Calculus Lectures: Volumes i

by Kathy Davis

Copyright ©2021 Kathy Davis

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2 Kathy Davis

Figure 1: Riemann Slices

When you do a Riemann sum for thisvolume, it looks like slices.

Figure 2: One Piece

One slice; we’ll approximate the volumeof this.

Figure 3: Areas

The volume of a slice will be the thick-ness times the area of the front.

Figure 4: Front View

The area A(y1) is a lot like the areayou’d get from a regular integral.

How

This lecture shows how to compute volumes by taking two integrals.And I’ll show it using Riemann Sums.

Say I want to know the volume of the loaf of bread above. It wasprobably baked in a pan with a rectangular base and vertical sides,so D will be a rectangle {(x, y) | a ≤ x ≤ b; c ≤ y ≤ d}. The topof the loaf was free to rise, and it’s described by the usual functionz = f (x, y). Now I slice in the usual way, with the knife cuttingparallel to the x-axis. Any line parallel to the x-axis has a constant y;that means I Riemannize in y: c = y0 < y1 < y2 · · · yn = d, Figure 1.Now I want to look at one slice, Figure 2. It’s the piece y0 ≤ y ≤ y1,with thickness ∆y1 = y1 − y0.

I approximate the volume of the slice as the area of the front, timesthe thickness of the slice. From Figure 3, that’s A(y1) · ∆y1. ButA(y1)? looks a lot like Figure 4: just a plain old definite integral inx,

A(y1) =∫ b

af (x, y1) dx

Now Riemannize in y (add up the individual approximations andtake the limit as all the ∆y’s go to zero), you get an equation for thevolume V

V = lim∆yn→0

∑n

A(yn) · ∆yn = lim∆yn→0

∫ b

af (x, y1) dx · ∆yn

=∫ d

cA(y) dy =

∫ d

c

∫ b

af (x, y) dxdy

calculus lectures 3

∫ d

c

∫ b

af (x, y) dxdy

is called a Type II or T2 double integral for the volume; the outsideintegral has limits of integration c, d for y, and corresponding to that,the outside is a dy. They have to match, and also the insides match inx. It’s probably no surprise that this will correspond to a T2 domain,which I’ve been leading up to for a few lectures.

If I made my slices parallel to the y axis (a very long sandwich), thatwould have given me

V =∫ b

aA(x) dx =

∫ b

a

∫ d

cf (x, y) dydx

a Type I or T1 double integral for the volume. As the saying goes,whichever way you slice it – the result is called Fubini’s Theorem: iff (x, y) is continuous,∫ b

a

∫ d

cf (x, y) dydx =

∫ d

c

∫ b

af (x, y) dxdy

Going back and forth T1 ←→ T2 is called interchanging integrals, andit’s part of what’s on Q8, the "final."

And now it’s time to do some actual computations. These aren’t onthe quiz, but they are preparation for doing the quiz computations.

Worked Problems

f (x, y) = x(1 + y); D = {(x, y) | 0 ≤ x ≤ 1; 1 ≤ y ≤ 2}

Problem: Find the volume above D, under f , using each of T1 and T2

double integrals.

I’ll start with T1; that’s where the x-stuff goes on the outside, so it’s:∫ 1

0

∫ 2

1x(1 + y) dydx

Double integrals are like second partial derivatives: do the insideone first. And, like ∂

∂y , x looks like a constant, so it factors out ofderivatives – and integrals:

∫ 2

1x(1+ y) dy = x

∫ 2

1(1+ y) dy = x

[y +

y2

2

]2

1= x

[2 +

42

]− x

[1 +

12

]= x

[4− 3

2

]= x

[82− 3

2

]=

5x2

Now the outside integral:

52

∫ 1

0x dx =

52

[x2

2

]1

0=

52

[12

]=

54

4 Kathy Davis

T2 :∫ 2

1

∫ 1

0x(1 + y) dxdy

Inside :∫ 1

0x(1+ y) dx = (1+ y)

∫ 1

0x dx = (1+ y)

[x2

2

]1

0= (1+ y)

12

Outside :∫ 2

1(1+ y)

12

dy =12

[y +

y2

2

]2

1=

12

[2 +

42

]− 1

2

[1 +

12

]=

12

[4− 3

2

]=

12· 5

2=

54

So T1 = T2 after all. But there’s something unusual about the com-putations: it seems like I was doing the same computation twice.There’s a secret reason for this: My f (x, y) was equal to g(x)h(y), so∫ b

a

∫ d

cg(x)h(y) dydx =

∫ b

ag(x)

∫ d

ch(y) dydx since

∫dy thinks g(x) is a constant.

But now,∫ d

ch(y) dy is a constant! So it pulls out of

∫ b

ag(x)

∫ d

ch(y) dydx

and now I get∫ b

a

∫ d

cg(x)h(y) dydx =

(∫ b

ag(x)dx

)(∫ d

ch(y) dy

)which is a cool trick I’ll use a lot. ONLY when f (x, y) = g(x)h(y).

Problem: Compute ∫ 1

0

∫ x2

0(x + y) dydx

Inside :∫ x2

0x+ y dy =

[xy +

y2

2

]x2

0=

[x · x2 +

(x2)2

2

]− [0] = x3 +

x4

2

Outside :∫ 1

0x3 +

x4

2dx =

[x4

4+

x5

10

]1

0=

[14+

110

]− [0] =

1440

=7

20

Now I’ll try interchanging integrals:∫ x2

0

∫ 1

0(x + y) dxdy

Inside :∫ 1

0(x + y) dx =

[x2

2+ xy

]1

0=

[12+ y]− [0] =

12+ y

Outside :∫ x2

0

12+ y dy =

[y2+

y2

2

]x2

0=

[x2

2+

x4

2

]− [0] =

12(x2 + x4)

One of the early inventors of Quantum Mechanics, Wolfgang Pauli,once looked at some work on a blackboard and famously said,"That’s not even wrong." The answer I just got is what’s called acategory mistake; it’s like asking ’What’s the capital of France?" andgetting the answer "taco."

That’s about how far off I am. So what went wrong with Fubini? Ican’t interchange integrals now?

calculus lectures 5

Figure 5: Integral As Domain

You can change the limits of integrationin a double integral over a domain.

Figure 6: Interchanging Integrals

The key to interchanging integrals is tointerchange domains.

Interchanging Integrals

I just tried Fubini and got an idiot answer:∫ x2

0

∫ 1

0(x + y) dxdy =

12(x2 + x4)

This is the real way to do Fubini:∫ x2

0

∫ 1

0(x + y) dxdy =

∫ 1

0

∫ 1

√y(x + y) dxdy

Inside∫ 1

√y(x + y) dx =

[12

x2 + xy]1

√y=

[12+ y]−[

12

y + y√

y]

=12(1 + y)− y

32

Outside∫ 1

0

12(1 + y)− y

32 dy =

[y2+

y2

4− 2

5y

52

]1

0

=

[12+

14− 2

5

]=

[34− 2

5

]=

15− 820

=720

But how am I supposed to know∫ 1

0

∫ x2

0f dydx ←→

∫ 1

0

∫ 1

√y

f dxdy ?

The answer is by interchanging domains, T1 ↔ T2:

Worked Problems

Problem a) Find the domain D for∫ 1

0

∫ x2

0 f dydx and then draw D.

In the outside integral∫ 1

0 . . . dx, the limits of integration tell me

0 ≤ x ≤ 1; similarly, the inside integral∫ x2

0 . . . dy tells me 0 ≤ y ≤ x2.So D = {(x, y) | 0 ≤ x ≤ 1; 0 ≤ y ≤ x2}. The integral was a T1

integral, and D is a T1 domain, Figure 5.

Problem b) Write D as a T2 domain.

I need D = {(x, y) | c ≤ y ≤ d; l(y) ≤ x ≤ r(y)}, see Figure 6.

i) c is the bottom of D, where y = x2 intersects x = 0: at (0, 0); c = 0.ii) d is the top of D, where y = x2 intersects x = 1, at (1, 1): d = 1.iii) x = l(y) is the left curve of D; that’s the curve y = x2, but I needx = l(y), so take y = x2 and solve for x: x = ±√y. But D is in thefirst quadrant, so l(y) =

√y.

iv) x = r(y) is the right curve of D; it’s the line x = 1 so r(y) = 1.

D = {(x, y) | 0 ≤ y ≤ 1;√

y ≤ x ≤ 1)}

6 Kathy Davis

Figure 7: Flipped Parabola

Changing y = x2 to y = −x2 flips theparabola upside-down.

Figure 8: D as T1

Mostly easy: the parabola is the top; theline is the bottom.

Problem c) Write the T1 integral as a as a T2 integral.

This is easy: change 0 ≤ y ≤ 1 to∫ 1

0 dy, and√

y ≤ x ≤ 1 to∫ 1√

y dx;in T2, y is on the outside, so I get

T2 =∫ 1

0

∫ 1

√y(x + y) dxdy

If you can see the answers right away, you can show less work, butyou still have to show YOU did the work. But, having a fixed proce-dure is good when it’s hard to see the answer right away!

Important Ideas:0) If D isn’t a rectangle, there are variables in the limits of integration.i) If the inside integral has variables in its limits of integration, inter-changing integrals by just switching the integral signs won’t work.ii) The way to interchange integrals is to use the limits of integrationto find a domain, then interchange the domains T1 ↔ T2, and thenplace those into the new limits of integration..iii) If I want a numerical answer for a volume, I can’t have variablesin the limits of integration for the outside integral.

Words to live by. Or, at least, to pass the quizzes.

Problem a) Let D be bounded by y = −x2, y = −1; draw D.

Solution: see Figure 7.

Problem b) Write D as a T1 domain.

Solution: I need D = {(x, y) | a ≤ x ≤ b; b(x) ≤ y ≤ t(x)}. a, b arethe points where y = −x2 intersects y = −1:

y = −1 = −x2 → x2 = 1 → x = ±1→ a = −1, b = 1

Figure 8 shows the top and bottom curves are the ones in the thebounds given to me, "bounded by y = −x2, y = −1". So,D = {(x, y) | − 1 ≤ x ≤ 1; −1 ≤ y ≤ −x2}.

Problem c) If f (x, y) = y, write a T1 integral for the volume under f ,above D, and compute the volume.

Solution: usually I’ll just say: "find∫

D

∫f · dA". Nw I’ll skip to the

next page, so everything is right in front of me.

calculus lectures 7

D = {(x, y) | − 1 ≤ x ≤ 1; −1 ≤ y ≤ −x2}

To get the limits of integration, just take the x, y limits from D:∫D

∫f · dA =

∫ 1

−1

∫ −x2

−1y dydx

Inside∫ −x2

−1y dy =

[y2

2

]−x2

−1=

[x4

2

]−[

12

]=

12(x4 − 1)

Outside∫ 1

−1

12(x4− 1) dx = 2 ·

∫ 1

0

12(x4− 1) dx since f is an even function

=

[x5

5− x]1

0=

[15− 1]− [0] = −4

5

Figure 9: D as T2

This time, I have to split the parabolainto a left and a right piece.

Problem d) Now do everything all over again as T2. Less snarky:Write D as a T2 domain.

See Figure 9. c, d are the top and bottom parts of the domain; thebottom was handed to me as y = −1 and the top is the vertex of theparabola, at y = 0. So c = −1, d = 0.

For the left, right pieces, the parabola y = −x2 is a one-piece deal,and I have to split it. I’ve run into this before, with circles: all I haveto do is solve for x by taking a square root: y = −x2 and then solve:x2 = −y, x = ±√−y. Taking account which ± puts us in whichquadrant, l(y) = −√−y, r(y) = +

√−y:

D = {(x, y) | − 1 ≤ y ≤ 0; −√−y ≤ x ≤

√−y}

Problem d) If f (x, y) = y, use a T2 integral to find the volume underf , above D, and compute the volume.∫

D

∫f · dA =

∫ 0

−1

∫ √−y

−√−yy dxdy

Inside∫ √−y

−√−yy dx = y

∫ √−y

−√−y1 dx = 2y

∫ √−y

01 dx = 2y

√−y

Outside∫ 0

−12y√−y dy and I′m stuck. But try this∫ 0

−12y√−y dy = −2

∫ 0

−1−y√−y dy; u = −y so = −2

∫ 0

1u√

u (−du)

= −2∫ 1

0u

32 du = −2

[25

u52

]1

0= −2

[25

]= −4

5

So again T1 = T2, which I knew. There’s a point to this? T1 was a loteasier to do than T2. Since the two of them are equal, I can alwayschoose the easiest one. Except on quizzes :(

8 Kathy Davis

Figure 10: Sector of Circle as T2

I saw last Friday that the sector isn’teven T1. Another reason why I needboth T! and T2.

Now I want to do a problem that leads me to the next topic in thecourse. Last Friday, I did a sector of a circle, Figure 10. It’s given byD = {(x, y) | x2 + y2 ≤ 1; 0 ≤ y ≤ x}. It isn’t T1, but it is T2:

{(x, y) | 0 ≤ y ≤ 1√2

; y ≤ x ≤√

1− y2}

Problem a) Write∫

D

∫y · dA as a T2 integral.

Solution: as before, just take the limits from the T2 domain:∫D

∫y · dA =

∫ 1√2

0

∫ √1−y2

yy dxdy

Problem b) Now compute the integral.

Solution: This isn’t easy. And that’s the take-away that motivates thenext lecture, where it will become lots easier. Anyway:

Inside∫ √1−y2

yy dx = y

∫ √1−y2

y1 dx = y

[√1− y2 − y

]

Outside∫ 1√

2

0

[y√

1− y2 − y2]

dy =∫ 1√

2

0y√

1− y2 dy−∫ 1√

2

0y2 dy

The second integral is just[y3

3

] 1√2

0=

(13

)(1√2

)3=

(13

)(12

) 32

In the first integral, make u = 1− y2; du = −2y dy and y dy = − 12 du.

When y = 0, u = 1; when y = 1√2

, u = 12 . So, the integral is

∫ 12

1

√u (−1

2)du =

12

∫ 1

12

√u du =

12

[23

u32

]1

12

=13

[1− (

12)

32

]Together, the value is

13

[1−

(12

) 32]−(

13

)(12

) 32=

13

[1− 2

(12

)1 (12

) 12]=

13

[1− 1√

2

]

Compute∫ 1

0

∫ 1

xe−y2

dydx

This is going to be tricky:∫

e−y2dy isn’t equal to any other function.

I could try Taylor series :) :)

ex =∞

∑n=0

xn

n!; e−y2

=∞

∑n=0

(−y2)n

n!=

∞

∑n=0

(−1)n y2n

n!∫e−y2

dy =∞

∑n=0

(−1)n∫ y2n

n!dy =

∞

∑n=0

(−1)n y2n+1

(2n + 1)n!

The good news is that this works; I put in an hour of work, and gotthe right answer.

calculus lectures 9

Figure 11: Domain As T1

The domain turns out to be a triangle

Figure 12: Now It’s T2

Still a triangle

Interchanging integrals is easier; it’s almost like this integral is theposter child for the Interchanging Foundation. As always, I have toconvert the limits of integration into a T1 domain, then change to a T2

domain, then convert to an integral, and finally, solve it.∫ 1

0

∫ 1

xe−y2

dydx −→ D = {(x, y) | 0 ≤ x ≤ 1; x ≤ y ≤ 1}

Figure 11 shows what it looks like. Figure 12 rethinks it as a T2 do-main.

i) c is the bottom point of D: y = x intersects x = 0, at (0, 0), so c = 0.ii) d is the top point of the triangle: y = x intersects x = 1, at (1, 1);d = 1.iii) x = l(y) is the left line of the triangle, x = 0: l(y) = 0.iv) x = r(y) is the right line of the triangle, y = x → x = y: r(y) = y.

D = {(x, y) | 0 ≤ y ≤ 1; 0 ≤ x ≤ y)} −→∫ 1

0

∫ y

0e−y2

dxdy

Inside :∫ y

0e−y2

dx = e−y2∫ y

01 dx = e−y2

(y− 0)

Outside :∫ 1

0ye−y2

dx

Now a u-subbie works: let u = −y2; du = −2y dy and y dy = − 12 du.

When y = 0, u = 0; when y = 1, u = −1. So now I have

∫ 1

0ye−y2

dx =∫ −1

0eu(−1

2) du = (−1

2)∫ −1

0eu du =

12

∫ 0

−1eu du

=12[eu]0−1 =

12

[e0]− 1

2

[e−1]=

12

[1− 1

e

]I want to finish the section with an example: the great Miyagi earth-quake and tsunami of March 2011. It killed over 10,000 people,caused the meltdown of the three Fukushima Dai Ichi nuclear re-actors, which led to radioactive contamination on a massive scale,and resulted in over a trillion dollars of damage. In some sense, thecountry was lucky; Miyagi has a relatively small population.

Because the 3/11 quake was so devastating, emergency planners hadto rethink what would happen if a quake and tsunami hit Tokyo.Even though Tokyo has 13 million inhabitants, the buildings arewell protected against quakes, so Emergency Preparedness estimatesagain about 10,000 deaths but 100,000 injuries from the quake.

10 Kathy Davis

Figure 13: Transportation

Tokyo is almost completely dependenton its rail system. Here, the destructionof transportation in Miyagi Prefecture.

Figure 14: Outside Is Safe

In an earthquake, people know to leavetheir buildings, in case of collapse. Eventhough Tokyo only felt a slight shock.

Figure 15: Outside Is Not Safe

With no effective warning and nowhereto go – it’s over.

Figure 16: Is It Enough?Notice the size; the arrow points to onehuman-unit height. How much wouldthis hold?

The first problem is getting care to the injured: Figure 13 shows whatthe quake did to Miyagi transportation system: injured people willlikely stay injured-in-place.

The second problem is people: in a quake, Tokyo-ites know to headto the streets, Figure 14. This didn’t help communication: 13 millionpeople tried to call loved ones, shutting down the cellular network –even though Tokyo was not hit!

Next comes the tsunami, Figure 15. Large concrete buildings tumbledover. Hundred-ton ships crashed on top of schools.

If 13 million people are on the streets of Tokyo, with no communi-cation and nowhere to go, there would be 13 million dead peopleafter the tsunami sweeps them away. The entire business, social andintellectual elite of Japanese society would die. Trillions of dollars ofdamage; major businesses would lose their data, financial marketswould be destroyed: the heart of the country would be in ruins.

People who work in disaster planning know: a tsunami cannot beallowed to hit Tokyo. How much water would come? Where willit go? The modern answer is shown in Figure 16. I can just makeout the figure of an hydraulic engineer; in back, the sewer system tochannel water; in front, peering into a vast "G-can." How much willthis hold? Will it be enough? Below, the size of one Gcan, from theSairyu(dragon) water project: