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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?
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(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.
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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?
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(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?
(c) Evaluate∫∫∫
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?
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(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
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(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 .
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Plot for 3c
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(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
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(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];
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(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.
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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
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(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
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(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
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6. For the region D defined by −x ≤ y ≤ 0; 0 ≤ z ≤ x; 0 ≤ x ≤ 5:
(a) Plot D.
Plot for 6a
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(b) Evaluate∫∫∫
D1dV .
(c) Evaluate∫∫∫
D2x+ yzdV .
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(d) Now consider D′ which is a subregion of D with added constraint that z − y ≤ x.
Plot D′.
Plot for 6d
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(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 .
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(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 .
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(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.
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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
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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
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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.
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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 .
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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 .
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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
The inner integral is ∫ 0
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
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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
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2(c). Answer: 0 . We evaluate∫∫∫D
f(x, y, z)dV =
∫∫R
(∫ 4
z=0
xdz
)dA
The inner integral is ∫ 4
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
The inner integral is ∫ 0
y=−84xdy = [4xy]0y=−8 = 32x
The outer integral is∫ 2
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.
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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.
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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.
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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
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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
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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
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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.
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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.
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4. Integrals over boxes:
4(a). Answer:1
3. This is
∫∫∫D
x2dV =
∫∫R
(∫ 1
z=0
x2dz
)dA
The inner integral is∫ 1
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
The inner integral is ∫ 1
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
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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
The inner integral is ∫ 2
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
The inner integral is∫ 0
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.
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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
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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
The inner integral is ∫ 8
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
The outer integral is∫ 2
x=−2(32x+ 384)dx =
[16x2 + 384x
]2x=−2 = 64 + 768− (64− 768) = 1536
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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
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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
c©2019 MathTutorDVD.com 41
= [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
c©2019 MathTutorDVD.com 42
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
c©2019 MathTutorDVD.com 43
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)
c©2019 MathTutorDVD.com 44
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
c©2019 MathTutorDVD.com 45
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
The inner integral is∫ 1
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
Finally, the outer integral is∫ 2
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
The inner integral is∫ 0
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
c©2019 MathTutorDVD.com 46
Finally, the outer integral is∫ 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
c©2019 MathTutorDVD.com 47
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.
c©2019 MathTutorDVD.com 48
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.
c©2019 MathTutorDVD.com 49
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
The next integral is∫ 0
y=−x2x2 +
x2y
2dy =
[2x2y +
x2y2
4
]0y=−x
= 0− 2x2(−x)− x2(−x)2
4= 2x3 − x4
4
Finally, the outer integral is∫ 5
x=0
2x3 − x4
4dx =
[x4
2+x5
20
]5x=0
=625
2− 3125
20=
3125
20=
625
4
c©2019 MathTutorDVD.com 50
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.
c©2019 MathTutorDVD.com 51
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
The next integral is∫ 0
y=−xy + xdy =
[y2
2+ xy
]0y=−x
= 0−(x2
2− x2
)=x2
2
Finally, the outer integral is∫ 5
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.
c©2019 MathTutorDVD.com 52
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
The next integral is∫ 0
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
Finally, the outer integral is∫ 5
x=0
x3
6dx =
[x4
24
]5x=0
=625
24
c©2019 MathTutorDVD.com 53
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
Finally, the outer integral is∫ 1
x=0
2x4
3+ 2x2dx =
[2x5
15+
2x3
3
]1x=0
=2
15+
2
3− 0 =
12
15=
4
5
c©2019 MathTutorDVD.com 54
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
Finally, the outer integral is∫ 2
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
c©2019 MathTutorDVD.com 55
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
c©2019 MathTutorDVD.com 56
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
c©2019 MathTutorDVD.com 57
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
c©2019 MathTutorDVD.com 58
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
Finally, the outer integral is∫ 2
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
c©2019 MathTutorDVD.com 59