1

7

Method 2 Treating as a type II region.

ZZ

px y2 dA = Z1

0

Zy 12y4

px y2 dxdy

=Z10

2423 x

3

2 xy2

y1

2

y4

35dy

=

Z10

2

3y3

4 y 52

2

3y6 y6

dy

=

Z10

2

3y3

4 y 52 + y6

3

dy

=

8

21y7

4 27

y7

2 y7

21

1

0

=

8

21 2

7+

1

21

=

1

7

Example 349 Evaluate

ZZ

cos x2

2 dA whereD is the region bounded byx = 1,

y = 0 and y = x.


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Figure 4.20: Region bounded by x = 0, x = 1, y = 0 and y = x

The region is shown in gure 4.20. Once again, we see that this region can betreated as a type I or type II region.

Method 1 Treating as a type I region. The corner points of the region are(0; 0), (1; 0) and(1; 1). The region is determined by = f(x; y)g0 x 1 and 0 y xgIn this case:

ZZ

cosx2

2dA =

Z10

Zx0

cosx2

2dydx

=

Z1

0

y cos

x2

2

x0

dx

=

Z10

x cosx2

2dx

We nish with a substitution. If we letu = x2

2 then du = xdx. Also,when x = 0, u = 0 and when x = 1, u = 2 . So, we have

ZZ

cosx2

2dA =

1

Z2

0

cos udu

=1

(sin u)j

2

0

=1

Method 2 Treating as a type II region. The region is determined by =


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f(x; y) j 0 y 1 and y x 1g. SoZZ

cosx2

2dA =

Z10

Z1y

cosx2

2dxdy

We cannot evaluate the inside integral as we do not know an antiderivative

with respect to x ofR1y

cos x2

2dx.

Example 350 Calculate the volume of the solid within the cylinder x2 + y2 =b2, between the planes y + z = a and z = 0, given that a

b > 0.

The region is shown in gure 350. The region D inR2 by which the solid isbounded is D =

(x; y) j x2 + y2 b2. The solid is also bounded by the planes

z = 0 and z = y + a. Therefore, its volume is given by the integralZZD

(a y) dA =ZZD

adA ZZD

ydA

The second integral is easier to evaluate than it seems. The regionD can bewritten as

D =n

(x; y) j b x b and p

b2 x2 y p

b2 x2o

Thus, ZZD

ydA =

Zbb

Zpb2x2pb2x2

ydydx

You will recall that integrating an odd function over a symmetric integral gives0, in other words, if f(x) is an odd function, then

Raa f(x) dx = 0. Thus,


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Rpb2x2

pb2x2 ydy = 0. It follows that

ZZD

(a y) dA =ZZD

adA

= a

ZZD

dA

= aA (D) by one of the properties above

= ab2

4.3.6 Switching the Order of Integration

Evaluating a double integral over a region amounts to evaluating one or moreiterated integrals. As we know, there are two kinds of iterated integrals. Onein which the dx integral is the inner integral and the dy integral is the outerintegral. The other kind is the reverse. When the region of integration is arectangle, Fubinis theorem tells us that the two integrals are the same. If R =

[a; b] [c; d]. ThenZZR

f(x; y) dA =Rba

Rdc

f(x; y) dydx =Rdc

Rba

f(x; y) dxdy.

When the region of integration is not a rectangle, we may not have a choice.We may be forced to use one of the iterated integrals as suggests example349. That example also suggests that given an iterated integral, to evaluate itmight require switching the order of integration. This cannot be done by justreversing the order of the integrals. It involves changing the type of region we

are integrating over. We illustrate this with an example.

Example 351 Switch the order of integrationR10

R1y

cos x2

2dxdy.

First, we need to identify the region, decide what type of region it is. Then,we rewrite it as a region of the other type and write the integral in terms ofthat region. From the limits of integration, we see that the region is D =

f(x; y) : 0 y 1 and y x 1g. Thus, this is a type II region. ThusR10

R1y

cos x2

2 dxdy =ZZD

cos x2

2dA where D is as indicated and is shown in gure 4.20. To rewrite

as a type I region, we need to express x between two constants and y betweentwo functions. From gure 4.20, we see that D can also be written as D =

f(x; y) : 0 x 1 and 0 y xg. Therefore, ZZD

cos x2

2 dA =R10Rx0 cos x

2

2 dydx.

Hence Z10

Z1y

cosx2

2dxdy =

Z10

Zx0

cosx2

2dydx


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4.3.7 Summary

We list important points covered in this section.

Since not every region in the plane is a rectangle, we generalized what wehad learned in the previous section to more general regions, closed andbounded regions. If we call D such a region, then we learned to computeZZD

f(x; y) dA.

It is very important to understand that for multiple integrals, the region ofintegration, D, plays a very important role. I cannot emphasize enough theimportance of making sure one knows the region of integration well before

attempting to compute ZZD

f(x; y) dA. Every multiple integral should

start with drawing D, the region of integration, and writing a concisemathematical description of it.

We actually consider two types of regions. Type I regions and type IIregions. Unlike rectangular regions, the order of integration cannot beswitched easily, at least not without some work.

Type I regions: D = f(x; y) j a x b and g1 (x) y g2 (x)g. In thiscase

ZZD

f(x; y) dA =

b

Za

g2(x)

Zg1(x)

f (x; y) dydx

Note that the limits of integration of the outer integral are always con-stants.

Type II regions: D = f(x; y) j c y d and h1 (y) x h2 (y)g. In thiscase ZZ

D

f(x; y) dA =

dZc

h2(x)Zh1(x)

f(x; y) dxdy

Note that the limits of integration of the outer integral are always con-stants.

GivenZZD

f(x; y) dA, students must be able to determine if D is a type

I or II region and write the double integral as the corresponding iteratedintegral.


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Given an iterated integral, that is one of

b

Za

g2(x)

Zg1(x)

f(x; y) dydx or

d

Zc

h2(x)

Zh1(x)

f(x; y) dxdy,

students must be able to nd the region D such that the iterated integral

is the same as

ZZD

f(x; y) dA.

When D, the region of integration, can be written as both a type I and atype II region, it is possible to switch the order of integration. Sometimesit is necessary to do so. To reverse the order of integration of an iteratedintegral, perform the following steps:

Given an iterated integral, nd the region D that it corresponds toand identify the type of region D is.

Write D as a region of the other type. Write the iterated integral corresponding to the newly written region.

If the graph of z = f(x; y) is above the xy-plane thenZZD

f(x; y) dA is

the volume of the solid with cross section D, between the xy-plane andz = f(x; y).

ZZD

dA gives the area of the region D.

4.3.8 Assignment1. Do odd # 1 - 27, 29, 35, 37 at the end of 12.2 in your book.

2. Do odd # 1, 3, 5, 9, 11 at the end of 12.3 in your book.

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