Calculus Lectures: Volumes I

by Kathy Davis

Copyright ©2020 Kathy Davis

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Written November 2020

2 Kathy Davis

The purpose of computationis understanding, not numbers.

- Richard Hamming, (d1998)

Figure 1: Name That Building

Very Odd: windows on the lower floors;no windows above, and a slanted roof.

Figure 2: Our Volumes

The type of 3D volumes I’ll work, a bitlike Figure 1.

Figure 3: More Volumes

Another example.

Figure 4: Drawing Domains

The simplest example is this rectangle.

Why?

Check out Figure 1: it’s the Cook County Courthouse (in Chicago;I taught at the University of Chicago before UT). Lower floors arecourt-rooms, higher floors jail cells (no windows). You can’t escapeby jumping out a window, and no helicopter can land on the roof.

An environmental engineer (working with the architecture firm thatdesigned this building) would need to decide what the HVAC (heat-ing, air-conditioning, disinfecting) requirements are. And to do that,they’d need to know the volume of air. How do they compute that?

I’m going to look at volumes for a very limited number of objects;they will all be volumes under a surface z = f (x, y) and above aDomain D in the xy-plane. Figure 2 shows a cut cylinder: it’d belike a can of beans, except someone took a chain saw and cut offthe top in a slanting way (trained professional, do not attempt athome). The function f (x, y) gives the top (it would be something likez = 3− x− y); the domain D is the D in the xy plane, and it’s the partinside a circle, because bottoms of cans are circular.

Figure 3 is another typical example; the false-colored surface is givenas z = 3 + y + x2; the domain D is a rectangle.

My lectures discussed functions z = f (x, y), but said nothing aboutdomains D. So the first issue is how to describe, mathematically,those domains. I’ll work through some examples; more in the 14Uproblems.

The domain in Figure 3 is a rectangle; here’s what that looks like:

D = {(x, y) | 1 ≤ x ≤ 2; 1 ≤ y ≤ 3}

This is how I’ll describe many domains; "{(x, y)" is pronounced "thecollection of all pairs in the plane." The vertical bar "|" is pronounced"where" or "such that". And I know how to say "x is between 1 and 2

and y is between 1 and 3".

Usually, a formula like this is just given, and I have to work with it:I’d have to to draw the domain, Figure 4. Here’s how I did it.

I converted 1 ≤ x ≤ 2 into two equations: x = 1 and x = 2 and drewthose with dotted lines. I did the same for y = 1 and y = 3. NowI have a rectangle, and the domain D is what’s inside. D is the partinside because D was described with equations like 1 ≤ x ≤ 2. Thatmeans x, y are confined to a small region: the inside.

calculus lectures 3

Figure 5: Triangle

An issue: which triangle is it?

Figure 6: This Triangle

The bottom triangle in Figure 5 doesn’thave a bottom!

Figure 7: Sector Of A Circle

A generic sector of a circle. With extralines.

Figure 8: Our Sector

Not theirs. Sectors are easier to graphin polar co-ordinates!

Figure 9: A Bounded Region

Something different – I don’t give youD = { . . ..

Worked Examples

Sketch D = {(x, y) | 0 ≤ x ≤ 1; x ≤ y ≤ 1}

As before, there are lines x = 0 and x = 1 (that first is hard to drawbecause it’s right on the y-axis), and y = x and y = 1. Figure 5 lookslike there are two triangles, a top and a bottom, but that’s an illusion.A triangle has three sides, which we are drawing in blue. But thetriangle with one ’?’ doesn’t have a blue bottom! I have to be carefulhere: the correct domain is the ??, because it is contained by threesides. It’s drawn in Figure 6.

This next one is even sneakier, so it’s very quiz-like:

D = {(x, y) | x2 + y2 ≤ 1; 0 ≤ y ≤ x}

x2 + y2 = 1 is a circle; since the equation has a ≤ in it, that means it’sthe inside of a circle (called a disc). For the second pair of inequali-ties, the two equations are x = 0 and y = x, all drawn in Figure 7 (thecomputer program I use won’t draw dotted lines on circles, so it’s asolid line). The two lines x = 0 and y = x divide the inside of thecircle into four pieces, and again I have to figure out which one is D.

This time, the inequalities tell me which it is. Since x ≥ 0, D has to beentirely inside either the first or fourth quadrants, so regions II andIII are wrong.Then y ≥ x ≥ 0 so y has to be positive, and thereforeD has to be in either the first or second quadrant. The only regionsatisfying both the x and y restrictions is region I, in Figure 8.

And now for something completely different: Sketch the regionbounded by y = x2 and y = x + 2. Then write it as D = { . . .}.

The sketch is easy; I can draw a parabola and a straight line, below.

But to do D = { . . .}, I need inequalities like ? ≤ x ≤ ??; ditto fory. Here, the y is easiest; y goes between the parabola on the bottomand the line on the top. Or, just take the equations for y and put themback into an inequality: x2 ≤ y ≤ x + 2. It’s the x that’s hard.

4 Kathy Davis

Figure 10: A Bounded Region

Something different – what isD = { . . .}.

Figure 11: It’s Back . . .This one really is different. We talkedabout graphs like this when we didparametric equations. Now we reallyhave to deal with them.

Figure 12: D As T1

The what and where of a T1 descrip-tion.

Figure 13: D As T2

The what and where of a T2 descrip-tion.

What I used to do was look for was vertical x = lines; now I’ll haveto think. a ≤ x ≤ b would tell me how far left or right x can go inFigure 10. Those are indicated by the green dots in the figure; I’ll justsee what the co-ordinates are!

A better way for this kind of problem is to notice the dots are placedwhere the line hits the parabola: x + 2 = x2. It’s a quadratic, so I’lljust solve for x:

x2 = x + 2 ↔ x2 − x− 2 = 0 ↔ (x + 1)(x− 2) = 0 so x = −1, x = 2.

D = {(x, y) | − 1 ≤ x ≤ 2; x2 ≤ y ≤ x + 2}

The D = {(x, y) | − 1 ≤ x ≤ 2; x2 ≤ y ≤ x + 2} is a mathematicaldescription of some curves; this assumes that the curves are all givenby y = f (x). Well, of course! What isn’t?

What isn’t is Figure 11; showing the domain D bounded by curvesx = y2, y = 1. And, from the lecture on parametric equations, I knowthere’s no easy way out of this; I have to accept that some curvessimply aren’t y = f (x).

This means I’ll have to rethink how I describe domains. And I needterminology for that. D = {(x, y) | − 1 ≤ x ≤ 2; x2 ≤ y ≤ x + 2}is called a Type I or T1 description of D. In general, this is what a T1

description looks like:

D = {(x, y) | a ≤ x ≤ b; b(x) ≤ y ≤ t(x)}

What this means is shown in Figure 12. Sorting this out:1) a is the left-most x in D; b is the right-most x in D2) y = b(x) is the bottom curve of D; y = t(x) is the top curve of D.

Side note: My old exams or quizzes use different notation: the oldgB(x) is now b(x); the old gT(x) is now t(x). With that sorted out,what am I supposed to do with the domain bounded by the curvesx = y2, y = 1, Figure 11? Because there’s no y = f (x), it isn’tD = {(x, y) | a ≤ x ≤ b; b(x) ≤ y ≤ t(x)}.

A Type II or T2 description of D is given by

D = {(x, y) | c ≤ y ≤ d; l(y) ≤ x ≤ r(y)}

What this means is shown in Figure 13. Again:1) c is the bottom-most y in D; d is the top-most y in D2) x = l(y) is the left curve of D; r = r(y) is the right curve of D.

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Figure 14: D As T2

Repeat!

Figure 15: Both

This time, T1 and T2.

Figure 16: Trouble!T1, T2 or . . . ?

Worked Problems

Problem: write the domain bounded by the curves x = y2, y = 1,Figure 14, as a T2 domain.

x = l(y) and x = r(y) are easy from Figure 14: x = 1 so r(y) = 1, andx = y2 so l(y) = y2. As for c, d, they’re the points where the curvesx = y2 and x = 1 intersect, so y2 = x, x = 1 gives y2 = 1, just asbefore, a quadratic, with two solutions: y = ±1. So:

D = {(x, y) | − 1 ≤ y ≤ 1; y2 ≤ x ≤ 1}

Problem: write the domain bounded by the curves x = −y2, y = x,Figure 15, as T1 and T2 domains.

To start, I’ll get a, b for T1 and c, d for T2. Figure 15 shows thesenumbers are the co-ordinates of the places where the two curvesintersect: y = x intersects x = −y2:

y = −y2 ↔ y2 + y = 0↔ y(y+ 1) = 0 so y = 0, y = −1. But y = x so, (−1,−1), (0, 0)

The T2 equations are the easiest, because – well, the left and rightcurves x = l(y) and x = r(y) are right there: y = x → l(y) = y. And,x = −y2 → r(y) = −y2.

D(T2) = {(x, y) | − 1 ≤ y ≤ 0; y ≤ x ≤ −y2}

For T1, the top curve is y = x → t(x) = x. The bottom curve isx = −y2, but this needs some rewriting:

x = −y2 ↔ y2 = −x so y = ±√−x.

Since −1 ≤ y ≤ 0, my y = b(x) equation has to have negatives: Iwant the negative square root, y = −

√−x so b(x) = −

√−x.

D(T1) = {(x, y) | − 1 ≤ x ≤ 0; −√−x ≤ y ≤ x}

The√−x drives some people up the wall, but then −1 ≤ x ≤ 0, or

1 ≥ −x ≥ 0 so I’m taking square roots of positive numbers. Or, justtry it: x = −1 is in D, so

√−x =

√−(1) =

√1. It’s magic! It all

works out.

What would I do with the domain bounded by y = x, y = x3,shown in Figure 16? It’s neither T1, nor T2, though the parts aboveand below the x-axis are each, separately, both T1 and T2.

The general technique for this kind of domain is to break it into twoseparate domains, and, whatever I intend to do, I do in each domainby itself. But that’s a topic for a later lecture; now it’s time to try the14U.