Calculus 3 Tutor, Volume II Worksheet 1 Triple Integrals · Calculus 3 Tutor, Volume II Worksheet 1...

Calculus 3 Tutor, Volume II

Worksheet 1

Triple Integrals

Worksheet for Calculus 3 Tutor, Volume II, Section 1:

Triple Integrals

1. For the region D in the diagram below:

Plot for 1a

(a) What kind of shape is D? What is the volume of D?

c©2019 MathTutorDVD.com 1

(b) To integrate over D, mark the volume element dV , the base region R, the area

element dA, and the height element dz.

(c) Evaluate∫∫∫

D1dV .

(d) If f(x, y, z) is constant with respect to z, it can be written as f(x, y). Write∫∫∫Df(x, y)dV in terms of dA and R.


2. For the region D in the diagram below:

Plot for 2a

(a) Evaluate∫∫∫

D2dV . How does this integral compare to the volume of D?


(b) For the function f(x, y, z) = x, mark the parts of D where f is positive and the

parts of D where f is negative. Is there anything to note?


Df(x, y, z)dV for f(x, y, z) = x. Explain your results in terms of the

distribution of positive and negative areas of f .

(d) For the function g(x, y, z) = z− 3, mark the parts of D where g is positive and the

parts of D where g is negative. Is there anything to note?


(e) Evaluate∫∫∫

Dg(x, y, z)dV for g(x, y, z) = z − 3. Explain your results in terms of

the distribution of positive and negative areas of g.

3. Let D be the region in the diagram below. In this problem we will approximate∫∫∫Df(x, y, z)dV where f(x, y, z) = 1

8(x2y + yz3).

Plot for 3a


(a) What is the volume of D?

(b) Approximate∫∫∫

Df(x, y, z) by multiplying the “average” value of f(x, y, z) by the

volume of D. You can find the “average” value of f(x, y, z) by evaluating f(x, y, z)

in the middle of the region D.

(c) Obtain a better approximation for∫∫∫

Df(x, y, z)dV as follows. Split the region D

into D1 and D2 as in the following diagram. Approximate∫∫∫

D1f(x, y, z)dV by

multiplying the “average” value of f(x, y, z) in D1 by ‖D1‖; that is, the value of

f(x, y, z) in the middle of the region D1, multiplied by the volume of D1. Approx-

imate∫∫∫

D2f(x, y, z)dV in the same way. Combine these two approximations to

obtain an approximation for∫∫∫

Df(x, y, z)dV .


Plot for 3c


(d) Obtain an even better approximation for∫∫∫

Df(x, y, z)dV by splitting D into the

eight equal regions D1, ..., D8 in the following diagram and adding the approxima-

tions of∫∫∫

Di

f(x, y, z)dV for all i.

Plot for 3d


(e) Evaluate∫∫∫

Df(x, y, z)dV using calculus. How do your results compare to the

successive approximations?

4. Evaluate the following integrals over boxes. Note that a box D written in the form

[x1, x2]× [y1, y2]× [z1, z2] is defined on x1 ≤ x ≤ x2; y1 ≤ y ≤ y2; z1 ≤ z ≤ z2.

(a)∫∫∫

Dx2dV where D is [0, 1]× [0, 1]× [0, 1];

(b)∫∫∫

D(x+ y + z)dV where D is [−1, 1]× [−1, 0]× [0, 2];


(c)∫∫∫

Dxyez where D is [0, 3]× [0, ln 2]× [− ln 2, ln 3];

(d)∫ 2

x=−2

∫ 5

y=−3

∫ 8

z=4x+ 2zdV ;

(e)∫ 1

y=0

∫ 1

x=−1

∫ 0

z=−1 x2y + yzdV ;

(f)∫ 2

z=1

∫ 3

y=2

∫ 4

x=31

(x+y+z)2dV . Hint: to evaluate

∫ln zdz, the antiderivative is closely

related to z ln z. What is the antiderivative, exactly? You also may use integration

by parts.


5. For the following regions D, evaluate∫∫∫

DfdV :

(a) Here, D is defined over a rectangular region R, and the top of D is a plane that

includes the points (0, 0, 4); (2, 0, 2); and (2, 1, 2). Let f(x, y, z) = xy + xz + yz.

Plot for 5a


(b) Here, D is the volume between two triangles in parallel planes. The first triangle

includes (0, 0, 0); (2, 0,−2); and (0, 2,−2). Let f(x, y, z) = xz.

Plot for 5b


(c) Here, D is two boxes stacked one on top of the other and f(z) = xy + xz + yz.

The bottom box has height 1.

Plot for 5c


6. For the region D defined by −x ≤ y ≤ 0; 0 ≤ z ≤ x; 0 ≤ x ≤ 5:

(a) Plot D.

Plot for 6a


(b) Evaluate∫∫∫

D1dV .


D2x+ yzdV .


(d) Now consider D′ which is a subregion of D with added constraint that z − y ≤ x.

Plot D′.

Plot for 6d


(e) Find the volume of D′ by evaluating∫∫∫

D′ 1dV .

(f) Evaluate∫∫∫

D′ zdV .

7. Evaluate the following integrals:

(a)∫ 1

x=0

∫ x

y=−x

∫ y2+2

z=1

xdV .


(b)∫ 2

x=1

∫ x

y=x2

∫ y2

z=0

z2dV .

(c)∫ π

2

x=0

∫ π3

y=0

∫ π4

z=0

sin(2x+ 3y − z)dV .

(d)∫ 2

x=−1

∫ 2

y=−1

∫ 1

z=−1x5yex

2yzdV .


(e)∫ π

x=0

∫ x

y=0

∫ sin y

z=0

zdV .

(f)∫ 2

x=1

∫ x

y=1

∫ y

z=1

∫ z

w=1

wdV . Note that integration in four or more dimensions works

the same way as integration in three dimensions.


Answer key.

1. For the box-shaped region D:

1(a). Answer: D is a rectangular prism (or box) with volume 100 . This shape is a

rectangular prism, or a box. In the x-direction, it runs from x = −6 to x = 4, so its

length is 10. In the y-direction, it runs from y = 1 to y = 6, so its width is 5. Finally,

in the z-direction, it runs from z = 0 to z = 2, so its height is 2. The volume of a

prism is

l · w · h

so the volume of this prism is

10 · 5 · 2 = 100


1(b). The relevant elements in D are as follows:

Plot for 1b

To evaluate a triple integral over dV , we first need to integrate over the height.

We need to find the contribution of the integral over a small patch of area dA,

in R. We find the contribution to the integral over dA by integrating over all the

different height values; that is, over the different values dz. We have that

dV = dzdA

That is, the contribution from a small value inside D is the contribution from the

base element times the contribution from the height element. The base element


is embedded in the base region, R, which here is shaded in pink. We have that∫∫∫D

fdV =

∫∫R

(∫fdz

)dA

where∫fdz is a function of x and y, varying over the different squares dA. We

first integrate f over the height in a given element dA, and then integrate over all

the base elements dA.


1(c). Answer: 100 . We have∫∫∫D

1dV =

∫∫ (∫ 2

z=0

1dz

)dA

In any square dA, the height z runs from 0 to 2 and dz is an infinitesimal between

0 and 2. Then, in any square dA, the contribution of 1 is given by∫ 2

z=01dz. We

have ∫ 2

z=0

1dz = [z]20 = 2

Then, ∫∫∫D

1dV =

∫∫R

2dA

As we learned in the first part of this course,∫∫R

2dA =

∫ 4

x=−6

∫ 6

y=1

2dydx =

∫ 4

x=−6

(∫ 6

y=1

2dy

)dx

The inner integral is ∫ 6

y=1

2dy = [2y]61 = 12− 2 = 10

The outer integral is then∫ 4

x=−610dx = [10x]4−6 = 10(4)− 10(−6) = 100

The integral of 1 over dV produces the volume, because each element dV adds

a contribution equal to the volume of dV . The sum of each of these infinitemisal

volumes is the total volume of V .


1(d). Answer:∫∫∫

D

f(x, y)dV =

∫∫R

2f(x, y)dA . We evaluate an integral over D by

taking a height integral from the top to the bottom, and then by taking a two-

dimensional integral over the area. We have∫∫∫D

f(x, y)dV =

∫∫R

(∫ z2

z1

f(x, y)dz

)dA

At any point (x, y), the height z included in the box runs from z = 0 to z = 2, so

the inner integral is ∫ 2

z=0

f(x, y)dz

Since f(x, y) is constant with respect to z, this is∫ 2

z=0

f(x, y)dz = f(x, y)

(∫ 2

z=0

1dz

)= 2f(x, y)

leaving us with∫∫

R

2f(x, y)dA .


2. For this box-shaped region D:

2(a). Answer: 256 . At any point (x, y) in R which is the base of D, the height z runs

from 0 to 4. Then, ∫∫∫D

2dV =

∫∫R

(∫ 4

z=0

2dz

)dA

We have ∫ 4

z=0

2dz = [2z]40 = 8

Then, ∫∫R

8dA =

∫ 2

x=−2

∫ 0

y=−88dydx


y=−88dy = [8y]0−8 = 64

The outer integral is∫ 2

x=−264dx = [64x]2x=−2 = 128− (−128) = 256

This is twice the volume of D, which is

l · w · h = 8 · 4 · 4 = 128

The volume of D is

V =

∫∫∫D

1dV

and ∫∫∫D

2dV = 2

∫∫∫D

1dV = 2V


2(b). We mark the parts of D where f is negative in red, and the parts of D where f is

positive we keep shaded in blue. We note that the volume of the red regions of

D is the same as the volume of the blue regions of D, because D is symmetric

about the yz-plane.

Plot for 2b


2(c). Answer: 0 . We evaluate∫∫∫D

f(x, y, z)dV =

∫∫R

(∫ 4

z=0

xdz

)dA


z=0

xdz = [xz]4z=0 = 4x− 0 = 4x

since x is constant with respect to z. Then we have∫∫R

4xdA =

∫ 2

x=−2

∫ 0

y=−84xdydx


y=−84xdy = [4xy]0y=−8 = 32x


x=−232xdx =

[16x2

]2x=−2 = 64− (64) = 0

This is not a surprising result since the red and blue regions are equal in volume.

Not only is D symmetric about the yz-plane, but f is odd with respect to the

yz-plane. Then, the contributions from the region in D where f is positive are

exactly equal to the contributions from the region in D where f is negative, and

the regions exactly match up so that all the contributions cancel to zero.


2(d). The function g(x, y, z) = z − 3 is positive above the plane z = 3, and negative

below the plane z = 3. So we shade the part below z = 3 red, and the part above

z = 3 stays blue:

Plot for 2d

The red region appears much larger than the blue region, which seems to sug-

gest that∫∫∫

DgdV will be negative.


2(e). Answer: −128 . We evaluate∫∫∫D

g(x, y, z)dV =

∫∫R

(∫ 4

z=0

(z − 3)dz

)dA

The inner integral is∫ 4

z=0

(z − 3)dz =

[z2

2− 3z

]4z=0

=42

2− 3(4) = −4

The outer integral is then∫∫R

−4dA = −4∫ 2

x=−2

∫ 0

y=−81dydx = −4(4 · 8) = −128

As expected, this integral is negative since the region where g is negative is

larger, so the sum of contributions to g from the pink region is more negative than

the sum of contibutions to g from the blue region is positive.


3. Approximations in triple integrals:

3(a). Answer: the volume of D is 64 . The shape D is a cube of side length 4, so its

volume is

43 = 64


3(b). Answer: 192 . The first-order approximation for∫ baf(x)dx is∫ b

a

f(x)dx ≈ (b− a)f(b+ a

2

)The integral is the area under a curve; so the value of the integral is the length

under the curve times the average height of the curve. In this approximation,

the area under the curve is approximated by a rectangle. The same concept is

true in 3D calculus. The 3D integral is the sum of small contributions from f at

dV , where the sum of all the volumes dV is just the volume V of D. What if dV

gets so large that it is the same as the volume of D? In this case, the integral is

approximated by the volume of D times the “average” value of f , which is when

f is evaluated in the middle of the region. That is,∫∫∫D

fdV ≈ ‖D‖ f(c(D))

where ‖D‖ is the volume of D and c(D) is the center of D. For the case of a

rectangular prism,∫ x2

x1

∫ y2

y1

∫ z2

z1

f(x, y, z)dV ≈ (x2 − x1) (y2 − y1) (z2 − z1) f(x2 + x1

2,y2 + y1

2,z2 + z1

2

)Here,

‖D‖ = 64

The center of the cube is (2, 2, 2), and

f(2, 2, 2) =1

8(22 · 2 + 2 · 23) = 1

8(8 + 16) = 3

Then, the first-order approximation to the integral is

3 · 64 = 192


3(c). Answer: 288 . We could approximate∫ baf(x)dx, which is the area under a curve,

by breaking the area under the curve into two rectangles: from a to a+b2

, and from

a+b2

to b. Then we would multiply the first length by f(3a+b4

), and the second by

f(a+3b4

), which is the average value taken at the midpoint of the range. Similarly,

we can approximate∫∫∫

Df(x, y, z)dV by∫∫∫

D

f(x, y, z)dV =

∫∫∫D1

f(x, y, z)dV +

∫∫∫D2

f(x, y, z)dV

≈ ‖D1‖ f(c(D1)) + ‖D2‖ f(c(D2))

where c(D1), c(D2) are the centers of the regions D1 and D2. In this diagram,

both regions D1 and D2 have volume equal to 32. The region D1 has center

c(D1) = (2, 2, 1)

and the region D2 has center

c(D2) = (2, 2, 3)

Then, the “average” value of f in D1 is

f(2, 2, 1) =1

8

(22 · 2 + 2 · 13

)=

1

8(8 + 2) =

5

4

and the approximate integral of f over D1 is

5

4· 32 = 40

The “average” value of f in D2 is

f(2, 2, 3) =1

8

(22 · 2 + 2 · 33

)=

1

8(8 + 54) =

31

4

so the approximate integral over D2 is

31

4· 32 = 248

The total from D1 and D2, which is the second-order approximation, is

248 + 40 = 288


3(d). Answer: 305 . Here, each of the eight subregions is a cube with volume 8.

Numbering the subregions like the octants, we have:

Region no. Center “Average” (f evaluated at center) region vol. Contribution

1 (3, 3, 3) 1088

8 108

2 (1, 3, 3) 848

8 84

3 (1, 1, 3) 288

8 28

4 (3, 1, 3) 368

8 36

5 (3, 3, 1) 308

8 30

6 (1, 3, 1) 68

8 6

7 (1, 1, 1) 28

8 2

8 (3, 1, 1) 108

8 10

Adding the contributions together, we obtain 304 as this approximation. This is

fairly close to the value of the last approximation, suggesting that this is a roughly

accurate approximation.


3(e). Answer:1024

3. Fortunately, this is a relatively simple integral to evaluate di-

rectly, allowing us to test the approximations.. We evaluate∫∫∫D

fdV =

∫ 4

x=0

∫ 4

y=0

∫ 4

z=0

1

8

(x2y + yz3

)dzdydx

The innermost integral is∫ 4

z=0

1

8

(x2y + yz3

)dz =

[1

8x2yz +

1

32yz4]4z=0

=1

2x2y +

1

32· 256y =

1

2x2y + 8y

With respect to y, the integral is∫ 4

y=0

1

2x2y + 8ydy =

[1

4x2y2 + 4y2

]4y=0

= 4x2 + 64

Finally, with respect to x, the integral is∫ 4

x=0

4x2 + 64dx =

[4

3x3 + 64x

]4x=0

=4

3(64) + 256 =

1024

3

This is about 341, which is in the same range as our approximations. Note that the

finer approximations are closer to the answer than the coarser approximations.


4. Integrals over boxes:

4(a). Answer:1

3. This is

∫∫∫D

x2dV =

∫∫R

(∫ 1

z=0

x2dz

)dA


z=0

x2dz = x2(∫ 1

z=0

1dz

)= x2 [z]10 = x2

since x2 is constant with respect to z. Then,∫∫R

x2dA =

∫ 1

x=0

∫ 1

y=0

x2dydx


y=0

x2dy =[x2y]1y=0

= x2

since x is also constant with respect to y. Then we are left with∫ 1

x=0

x2dx =

[x3

3

]10

=1

3


4(b). Answer: 2 . We can write∫∫∫D

(x+ y + z)dV =

∫ 1

x=−1

∫ 0

y=−1

∫ 2

z=0

(x+ y + z)dzdydx

This combines the steps of writing

dV = dzdA

and then

dA = dydx

In fact, we can simply write

dV = dzdydz

We then evaluate the innermost integral first, parenthesizing as follows:∫ 1

x=−1

∫ 0

y=−1

∫ 2

z=0

(x+ y + z)dzdydx =

∫ 1

x=−1

(∫ 0

y=−1

(∫ 2

z=0

(x+ y + z)dz

)dy

)dx


z=0

(x+ y + z)dz =

[xz + yz +

z2

2

]20

Since x and y are constants with respect to z, their antiderivatives are xz and yz

respectively. This evaluates to

2x+ 2y + 2

Then, we have ∫ 1

x=−1

∫ 0

y=−1(2x+ 2y + 2)dydx


y=−1(2x+ 2y + 2)dy =

[2xy + y2 + 2y

]0y=−1 = 0− (−2x+ 1− 2) = 2x+ 1

Then, the outer integral is∫ 1

x=−1(2x+ 1)dx =

[x2 + x

]1x=−1 = 1 + 1− (1− 1) = 2

It makes sense that the answer would be positive, since x + y + z is positive in

the region D more than it is negative.


4(c). Answer:45 (ln 2)2

8. We have

∫∫∫D

xeyz =

∫ 3

x=0

∫ ln 2

y=0

∫ ln 3

z=− ln 2

xyezdzdydz

The innermost integral is∫ ln 3

z=− ln 2

xyezdz = xy

∫ ln 3

z=− ln 2

ezdz = xy [ez]ln 3z=− ln 2 = xy

(eln 3 − 1

eln 2

)= xy

(3− 1

2

)=

5xy

2

Then, we have ∫∫∫D

xeyzdV =

∫ 3

x=0

∫ ln 2

y=0

5xy

2dydx

The next integral is∫ ln 2

y=0

5xy

2dy =

5x

2

∫ ln 2

y=0

ydy =5x

2

[y2

2

]ln 2

y=0

=5x

4· (ln 2)2

Then, the outermost integral is∫ 3

x=0

5x

4· (ln 2)2 dx =

5 (ln 2)2

4

[x2

2

]3x=0

=5 (ln 2)2

4· 92=

45 (ln 2)2

8


4(d). Answer: 1536 . We write∫ 2

x=−2

∫ 5

y=−3

∫ 8

z=4

x+ 2zdV =

∫ 2

x=−2

(∫ 5

y=−3

(∫ 8

z=4

x+ 2zdz

)dy

)dx


z=4

(x+ 2z)dz =[xz + z2

]8z=4

since x is a constant with respect to z. This is

8x+ 64− (4x+ 16) = 4x+ 48

The second integral is∫ 5

y=−3(4x+ 48) dy = (4x+ 48)

∫ 5

y=−31dy = (4x+ 48)(8) = 32x+ 384


x=−2(32x+ 384)dx =

[16x2 + 384x

]2x=−2 = 64 + 768− (64− 768) = 1536


4(e). Answer: −1

6. This is

∫ 1

y=0

(∫ 1

x=−1

(∫ 0

z=−1

(x2y + yz

)dz

)dx

)dy

The fact that the outer integral is in terms of y instead of x does not affect the

approach. It only means that the region R is taken as a horizontally simple region

and not considered as a vertically simple region (although it is both horizontally

and vertically simple). The innermost integral is∫ 0

z=−1

(x2y + yz

)dz =

[x2yz +

yz2

2

]0z=−1

= 0−(−x2y + y

2

)= x2y − y

2

The second integral is∫ 1

x=−1

(x2y − y

2

)dx =

[x3y

3− xy

2

]1x=−1

=y

3− y

2−(−y3+y

2

)=

2y

3− y = −y

3

Then the outermost integral is∫ 1

y=0

−y3dy =

[−y

2

6

]1y=0

= −1

6


4(f). Answer: 78 ln 2− 12 ln 3− 21 ln 7 . Here, the integral over z is the outermost

integral. This doesn’t affect the approach; it only means that the volume elements

are taken in reference to x and y before z. The inner integral is∫ 4

x=3

1

(x+ y + z)2dx

In this integral, y and z are held constant so that x can be varied. This is∫ 4

x=3

1

(x+ A)2dx =

[− 1

x+ A

]4x=3

=

[− 1

x+ y + z

]4x=3

=1

y + z + 3− 1

y + z + 4

because A = y + z is a constant with respect to x.The next integral is∫ 3

y=2

(1

y + z + 3− 1

y + z + 4

)dy

Here, z + 3 and z + 4 are constants, and∫1

y +Bdy = ln(y +B)

up to a constant so this is

[ln(y + z + 3)− ln(y + z + 4)]3y=2 = ln(z + 6)− ln(z + 7)− (ln(z + 5)− ln(z + 6))

= 2 ln(z + 6)− ln(z + 7)− ln(z + 5)

Now, the integral of ln z is ∫(ln z)dz = z ln z − z + C

which can be proved using integration by parts or guess-and-check observation.

It follows by u-substitution that∫ln(z + a)dz = (z + a) ln(z + a)− (z + a) + C

The outer integral is then∫ 2

z=1

(2 ln(z + 6)− ln(z + 7)− ln(z + 5)) dz

= [2(z + 6) ln(z + 6)− 2(z + 6)− (z + 7) ln(z + 7) + (z + 7)− (z + 5) ln(z + 5) + (z + 5)]2z=1


= [2(z + 6) ln(z + 6)− (z + 7) ln(z + 7)− (z + 5) ln(z + 5)]2z=1

because all the terms not involving logarithms cancel. This is then

16 ln(8)− 9 ln(9)− 7 ln(7)− (14 ln(7)− 8 ln(8)− 6 ln(6))

= 48 ln 2− 18 ln 3− 21 ln 7 + 24 ln 2 + 6 ln 2 + 6 ln 3 = 78 ln 2− 12 ln 3− 21 ln 7


5. Evaluating∫∫∫

Dxy + xz + yzdV :

5(a). Answer:44

3. Before integrating, we need to find the boundaries of D in alge-

braic terms. The region R on which D has a z-dimension defined is the rectangle

0 ≤ x ≤ 2; 0 ≤ y ≤ 1

For x, y in this range, we have that z ≥ 0. We can determine the equation of the

plane that defines the maximum value of z on this region. This plane includes the

slanted blue lines, z = 4− x; y = 0 and z = 4− x; y = 1. Therefore the plane that

defines the top of this region is

z = 4− x

For any x, y in R, the region D is defined by

0 ≤ z ≤ 4− x

If y changes, the height of the region remains the same. Then, we can evaluate:∫∫∫D

xy + xz + yzdV =

∫ 2

x=0

∫ 1

y=0

∫ 4−x

z=0

(xy + xz + yz)dzdydz

The inner integral is∫ 4−x

z=0

(xy+xz+yz)dz =

[xyz +

xz2

2+yz2

2

]4−xz=0

= xy(4−x)+ x(4− x)2

2+y(4− x)2

2

= 4xy − x2y + x3

2− 4x2 + 8x+

x2y

2− 4xy + 8y =

x3

2− x2y

2− 4x2 + 8x+ 8y

The next integral, with respect to y, is∫ 1

y=0

x3

2− x2y

2− 4x2 + 8x+ 8ydy =

[x3y

2− x2y2

4− 4x2y + 8xy + 4y2

]1y=0

=x3

2− x2

4− 4x2 + 8x+ 4 =

x3

2− 17x2

4+ 8x+ 4

Finally, the last integral, in x, is∫ 2

x=0

x3

2−17x2

4+8x+4dx =

[x4

8− 17x3

12+ 4x2 + 4x

]2x=0

=16

8−17 · 8

12+16+8 = 26−34

3=

44

3


5(b). Answer: −20

3. Once again, we need to express D in algebraic terms before

we can compute any integrals. First we need to find the boundaries of the top

triangle. It includes the blue line between (0, 0, 0) and (2, 0,−2). This is the line

z = −x, y = 0. It also includes the line between (0, 0, 0) and (0, 2,−2). This is the

line z = −y;x = 0. The only plane that includes both of these two line segments

is

z = −x− y

Therefore, the top triangle is a cutout of z = −x − y. At any x, y, the region D

is defined between z = −x − y − 2 and z = −x − y since the top and bottom

triangles are separated by exactly two units of z. Finally, the region in the plane

is the triangle between (0, 0); (0, 2); and (2, 0). In this triangle,

0 ≤ y ≤ 2− x

and

0 ≤ x ≤ 2

The integral is then ∫ 2

x=0

∫ 2−x

y=0

∫ −x−yz=−x−y−2

xzdzdydz

The innermost integral is∫ −x−yz=−x−y−2

xzdx =

[xz2

2

]−x−yz=−x−y−2

=x(x+ y)2

2− x(x+ y + 2)2

2

By the sum of two squares,

x(x+ y + 2)2

2=x(x+ y)2 + 4x(x+ y) + 4x

2

The first term cancels, leaving us with

−2x(x+ y)− 2x = −2xy − 2x2 − 2x

The second integral is∫ 2−x

y=0

−2xy−2x2−2xdy =[−xy2 − 2x2y − 2xy

]2−xy=0

= −x(2−x)2−2x2(2−x)−2x(2−x)


This expands to

−x3 + 4x2 − 4x− 4x2 + 2x3 − 4x+ 2x2 = x3 + 2x2 − 8x

Finally, the outer integral is∫ 2

x=0

x3 + 2x2 − 8xdx =

[x4

4+

2x3

3− 4x2

]2x=0

=16

4+

16

3− 16 =

16

3− 36

3= −20

3


5(c). Answer: 23 . We have∫∫∫D

fdV =

∫∫∫D1

fdV +

∫∫∫D2

fdV

where D1, D2 are the two boxes. In other words, we need to integrate separately

over each box. The upper box is

0 ≤ x ≤ 2; 0 ≤ y ≤ 2; 0 ≤ z ≤ 1

and the lower box is

0 ≤ x ≤ 4; 0 ≤ y ≤ 3;−1 ≤ z ≤ 0

The integral over the upper box is∫ 2

x=0

∫ 2

y=0

∫ 1

z=0

(xy + xz + yz)dzdydz


z=0

xy + xz + yzdz =

[xyz +

xz2

2+yz2

2

]1z=0

= xy +x

2+y

2

The next integral is then∫ 2

y=0

xy +x

2+y

2=

[xy2

2+xy

2+y2

4

]2y=0

= 2x+ x+ 1 = 3x+ 1


x=0

3x+ 1dx =

[3x2

2+ x

]2x=0

= 6 + 2 = 8

The integral over D1 is 8. Now we must integrate over D2. We have∫∫∫D2

fdV =

∫ 4

x=0

∫ 3

y=0

∫ 0

z=−1xy + xz + yzdzdydx


z=−1xy+xz+yzdz =

[xyz +

xz2

2+yz2

2

]0z=−1

= 0−(−xy + x

2+y

2

)= xy− x

2− y2

The next integral is∫ 3

y=0

xy − x

2− y

2dy =

[xy2

2− xy

2− y2

4

]3y=0

=9x

2− 3x

2− 9

4= 3x− 9

4



x=0

3x− 9

4dx =

[3x2

2− 9x

4

]4x=0

= 3(8)− 9 = 15

This is the total for D2. The total for the full region D is then

8 + 15 = 23


6. For the region D defined by −x ≤ y ≤ 0; 0 ≤ z ≤ x; 0 ≤ x ≤ 5:

6(a). The plot of D is as follows:

Plot for 6a

This is a pyramid. There are several ways that we can determine how to draw this

shape. At x = 5, the “footprint” of y and z traces out a square with−5 ≤ y ≤ 0; 0 ≤

z ≤ 5. At any value of x, the cross-section in y and z is a square. We can also

draw the triangles 0 ≤ z ≤ x; 0 ≤ x ≤ 5 with y = 0; and −x ≤ y ≤ 0; 0 ≤ x ≤ 5

with z = 0. Then we can complete the shape around these two triangles.


6(b). Answer:125

3. To find the volume of D, which is given by this integral, we take

∫∫∫D

1dV =

∫ 5

x=0

∫ 0

y=−x

∫ x

z=0

1dzdydx

The innermost integral is ∫ x

z=0

1dz = [z]xz=0 = x

The next integral is ∫ 0

y=−xxdy = [xy]0y=−x = 0− (−x2) = x2

The outermost integral is ∫ 5

x=0

x2dx =

[x3

3

]5x=0

=125

3

This matches the geometric result that the volume of a square pyramid is s3

3.


6(c). Answer:625

4. This integral is evaluated the same way:

∫ 5

x=0

∫ 0

y=−x

∫ x

z=0

2x+ yzdzdydx

can be evaluated by first taking the innermost integral:∫ x

z=0

2x+ yzdz =

[2xz +

yz2

2

]xz=0

= 2x2 +x2y

2


y=−x2x2 +

x2y

2dy =

[2x2y +

x2y2

4

]0y=−x

= 0− 2x2(−x)− x2(−x)2

4= 2x3 − x4

4


x=0

2x3 − x4

4dx =

[x4

2+x5

20

]5x=0

=625

2− 3125

20=

3125

20=

625

4


6(d). The plot of D′ is as follows:

Plot for 6d

At x = 5, we have z−y ≤ 5, which is the region below the blue line x = 5, z−y = 5.

At any x, we have that the region below z − y = x (which is the outermost blue

face) is included. The cross-section of this shape in x is a triangle that shrinks in

size as x decreases.


6(e). Answer:125

6. We have

∫∫∫D′

1dV =

∫ 5

x=0

∫ 0

y=−x

∫ y+x

z=0

1dzdydx

Instead of running from 0 to x, z runs from 0 to y + x. This is a smaller range of z

since y is always negative. If y = 0, the range of z is exactly the same (because

D′ is the same as D in the xz plane). The inner integral is∫ y+x

z=0

1dz = y + x


y=−xy + xdy =

[y2

2+ xy

]0y=−x

= 0−(x2

2− x2

)=x2

2


x=0

x2

2dx =

[x3

6

]5x=0

=125

6

The base area (in y and z when x = 5) of D′ is half the base area of D, and the

volume of D′ is half the volume of D.


6(f). Answer:625

24. We take

∫ 5

x=0

∫ 0

y=−x

∫ y+x

z=0

zdzdydx

The inner integral is∫ y+x

z=0

zdz =

[z2

2

]y+xz=0

=(y + x)2

2=y2

2+x2

2+ xy


y=−x

y2

2+x2

2+ xydy =

[y3

6+x2y

2+xy2

2

]0y=−x

= 0−(−x

3

6− x3

2+x3

2

)=x3

6


x=0

x3

6dx =

[x4

24

]5x=0

=625

24


7. Evaluate the following integrals:

7(a). Answer:4

5. We have

∫ 1

x=0

∫ x

y=−x

∫ y2+2

z=1

xdV =

∫ 1

x=0

∫ x

y=−x

∫ y2+2

z=1

xdzdydx

The inner integral is∫ y2+2

z=1

xdz = [xz]y2+2z=1 = x(y2 + 2)− x = xy2 + x

The next integral is∫ x

y=−xxy2 + xdy =

[xy3

3+ xy

]xy=−x

=x4

3+ x2 −

(−x

4

3− x2

)=

2x4

3+ 2x2


x=0

2x4

3+ 2x2dx =

[2x5

15+

2x3

3

]1x=0

=2

15+

2

3− 0 =

12

15=

4

5


7(b). Answer:10795

7168. We write

dV = dzdydz

The inner integral is ∫ y2

z=0

z2dz =

[z3

3

]y2z=0

=y6

3

The next integral is ∫ x

y=x2

y6

3dy =

[y7

21

]xy=x

2

=x7

21− x7

21 · 27


x=1

x7

21− x7

21 · 27dx =

[x8

168− x8

168 · 27

]2x=1

=28

168− 28

168 · 27−(

1

168− 1

168 · 27

)This is

256

168− 2

168−(

1

168− 1

168 · 128

)=

253

168+

1

168 · 128=

32385

21504=

10795

7168


7(c). Answer:2

3−√2

3. We have

dV = dzdydx

The inner integral is ∫ π4

z=0

sin(2x+ 3y − z)dz

Here, 2x + 3y is a constant, so the antiderivative of sin(C − z) is cos(C − z)

because the negative sign that we get when differentiating cosine cancels with

the negative sign from −z in the chain rule. So this is

[cos(2x+ 3y − z)]π4z=0 = cos(2x+ 3y − π

4)− cos(2x+ 3y)

The next integral is ∫ π3

y=0

cos(2x+ 3y − π

4)− cos(2x+ 3y)dy

The antiderivative of cos(3y + C) is 13sin(3y + C), as a u-substitution will reveal.

This integral is then [1

3sin(2x+ 3y − π

4)− 1

3sin(2x+ 3y)

]π3

y=0

=1

3sin(2x+

3π

4)− 1

3sin(2x+ π)− 1

3sin(2x− π

4) +

1

3sin(2x)

We have

sin(2x+ π) = − sin(2x)

and likewise

sin(2x− π

4) = − sin(2x+

3π

4)

so this simplifies to2

3sin(2x+

3π

4) +

2

3sin(2x)

Then, we integrate once again:∫ π2

x=0

2

3sin(2x+

3π

4) +

2

3sin(2x)dx =

[−1

3cos(2x+

3π

4)− 1

3cos(2x)

]π2

x=0

= −1

3cos

7π

4−1

3cosπ+

1

3cos

3π

4+1

3cos 0 = −1

3

(√2

2

)+1

3+1

3

(−√2

2

)+1

3=

2

3−√2

3


7(d). Answer:1

4e8 − 1

4e−8 − 1

2e4 +

1

2e−4 − 1

4e2 +

1

4e−2 +

1

2e− 1

2e−1 . After writing dV =

dzdydx, we have that the inner integral is∫ 1

z=−1x5yex

2yzdz = x5y

∫ 1

z=−1ex

2yzdz

The antiderivative of eAz is 1AeAz + C so this is

x5y

[1

x2yex

2yz

]1z=−1

= x3(ex

2y − e−x2y)

The next integral is ∫ 1

y=−1x3ex

2y − x3e−x2ydy

This is

x3[1

x2ex

2y +1

x2e−x

2y

]2y=−1

= x(e2x

2

+ e−2x2 − ex2 − e−x2

)Finally, the outer integral is∫ 2

x=−1xe2x

2

+ xe−2x2 − xex2 − xe−x2

dx

We let

u = x2

so

dx =1

2xdu

Then, this is∫ 2

x=−1x(e2u + e−2u − eu − e−u

)· 12xdu =

∫ 2

x=−1

1

2

(e2u + e−2u − eu − e−u

)du

=

[1

4e2u − 1

4e−2u − 1

2eu +

1

2e−u]2x=−1

Since u = x2 this is1

4e8 − 1

4e−8 − 1

2e4 +

1

2e−4 − 1

4e2 +

1

4e−2 +

1

2e− 1

2e−1


7(e). Answer:π2

8. Since dV = dzdydx the inner integral is

∫ sin y

z=0

zdz =

[z2

2

]sin yz=0

=sin2 y

2

We have that

cos 2y = 1− 2 sin2 y

sosin2 y

2=

1

4− cos 2y

4

The next integral is∫ x

y=0

1

4− cos 2y

4dy =

[y

4− sin 2y

8

]xy=0

=x

4− sin 2x

8− 0− 0

The outer integral is∫ π

x=0

x

4− sin 2x

8dx =

[x2

8+

cos 2x

16

]πx=0

=π2

8+

1

16− 0− 1

16=

π2

8


7(f). Answer:1

20. In four-space,

dV = dwdzdydz

We have that∫ 2

x=1

∫ x

y=1

∫ y

z=1

∫ z

w=1

wdV =

∫ 2

x=1

∫ x

y=1

∫ y

z=1

∫ z

w=1

wdwdzdydx

=

∫ 2

x=1

(∫ x

y=1

(∫ y

z=1

(∫ z

w=1

wdw

)dz

)dy

)dx

The inner integral is ∫ z

w=1

wdw =

[w2

2

]zw=1

=z2

2− 1

2

The next integral is∫ y

z=1

z2

2− 1

2dz =

[z3

6− z

2

]yz=1

=y3

6− y

2− 1

6+

1

2=y3

6− y

2+

1

3

Then, we take∫ x

y=1

y3

6− y2+1

3dy =

[y4

24− y2

4+y

3

]xy=1

=x4

24−x

2

4+x

3− 1

24+1

4− 1

3=x4

24−x

2

4+x

3− 1

8


x=1

x4

24−x

2

4+x

3−1

8dx =

[x5

120− x3

12+x2

6− x

8

]2x=1

=32

120− 8

12+4

6−2

8− 1

120+

1

12−1

6+1

8

=31

120− 7

12+

3

6− 1

8=−39120

+3

8=−1340

+15

40=

1

20


Calculus 3 Tutor, Volume II Worksheet 1 Triple Integrals · Calculus 3 Tutor, Volume II Worksheet 1...

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Transcript of Calculus 3 Tutor, Volume II Worksheet 1 Triple Integrals · Calculus 3 Tutor, Volume II Worksheet 1...