# GROUP WORK 1, SECTION 16.1 Back to the Park

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## 16 MULTIPLE INTEGRALS

### 16.1 DOUBLE INTEGRALS OVER RECTANGLES

TRANSPARENCIES AVAILABLE

#48 (Figures 4 and 5), #49 (Figures 7 and 8), #50 (Figure 11), #51 (Figures 12 and 13) SUGGESTED TIME AND EMPHASIS

12–1 class Essential Material POINTS TO STRESS

1. The definition and properties of the double integral.

2. The analogy between single and double integration.

3. Volume interpretations of double integrals.

QUIZ QUESTIONS

• Text Question: Compute 2

i=1

• Drill Question: If we partition [a, b] into m subintervals of equal length and [c, d] into n subintervals of equal length, what is the value ofA for any subrectangle Ri j?

b− a m

 d− c n



MATERIALS FOR LECTURE

• Review single variable integration, including Riemann sums. Then show how double integration extends the concepts of single variable integration.

• Use a geometric argument to directly compute

R(3 + 4x) d A over R = [0, 1] × [0, 1].

• Discuss what happens when f (x, y) takes negative values over the region of integration. Start with an odd function such as x3+ y5being integrated over R = [−1, 1] × [−1, 1]. Then discuss what happens when a function that is not odd has negative values, for example, z = 2 − 3x on R = [0, 1] × [0, 1]. Illustrate numerically. (This can also be done using the group work “An Odd Function”.)

WORKSHOP/DISCUSSION

• Do a problem involving numerical estimation, such as estimating  

x2+ 2y2

d A over 0 ≤ x ≤ 2, 0 ≤ y ≤ 2, using both the lower left corners and midpoints as sample points. Also approximate faveover

R.

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CHAPTER 16 MULTIPLE INTEGRALS

• The following example can be used to help solidify the idea of approximating an area. Consider a square pyramid with vertices at(1, 1, 0), (1, −1, 0), (−1, 1, 0), (−1, −1, 0) and (0, 0, 1). Derive an equation for the surface of the pyramid, using functions such as z = 1 − max (|x| , |y|), and then the equations of the planes containing each of the five faces (z = 0, y + z = 1, and so on). Approximate the volume using the Midpoint Rule for the following equal subdivisions. Note that m is the number of equal subdivisions in the x-direction and n is the number of equal subdivisions in the y-direction.

1. m = n = 2 (Approximation is 2) 2. m = n = 3 (Approximation is4427) 3. m = n = 4 (Approximation is32) 4. m = n = 5 (Approximation is3625)

Compare your answers with the actual volume 43 computed using the formula V = 13Bh.

• Consider 

4− y2d A, with R = [0, 3] × [−2, 2]. Use a geometric argument to compute the actual volume after approximating as above with m = n = 3 and m = n = 4. Show that the average value is

12 = π2, and that the point

 0,

4−14π2



≈ (0, 1.24) in R satisfies f

 0,

4−14π2



= π2.

GROUP WORK 1: Back to the Park

This group work is similar to Example 4 and uses the Midpoint Rule.

GROUP WORK 2: An Odd Function The goal of the exercise is to estimate 

[−2,2] × [−2,2]

x3+ y5

d A numerically. There are three different problem sheets, each one suggesting a different strategy to obtain sample points for the estimation. Groups seated near each other should get different problem sheets, without this fact necessarily being announced.

After all the students are finished, have them compare their results. The exact value of the integral is zero, by symmetry.

The answers for arbitrarily chosen points will vary.

Version 1:

Four regions: 22

( f (−2, 0) + f (−2, −2) + f (0, 0) + f (0, −2)) = −320 Nine regions:4

3

2 f 

−2,23 + f 

−2, −23

+ f (−2, −2) + f

23,23 + f 

23, −23 + f

23, −2 + f 

23,23 + f 

23, −23 + f 

23, −2 

= −6403 ≈ −213.33 Sixteen regions: 

12 /

f (−2, 1) + f (−2, 0) + f (−2, −1) + f (−2, −2) + f (−1, 1) + f (−1, 0) + f (−1, −1) + f (−1, −2) + f (0, 1) + f (0, 0)

+ f (0, −1) + f (0, −2) + f (1, 1) + f (1, 0) + f (1, −1) + f (1, −2)0

= −160

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Version 2:

Four regions: 22

( f (2, 0) + f (2, 2) + f (0, 0) + f (0, 2)) = 320 Nine regions:4

3

2 f 

2, −23 + f

2,23

+ f (2, 2) + f 2

3, −23

+ f2

3,23

+ f 2

3, 2 + f 

23, −23 + f

23,23 + f 

23, 2 

= 6403 ≈ 213.33 Sixteen regions: 

12 /

f (2, −1) + f (2, 0) + f (2, 1) + f (2, 2) + f (1, −1) + f (1, 0) + f (1, 1) + f (1, 2) + f (0, −1) + f (0, 0)

+ f (0, 1) + f (0, 2) + f (−1, −1) + f (−1, 0) + f (−1, 1) + f (−1, 2)0

= 160 Version 3:

Four regions: 22

( f (1, 1) + f (1, −1) + f (−1, 1) + f (−1, −1)) = 0 Nine regions:4

3

2 f 4

3, −43

+ f 4

3, 0

+ f 4

3,43 + f 

0, −43

+ f (0, 0) + f 

0,43 + f 

43, −43 + f 

43, 0 + f 

43,43 

= 0 Sixteen regions: 0 (by symmetry)

GROUP WORK 3: Justifying Properties of Double Integrals

Put the students into groups. Have them read the section carefully, and then have some groups try to justify Equation 7, some Equation 8, and some Equation 9 for nonnegative functions f and g. They don’t have to do a formal proof, but they should be able to justify these equations convincingly, either using sums or geometrical reasoning.

There is no handout for this activity.

GROUP WORK 4: Several Ways to Compute Double Integrals

The students may need help finding good points to use for Problem 1. The Midpoint Rule works best.

Let f (x, y) = 2 − x − y.

1.

1

2

2 f1

4,14

+ f 3

4,14

+ f 1

4,34

+ f 3

4,34

= 1 2. A (x) = 32 − x 3.2 0

3

2− x

d x = 1 HOMEWORK PROBLEMS

Core Exercises: 1, 5, 6, 7, 9, 13, 14

Sample Assignment: 1, 2, 4, 5, 6, 7, 9, 10, 12, 13, 14, 17

Exercise D A N G

1 ×

2 ×

4 ×

5 × ×

6 ×

7 ×

Exercise D A N G

9 × ×

10 × ×

12 ×

13 ×

14 ×

17 ×

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### GROUP WORK 1, SECTION 16.1 Back to the Park

The following is a map with curves of the same elevation of a region in Orangerock National Park:

Estimate (numerically) the average elevation over this region using the Midpoint Rule.

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### An Odd Function (Version 1)

In this exercise, we are going to try to approximate the double integral

[−2,2] × [−2,2]

x3+ y5

d A. We start by partitioning the region [−2, 2] × [−2, 2] into four smaller regions, then nine, then sixteen, like this:

Now we approximate the double integral as discussed in the text, picking one point in every smaller region.

To make things simple, just choose the lower left corner of every small region, like so:

Approximation for four regions:

Approximation for nine regions:

Approximation for sixteen regions:

When you are finished, try again using a point of your choice in each region.

Using Arbitrarily Chosen Points Approximation for four regions:

Approximation for nine regions:

Approximation for sixteen regions:

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### GROUP WORK 2, SECTION 16.1 An Odd Function (Version 2)

In this exercise, we are going to try to approximate the double integral

[−2,2] × [−2,2]

x3+ y5

d A. We start by partitioning the region [−2, 2] × [−2, 2] into four smaller regions, then nine, then sixteen, like this:

Now we approximate the double integral as discussed in the text, picking one point in every smaller region.

To make things simple, just choose the upper right corner of every small region, like so:

Approximation for four regions:

Approximation for nine regions:

Approximation for sixteen regions:

When you are finished, try again using a point of your choice in each region.

Using Arbitrarily Chosen Points Approximation for four regions:

Approximation for nine regions:

Approximation for sixteen regions:

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### An Odd Function (Version 3)

In this exercise, we are going to try to approximate the double integral

[−2,2] × [−2,2]

x3+ y5

d A. We start by partitioning the region [−2, 2] × [−2, 2] into four smaller regions, then nine, then sixteen, like this:

Now we approximate the double integral as discussed in the text, picking one point in every smaller region.

To make things simple, just choose the midpoint of every small region, like so:

Approximation for four regions:

Approximation for nine regions:

Approximation for sixteen regions:

When you are finished, try again using a point of your choice in each region.

Using Arbitrarily Chosen Points Approximation for four regions:

Approximation for nine regions:

Approximation for sixteen regions:

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### GROUP WORK 4, SECTION 16.1 Several Ways to Compute Double Integrals

Consider the double integral

(2 − x − y) d A, where R = [0, 1] × [0, 1].

1. Estimate the value of the double integral using two equal subdivisions in each direction.

2. Fix x such that 0≤ x ≤ 1. What is the area A (x) of the slice shown below?

A(x) y=0, z=2-x

y=1, z=1-x 2

1 1 y

x 2

z

3. Find the exact volume of the solid with cross-sectional area A(x) using single variable calculus.

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### 16.2 ITERATED INTEGRALS

SUGGESTED TIME AND EMPHASIS

12–1 class Essential material

POINTS TO STRESS 1. The meaning ofb

a

d

c f (x, y) dy dx for a positive function f (x, y) over a rectangle [a, b] × [c, d].

2. The statement of Fubini’s Theorem and how it makes computations easier.

3. The geometric meaning of Fubini’s Theorem: slicing the area in two different ways.

QUIZ QUESTIONS

• Text Question: Consider Figures 1 and 2 in the text. Why isb

a A(x) dx =d

c A(y) dy?

x

0 z

xa b

y A(x)

C

x

0 z

y

c y d

Answer: The two integrals express the volume of the same solid.

• Drill Question: Compute3 0 4

3 x2y d y d x.

MATERIALS FOR LECTURE

• Revisit the example

[−2,2] × [−2,2]

x3+ y5

d A using integrated integrals, to illustrate the power of the technique of iteration.

• Use an alternate approach to give an intuitive idea of why Fubini’s Theorem is true. Using equal intervals of lengthx and y in each direction and choosing the lower left corner in each rectangle, we can write the double sum in Definition 16.1.5 as the iterated sum

m i=1

n j=1 f 

xi j, yi j

x y = m

i=1

4 n

j=1 f xi j, yi j

x 5

y

which in the limit gives an iterated integral. Go through some examples such as

[−1,3] × [−1,3]x y d A and



[0,1] × [0,1](2 − x − y) d A to demonstrate this approach.

• Remind the students how, in single-variable calculus, volumes were found by adding up cross-sectional areas. Take the half-cylinder f (x, y) = 

1− y2, 0 ≤ x ≤ 2, and find its volume, first by the “old”

method, then by expressing it as a double integral. Show how the two techniques are, in essence, the same.

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CHAPTER 16 MULTIPLE INTEGRALS

WORKSHOP/DISCUSSION

• Have the students work several examples, such as

[0,1] × [0,1]y√

1− x2sin 2πy2

d A, which can be computed as1

0 f (x) dx 1

0 g(y) dy.

• Find the volume of the solids described by 

2,

2

× [−2,3]

2− x2

d A and



[−1,1] × [−1,1]

1+ x2+ y2 d A.

GROUP WORK 1: Regional Differences

If the students get stuck on this one, give them the hint that Problem 1(b) can be done by finding the double integral over the square, and then using the symmetry of the function to compute the area over R. The regions in the remaining problems can be broken into rectangles.

Answers: 1.(a)34 (b) 12 2.(a)−8 (b)−5 (c)−92

GROUP WORK 2: Practice with Double Integrals

It is a good idea to give the students some practice with straightforward computations of the type included in Problem 1. It is advised to put the students in pairs or have them work individually, as opposed to putting them in larger groups. The students are not to actually compute the integral in Problem 2; rather, they should recognize that each slice integrates to zero. Think about what happens when integrating with respect to x first.

1. (a) 245

6− 103

5−65

3+158

2 (b)−32ln 3+ 12+ 2 ln 2 (c) e2− 1 − e 2. True GROUP WORK 3: The Shape of the Solid

In Problem 3, the students could first change the order of integration to more easily recognize the “pup tent”

shape of the resulting solid.

1. Irregular tetrahedron, volume 32 2. Half a cylinder, volume 12π 3. “Pup tent”, volume 5 HOMEWORK PROBLEMS

Core Exercises: 1, 3, 6, 17, 23, 26, 32, 35

Sample Assignment: 1, 2, 3, 6, 9, 13, 15, 17, 20, 23, 26, 28, 32, 35

Exercise D A N G

1 ×

2 ×

3 ×

6 ×

9 ×

13 ×

15 ×

Exercise D A N G

20 ×

23 ×

26 ×

28 ×

32 × ×

35 ×

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### Regional Differences

1. Calculate the double integral

R(x + y) d A for the following regions R:

(a)

x

1_ 1 0 2

21_

1 y

R

(b) y

1 x 0

1

R

2. Calculate the double integral

R

x y− y3

d A for the following regions R.

(a)

x 0 1

2 y

R 1

_1

(b)

x 0 1

2 y

R 1

_1

(c)

x 0 1

2 y

R 1

_1

R

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### GROUP WORK 2, SECTION 16.2 Practice with Double Integrals

Compute the following double integrals:

1. (a)



[1,2] × [0,1]x

1+ y + x2d A

(b)



[0,1] × [1,2]

x

x+ yd x d y

(c)



[0,1] × [1,2]yex yd x d y

2. Is the statement

[0,1] × [0,1]cos 2π

y2+ x

d A= 0 true or false?

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### The Shape of the Solid

For each of the following integrals, describe the shape of the solid whose volume is given by the integral, then compute the volume.

1. 

[0,1] × [0,1](3 − 2x − y) d A

2. 3

−3

2

−2

4− y2d y d x

3. 1

−1

3

−2(1 − |x|) dy dx

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### 16.3 DOUBLE INTEGRALS OVER GENERAL REGIONS

SUGGESTED TIME AND EMPHASIS 1 class Essential Material

POINTS TO STRESS

1. The geometric interpretation ofb a

g(y)

f(y)d x d y andd c

k(x) h(x) d y d x.

2. Setting up the limits of double integrals, given a region over which to integrate.

3. Changing the order of integration.

QUIZ QUESTIONS

• Text Question: Sketch a region that is type II and not type I, and then sketch one that is both type II and type I.

• Drill Question: Is it true that1

0

1

x f (x, y) dy dx =1

0

1

y f (x, y) dx dy?

Answer: Yes, because the limits of integration describe the same region.

MATERIALS FOR LECTURE

• Clarify why we bother with the fuss of defining F(x, y) =

f (x, y) if (x, y) ∈ D

0 if(x, y) /∈ D instead of “just integrating over D”.

• Show that the order of integration matters when computing

ex2d A, where D is the region shown below.

0 2 4

1 x

y

y=3x+1

y=x+1

• Point out that

D1 d A gives the area of D, and thatb a

g(x)

f(x)1 d y d x gives the usual formula for the area between curves.

• Evaluate

Dh d A, where D is a circle of radius r and h is constant. Show how this gives the general formula for the volume of a cylinder. Ask the students to evaluate

Dh d A, where D is the parallelogram with vertices(1, 1), (2, 3), (5, 1), and (6, 3). Have them interpret their answer in terms of volume.

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WORKSHOP/DISCUSSION

• Let D be the region shown below. Set up

D f (x, y) d A both as a type I integral/

D f (x, y) dx dy0 and a type II integral/

D f (x, y) dy dx0 .

x y

x+1=(y-1)\$

D y=2x

_1 1 2 3

_2 2 4 6

• Evaluate

Dyx2d A, where D is the unit circle, both as a type I integral and as a type II integral.

• Show how to change order of integration in1 0 x1/3

x4 f (x, y) dy dx.

• Change order of integration for1 0

y1/3

(2/π) arcsin y f (x, y) dx dy.

GROUP WORK 1: Type I or Type II?

Before handing out this activity, remind the students of the definitions of type I and type II regions given in the text.

1. Both 2. Both 3. Type II 4. Neither 5. Type II 6. Type I 7. Both 8. Type II 9. Neither 10. Both GROUP WORK 2: Fun with Double Integrals

In this section, it is probably best to have the students get experience in setting up many double integrals, rather than have them spend a lot of time solving only one. During the homework, of course, both skills should be practiced.

1. (a) 3/4

0 x−x2

x/4 d y d x. This can also be done as two integrals in d x d y.

(b) 1

0

4x

x2 d y d x or1

0

y y4 d x d y (c) 1/2

0

4y

y2 d x d y. This can also be done as two integrals in d y d x.

(d) 1

0

2−√y

y d x d y. This can also be done as two integrals in d y d x.

2. A hemisphere

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CHAPTER 16 MULTIPLE INTEGRALS

GROUP WORK 3: Bounding on a Disk

You may or may not tell your students that there is no such thing as Nentebular science. The surface 1

x2+ y2+ 1 lends itself naturally to this type of analysis, as seen by its graph.

Before handing out this activity, review the method of finding the area of an annulus. Close the activity by pointing out that if we continued the process of dividing D into smaller and smaller rings, the upper and lower bounds would approach each other, converging on the actual area:



0

 5

0

1

r2+ 1r dr dθ = π ln 2 + π ln 13 ≈ 3.2581π Answers:

1. Answers will vary here. The upper bound should be no worse than 25π ≈ 78.5 (setting the function equal to the constant 1, its maximum) and the lower bound should be no worse than 25π26 ≈ 3.02 (setting the function equal to the constant 261, its minimum).

2. Upper: 4π (1) + (25π − 4π)1

5

= 415π≈ 25.8. Lower: 4π1

5

+ (25π − 4π)1

26

= 209130π ≈ 5.05.

3. Upper:π (1) + (9π − π)1

2

+ (25π − 9π)1

10

= 335 π≈ 20.7.

Lower: π1

2

+ (9π − π)1

10

+ (25π − 9π)1

26

= 249130π≈ 6.02.

HOMEWORK PROBLEMS

Core Exercises: 1, 5, 10, 13, 22, 32, 43, 45, 51

Sample Assignment: 1, 5, 8, 10, 13, 17, 20, 22, 25, 29, 32, 33, 40, 43, 45, 51, 53, 58, 59

Exercise D A N G

1 ×

5 ×

8 ×

10 ×

13 ×

17 ×

20 ×

22 ×

25 ×

29 ×

Exercise D A N G

32 ×

33 ×

40 ×

43 ×

45 ×

51 × ×

53 ×

58 × ×

59 ×

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### Type I or Type II?

Classify each of the following regions as type I, type II, both, or neither.

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### GROUP WORK 2, SECTION 16.3 Fun with Double Integration

1. Write double integrals that represent the following areas.

(a) The area enclosed by the curve y = x − x2 and the line y = x

4

(b) The area enclosed by the curves y=√4 x and√y = x

(c) The area enclosed by the curves y=√ x and√4y = x, and the line y = 12

(d) The area enclosed by the curves y= x2 and y = (x − 2)2, and the line y = 0

2. What solid region ofR3do you think is represented by1

−1

 √1 − x2

1 − x2

1− x2− y2d y d x?

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### Bounding on a Disk

One integral that is very important in Nentebular science is the Ossidoot integral:



D

1

x2+ y2+ 1d A

For these sorts of integrals the domain D is usually a disk. In this exercise, we are going to find upper and lower bounds for this integral, where D is the disk 0≤ x2+ y2≤ 25.

1. It is possible to get some crude upper and lower bounds for this integral over D without any significant calculations? Find upper and lower bounds for this integral (perhaps crude ones) and explain how you know for sure that they are true bounds.

2. One can get a better estimate by splitting up the domain as shown in the graph below, and bounding the integral over the inside disk and then over the outside ring. Using this method, what are the best bounds you can come up with?

3. Now refine your bounds by looking at the domain D as the union of three domains as shown.

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### 16.4 DOUBLE INTEGRALS IN POLAR COORDINATES

SUGGESTED TIME AND EMPHASIS

1 class Essential material. If polar coordinates have not yet been covered, the students should read Section 11.3.

POINTS TO STRESS

1. The definition of a polar rectangle: what it looks like, and its differential area r dr dθ 2. The idea that some integrals are simpler to compute in polar coordinates

3. Integration over general polar regions QUIZ QUESTIONS

• Text Question: Is the area of a polar rectangle R = {(r, θ) | a ≤ r ≤ b, α ≤ θ ≤ β} equal to (b − a) (β − α)?

• Drill Question: Convert1

−1 √1−y2

1−y2

x2+ y2

d x d y to polar coordinates.

0 1

0 r3dr dθ MATERIALS FOR LECTURE

• Start by reminding the students of polar coordinates, and ask them what they think the polar area formula will be: what will replace d x d y?

• Draw a large picture of a polar rectangle, and emphasize that it is the region between the gridlines r = r0, r = r0+ r, θ = θ0, andθ= θ0+ θ.

The following method can be used to show that Area(R) ≈ r r θ if r and θ are small:

1. Using the formula for the area of a circular sector, we obtain

Area(R) = 12(r + r)2 θ −12(r)2

= r r θ + 12(r)2

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2. We now take the limit asr, θ → 0, of Area(R)

rr θ and show that this limit is equal to 1.

r→0, θ→0lim

Area(R)

rr θ = lim

r→0, θ→0

rr θ + 12(r)2 θ rr θ

= lim

r→0, θ→01+r 2r = 1 3. The result follows, since whenr and θ are small,

Area(R) rr θ ≈ 1 or

Area(R) ≈ r r θ

• Show how to set up the two general types of polar regions. (The second type occurs less frequently, and may be omitted.) Then indicate to the students that polar coordinates are most useful when one has an obvious center of symmetry for the region R in the x y-plane.

• Point out that polar areas can be found by setting up (double) polar integrals of the function f (r cos θ, r sin θ) = 1 and compare this method to using the formula A =b

a

12r2dθ. Calculate 1. The area between the circles x2+ y2 = 1 and x2+ y2= 2 in the second quadrant.

2. The area inside the spiral r = θ where 0 ≤ θ ≤ π.

3. A=

(x, y) | −1 ≤ x ≤ 1, 0 ≤ y ≤√

1− x2 .

4. The area inside the first loop of the curve r = 2 sin θ cos θ.

WORKSHOP/DISCUSSION

• Illustrate that difficult problems can sometimes be simplified using geometry and symmetry by finding the area inside the circles x2+ y2= 1 and (x − 1)2+ y2= 1 (or r = 2 cos θ). First find the point of intersection

1,π3 (in polar coordinates). Next, break the double integral for the portion in the first quadrant into two pieces as shown in the figure. The lighter region is a sector of angle π3 of the unit disk (which contributes π6 to the area), the darker region is a simple polar integral, and the total area between the circles is double the area above the x-axis.

1 2 x

y

_1

_1 0

x@+y@=1 1 (x-1)@+y@=1

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CHAPTER 16 MULTIPLE INTEGRALS

• Present some different uses of polar integrals:

1. Compute

R

x2+ y22

d A, where R is the region enclosed by x2+ y2 = 4 between the lines y = x and y = −x.

0 1

_1

_2 2

x y

2 _2

2. Compute the volume between the cone z2= x2+ y2and the paraboloid z = 4 − x2− y2.

GROUP WORK 1: The Polar Area Formula

Perhaps start by having students guess the final answer to Problem 1. Notice that the students will have to be able to antidifferentiate cos2θ and sin2θ to complete this activity.

1. 2π

0 1

2(1 + cos θ)2 dθ = 2 2.π

0

1

2(1 + sin θ)212(1)2

dθ = π4 + 2

GROUP WORK 2: Fun with Polar Area Many students will write

0

θ

0 r dr dθ. Try to get them to see for themselves that they must now subtract the area of the region that is counted twice in this expression.

1. 92π3196π3 = 43π3 2.π

0 2

0 e−r2r dr dθ= π2 

1− e−4

GROUP WORK 3: Fun with Polar Volume

If the students get stuck on Problem 2, point out that the intersection can be shown to be x2+ y2= 34, z= 12, and hence the integral is over the region x2+ y234. If time is an issue, the students can be instructed to set up the integrals without solving them.

1. 

0

1

0

√1− r2+ 1

r dr dθ = 3 2. 

0 3/4

0

√1− r2− 1−√

1− r2 

r dr dθ= 48π 

37− 7√ 7

(23)

HOMEWORK PROBLEMS

Core Exercises: 2, 5, 11, 15, 22, 30, 33, 35

Sample Assignment: 2, 5, 8, 11, 15, 17, 20, 22, 25, 30, 31, 33, 35

Exercise D A N G

2 × ×

5 × ×

8 ×

11 ×

15 ×

17 ×

20 ×

22 ×

25 ×

30 ×

31 ×

33 × ×

35 ×

(24)

### GROUP WORK 1, SECTION 16.4 The Polar Area Formula

The following formula is used for finding the area of a polar region described by the polar curve r = f (θ), a ≤ θ ≤ b:

A=b a

12

/f (θ)02

1. Use this formula to compute the area inside the cardioid r = 1 + cos θ.

O

r=1+cos¬

2. Use the formula to find the area inside the cardioid r = 1 + sin θ and outside the unit circle r = 1.

O

r=1+sin¬

r=1

(25)

### Fun with Polar Area

1. Find the area of the region inside the curve r = θ, 0 ≤ θ ≤ 3π.

2. Rewrite2

−2 √4−x2

0 e−x2−y2d y d x as a polar integral and evaluate it.

(26)

### GROUP WORK 3, SECTION 16.4 Fun with Polar Volume

1. Find the volume of the region bounded above by the upper hemisphere of the sphere x2+y2+(z − 1)2= 1 and bounded below by the x y-plane.

2. Find the volume of the region bounded above by the sphere x2+ y2+ z2 = 1 and below by the sphere x2+ y2+ (z − 1)2 = 1.

(27)

### 16.5 APPLICATIONS OF DOUBLE INTEGRALS

SUGGESTED TIME AND EMPHASIS 1 class Optional material

POINTS TO STRESS

I recommend stressing one of the following topics:

1. Density, mass, and centers of mass (for an engineering- or physics-oriented course)

2. Probability and expected values (for a course oriented toward biology or the social sciences)

QUIZ QUESTIONS

• Text Question: What is a logical reason that the total area under a joint density curve should be equal to 1?

Answer: The total area is the probability that some outcome will occur.

• Drill Question: If a lamina has a uniform density and an axis of symmetry, what information do we then have about the location of the center of mass?

Answer: It is on the axis of symmetry.

MATERIALS FOR LECTURE

• Describe the general ideas behind continuous density functions, computations of mass, and centers of mass.

• Do one interesting mass problem. A good exercise is the mass over the unit disk if ρ (x, y) = |x| + |y|.

This reduces to 41

0

 √1−x2

0 (x + y) dy dx which, surprisingly, equals 83.

• Describe the general idea of the joint density function of two variables. Similarly, describe the concept of expected value. If time permits, show that f (x, y) = 1

 1

1+ x2+ y23/2 describes a joint density function.

(28)

CHAPTER 16 MULTIPLE INTEGRALS

WORKSHOP/DISCUSSION

• Define the centroid (x, y) of a plane region R as the center of gravity, obtained by using a density of 1 for the entire region. So x = 1

A(R)



x d x d y and y= 1 A(R)



y d x d y. Show the students how to find the centroid for two or three figures like the following:

x=2-y@ y=2-x@

1/Ï3 _1/Ï3

2/3

_1/3

Show the students that if x = 0 is an axis of symmetry for a region R, then x = 0, and more generally, that(x, y) is on the axis of symmetry. Point out that if there are two axes of symmetry, then the centroid (x, y) is at their intersection.

• Consider the triangular region R shown below, and assume that the density of an object with shape R is proportional to the square of the distance to the origin. Set up and evaluate the mass integral for such an object, and then compute the center of mass(x, y).

(29)

GROUP WORK 1: Fun with Centroids

Have the students find the centroids for some of the eight regions described earlier in Workshop/Discussion.

Perhaps give each group of students two different ones to work on, and have them present their answers at the board.

2. 8

,8 3.4

5, 0 4.

0,45 5.

0,23 6.2

3a,23b 7.1

3a,13b

8.(0, 0) GROUP WORK 2: Verifying the Bivariate Normal Distribution

Go over, in detail, Exercise 36 from Section 12.4, which involves computing the integral

−∞e−x2d x using its double integral counterpart. Since these functions are related to the normal distribution and the bivariate normal distribution, the students can then actually show that



−∞

1 σ√

2πe−(x/μ)2/(2σ) = 1, as it should be.

Show that this is also true for



−∞

1 σ√

2πe−((x−a)/μ)2/(2σ)by noting that replacing x by x−a just corresponds to a horizontal shift of the integrand. There is no handout for this activity.

GROUP WORK 3: A Slick Model

This activity requires aCASand is based on the results of Group Work 2.

1. K = 1

2. Both are 0. This is because the oil spread is radial. The expected values correspond to the center of mass, which is at the origin.

HOMEWORK PROBLEMS

Core Exercises: 1, 4, 7, 15, 19, 24, 27, 30

Sample Assignment: 1, 4, 5, 7, 13, 15, 16, 19, 21, 24, 27, 30, 33

Exercise D A N G

1 ×

4 ×

5 ×

7 ×

13 ×

15 ×

16 ×

19 ×

21 ×

24 ×

27 ×

30 × ×

33 × ×

(30)

### GROUP WORK 1, SECTION 16.5 Fun with Centroids

Find the centroids of the following figures.

1. 2.

3.

x=2-y@

4.

y=2-x@

5. 6.

7. 8.

1/Ï3 _1/Ï3

2/3

_1/3

(31)

### A Slick Model

An oil tanker has leaked its entire cargo of oil into the middle of the Pacific Ocean, far from any island or continent. The oil has spread out in all directions in a thin layer on the surface of the ocean. The slick can be modeled by the two-dimensional density function K exp



−2x2+ y2 w2



, wherew is a fixed constant and the origin of the x y-plane represents the location of the tanker. Assuming that none of the oil evaporates, the density function must account for all of the oil and hence can be interpreted as a probability distribution.

1. Supposew = 2. Find the value of K which ensures that K exp



−2x2+ y2 w2



is a probability distribution.

2. Find the expected valuesμx andμy of x and y in the probability distribution from Problem 1. Interpret your answer geometrically.

3. Find the radius of the circle centered at the origin which contains exactly 99% of the oil in the slick.

(32)

### 16.6 TRIPLE INTEGRALS

SUGGESTED TIME AND EMPHASIS 1 class Essential material

POINTS TO STRESS

1. The basic definition of a triple integral.

2. The various types of volume domain, and how to set up the volume integral based on each of them.

3. Changing the order of integration in triple integrals.

QUIZ QUESTIONS

• Text Question: Give an example of one region that is type I, type II, and type III.

• Drill Question: In the expression

E f (x, y, z) dV , what does the dV mean?

Answer: Either d x d y dz or a description of the geometric meaning of d V (the infinitesimal volume element) should be counted as correct.

MATERIALS FOR LECTURE

• One way to introduce volume integrals is by revisiting the concept of area, pointing out that area integrals can be viewed as double integrals (for example10

0 f (x) dx = 10 0 f(x)

0 d y d x) and then showing how some volume integrals work by an analogous process. Set up a typical volume integral of a solid S using double integrals and similarly transform it into a triple integral:

V =

Rf (x, y) d A =b a

h2(x)

h1(x) f(x, y) dy dx =b a

h2(x) h1(x)f(x,y)

0 dz d y d x Then “move” the bottom surface of S up to z = g (x, y), so S has volume V =b

a

h2(x)

h1(x) f(x,y)

g(x,y) dz d y d x.

If we now have a function k(x, y, z) defined on S, then the triple integral of k over S is



Sk(x, y, z) dV =b a

h2(x) h1(x)

 f(x,y)

g(x,y) k(x, y, z) dz dy dx

(33)

• Show the students why the region shown is type 1, type 2, and type 3, and describe it in all three ways.

z=1-x-y

(0, 1, 0) (0, 0, 1)

(1, 0, 0)

• Show how to identify the region of integration E shown for the volume integral

1

−1

 √1−x2

1−x2

 √1−x2−y2

0 f(x, y, z) dz dy dx, and then rewrite the integral as an equivalent iterated integral of the form



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

1

0 1

0

_1

0

_1

1 y

x z

WORKSHOP/DISCUSSION

• Do a sample computation, such as the triple integral of f (x, y, z) = z + xy2 over the volume V bounded by the surface sketched at right, in the

first octant. z=1-x@-y

• Set up a triple integral for the volume V of the piece of the sphere of radius 1 in the first octant, with different orders of integration.

(34)

CHAPTER 16 MULTIPLE INTEGRALS

• Sketch the solid whose volume is given by the triple integral

 √(5+1)/2

( 5+1)/2

 √1−y2 y2

3−x−y

0 dz d y d x. Such a sketch is shown at right.

5

x

y z

x+y+z=3

x@+y@=1 x=y@

• Compute

E

x2+ y21/2

d V , where E is the solid pictured at right, by first integrating with respect to z and then using polar coordinates in place of d x d y.

0 2 4

_2

2

y 2 x

z

GROUP WORK 1: The Square-Root Solid Answers:

1. 2.1

0

 √1−z2

1−z2

 √y2−z2

y2−z2d x d y dz 3.1

−1

 √1−y2

1−y2

√1

x2+y2d z d x d y 4.1

−1 √1−x2

1−x2

√1

x2+y2d z d y d x=

0 1

0 r2dr dθ= π3 The students don’t yet know how to convert the integral into cylindrical coordinates, but they can do polar integrals, and they can also use the formula for the volume of a cone.

GROUP WORK 2: Setting Up Volume Integrals Answers:

1. Symmetry gives 41

0

 √4−x2

1  √4−x2−y2

4−x2−y2dz d y d x+2

1

 √4−x2

0  √4−x2−y2

4−x2−y2dz d y d x

 or 81

0

 √4−x2

1

 √4−x2−y2

0 dz d y d x+2

1

 √4−x2

0

 √4−x2−y2

0 dz d y d x

 . 2. 3

3

 √4−x2

4−x2

 √4−x2−y2

1 d z d y d x 3. Symmetry gives 41

0

 √4−x2

1−x2

 √4−x2−y2

4−x2−y2dz d y d x+2

1

 √4−x2

0

 √4−x2−y2

4−x2−y2dz d y d x

 or 81

0  √4−x2

1−x2

 √4−x2−y2

0 dz d y d x+2

1  √4−x2

0  √4−x2−y2

0 dz d y d x

 .

(35)

4. Symmetry gives 41

0

 √4−x2

1  √4−x2−z2

4−x2−z2d y d z d x+2

1

 √4−x2

0  √4−x2−z2

4−x2−z2d y dz d x

 or 81

0

 √4−x2

1

 √4−x2−z2

0 d y dz d x+2

1

 √4−x2

0

 √4−x2−z2

0 d y dz d x



GROUP WORK 3: An Unusual Volume

This is a challenging group work for more advanced students. The idea is to show that just because a solid looks simple, the computation of its volume may be difficult. The line generated by P1 and P2 has equation z = −34x + 4, and hence this equation, interpreted in three dimensions, is also the equation of the plane S.

The integral V(E) =2

−2

 √4−(x−2)2

4−(x−2)2

−34x+ 4

d y d x requires polar coordinates to solve by hand (since the bounding circle has equation r = 4 cos θ, 0 ≤ θ ≤ π) and also requires the students to remember how to integrate cos2θ and cos4θ. The volume is 16π.

Note that the problem can be simplified by moving the solid so that the z-axis runs through the center of D.

Point out that a simple geometric solution can be obtained by replacing S by the horizontal plane z= 52, thus giving a standard cylinder.

HOMEWORK PROBLEMS

Core Exercises: 5, 11, 13, 19, 24, 25, 27, 34, 39

Sample Assignment: 2, 5, 8, 10, 11, 13, 17, 19, 21, 23, 24, 25, 26, 27, 30, 34, 39, 42, 46, 50, 51

Exercise D A N G

2 ×

5 ×

8 ×

10 ×

11 ×

13 ×

17 ×

19 ×

21 ×

23 ×

24 ×

Exercise D A N G

25 ×

26 ×

27 ×

30 ×

34 × ×

39 ×

42 ×

46 × ×

50 ×

51 ×

(36)

### GROUP WORK 1, SECTION 16.6 The Square-Root Solid

Consider the volume integral over the solid S given by V =1

−1

 √1 − x2

1 − x2

√1

x2+ y2dz d y d x.

1. Identify the solid S by drawing a picture.

2. Rewrite the volume integral as V =

Sd x d y dz.

3. Rewrite the volume integral as V =

Sdz d x d y.

4. Starting with the original iterated integral, compute the volume by any means at your disposal.

(37)

### Setting Up Volume Integrals

Set up volume integrals for the following solids:

1. S is the solid formed by hollowing out a square cylinder of side length 1 centered along the z-axis through the solid sphere x2+ y2+ z2= 4.

2. S is the part of the solid sphere x2+ y2+ z2= 4 lying above the plane z = 1.

3. S is the solid formed by drilling a cylindrical hole of radius 1 centered along the z-axis through the solid sphere x2+ y2+ z2= 4.

4. S is the solid formed by hollowing out a square cylinder of side length 1 centered along the y-axis through the solid sphere x2+ y2+ z2= 4.

(38)

### GROUP WORK 3, SECTION 16.6 An Unusual Volume

Consider the solid E shown below.

1. Find the equation of the plane S, parallel to the y-axis, which forms the top cap of E.

2. Set up a volume integral for E of the form V (E) =2

−2

 √4−(x−2)2

4−(x−2)2

−34+ 4 d x d y.

3. Compute V (E) by any means at your disposal.

Hint: Try polar coordinates.

(39)

### DISCOVERY PROJECT Volumes of Hyperspheres

Problems 1 and 2 review computations that the students may already know. Problem 4 is optional for this project, but it is highly recommended. To extend this project, students can be asked to find a book or article that discusses hyperspheres, and add some geometric discussion of these objects to their reports.

(40)

### 16.7 TRIPLE INTEGRALS IN CYLINDRICAL COORDINATES

TRANSPARENCY AVAILABLE

#52 (Figure 7)

SUGGESTED TIME AND EMPHASIS

12–1 class Essential material

POINTS TO STRESS

1. The cylindrical coordinate system as an extension of polar coordinates inR2. 2. The basic shapes of cylindrical solids.

3. The idea that the cylindrical coordinate system can be used to simplify equations and volume integrals of certain three-dimensional surfaces and solids.

QUIZ QUESTIONS

• Text Question: Does the region of Example 3 have an axis of symmetry? If so, what does it say about the choice of using cylindrical coordinates?

Answer: The solid is symmetric about the z-axis, which implies that cylindrical coordinates should be considered.

• Drill Question: Describe in your own words the surface given by the equation r = θ, 0 ≤ θ ≤ 6π in cylindrical coordinates.

Answer: It looks something like a rolled-up piece of paper.

MATERIALS FOR LECTURE

• Compute the intersection of the surfaces z = x2+ y2and z = x, first in rectangular coordinates, then in cylindrical coordinates. (Here is a case where the rectangular coordinates are the easiest to visualize, even though there is an x2+ y2term.)

• Point out that while lim

(x,y)→(0,0)

x y

x2+ y2 does not exist, lim

(x,y,z)→(0,0,0)

x yz

x2+ y2+ z2 does exist. To see this, use cylindrical coordinates to get x yz

x2+ y2+ z2 = r2z sinθ cos θ

r2+ z2 , and compare to r2z

r2+ z2, which approaches 0, as seen in Section 14.2.

• Convert a typical cylindrical volume integral of a solid S computed using double integrals into a triple integral:

V =

Rf (r, θ) r dr dθ =β

α

h2(θ)

h1(θ) f (r, θ) r dr dθ =β

α

h2(θ) h1(θ)

 f(r,θ)

0 r d z dr dθ

(41)

Move the bottom surface of S up to z= g (r, θ), as pictured at right.

The volume now becomes

β

α

h2(θ) h1(θ) f(r,θ)

g(r,θ) r dz dr dθ. The basic volume element is given in Figure 3 of the text. Conclude with the situation where we have h(r, θ, z) defined on S.

Then the triple integral of h on S is

β

α

h2(θ) h1(θ) f(r,θ)

g(r,θ) h(r, θ, z) r dz dr dθ.

WORKSHOP/DISCUSSION

• Describe in terms of cylindrical coordinates the surface of rotation formed by rotating z = 1/x about the z-axis, noting that there is an axis of symmetry. Point out that the equations come from simply replacing x by r .

• Develop a straightforward example such as the region depicted below:

The Capped Cone

Set this volume up as a triple integral in cylindrical coordinates, and then find the volume. (The computation of this volume integral is not that hard, and can be assigned to the students.) Conclude by setting up the volume integral of h(r, θ, z) = rz over this region.

GROUP WORK: A Partially Eaten Sphere

Notice that you are removing “ice cream cones” both above and below the x y-plane.

1. 36π− 2

0

3 0

6

−r

2r dz dr dθ = π

36− 10√ 6 2. 36π− 2

0

3

0

6

−r

2zr2sinθ dz dr dθ= 36π

(42)

CHAPTER 16 MULTIPLE INTEGRALS

HOMEWORK PROBLEMS

Core Exercises: 2, 3, 6, 9, 12, 15, 22, 27

Sample Assignment: 2, 3, 6, 8, 9, 12, 14, 15, 17, 20, 22, 24, 25, 27, 29

Exercise D A N G

2 ×

3 × ×

6 ×

8 ×

9 × ×

12 ×

14 ×

15 × ×

17 ×

20 ×

22 ×

24 × ×

25 ×

27 ×

29 × ×

(43)

### GROUP WORK , SECTION 16.7 A Partially Eaten Sphere

1. Compute the volume of the solid S formed by starting with the sphere x2+ y2 + z2 = 9, and removing the solid bounded below by the cone z2= 2

x2+ y2 .

2. Set up the triple integral

S yz d V in the same coordinate system you used for Problem 1.

(44)

### DISCOVERY PROJECT The Intersection of Three Cylinders

This discovery project extends the problem of finding the volume of two intersecting cylinders given in Ex- ercise 66 in Section 6.2. This is a very thought-provoking project for students with good geometric intuition.

It is worthwhile for students to work on it even if they don’t wind up with the correct answer. If the students are mechanically inclined, perhaps ask them to build a model of the relevant solid. A good solution is given in the Complete Solutions Manual.

Transparency 53 can be used with this project.

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