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MATH152 WEEK 2
6.2 Volumes by Disks, Washers and Slices
Let S be a solid that lies between x = a and x = b. If the cross-sectional area of S is acontinuous function A(x),
then the volume of S is:
V =
∫ b
a
A(x) dx.
This means to calculate V we can split the solid up into slices with cross-sectional area A.Two common shapes for the slice are disks and washers. The picture below shows how wecan start to obtain better approximations of the volulme of a sphere using 5 disks, 10 disksand then 20 disks.
1
The plot below shows a cross-section of the solid formed by rotating the region bounded byy = x2, x = 0, x = 1 and the x-axis about the y-axis. A typical slice perpendicular to they-axis has the shape of a washer, shown in blue.
−2 −1 1 2
1
2
3
4
The volume of the solid can then be calculated by integrating the area of this washer as yvaries from 0 to 1.
Examples:
(1) Show that the volume of a sphere of radius r is 43πr3.
Consider a circle of radius r centred at the origin, with equation x2 + y2 = r2. Re-volving this circle about the x-axis produces a sphere of radius r whose cross-section(perpendicular to the x-axis) is a disk of radius y =
√r2 − x2 and area
A(x) = πy2 = π(r2 − x2).
−1.5 −1 −0.5 0.5 1 1.5
−1
1
2
Thus the volume of the sphere is∫ r
−rA(x) dx =
∫ r
−rπ(r2 − x2) dx
= π
[r2x− x3
3
]r−r
= π
(r3 − r3
3−(−r3 − −r
3
3
))=
4
3πr3.
(2) Find the volume of the solid obtained by rotating the region bounded by y = x3,y = 8 and x = 0 about the y-axis.
A typical slice perpendicular to the y-axis has the shape of a disk with radius x = y1/3.The area of this disk is
A(y) = π(y1/3)2 = πy2/3.
−3 −2 −1 1 2 3
5
10
15
20
25
Thus the volume of the solid formed is∫ 8
0
A(y) dy =
∫ 8
0
πy2/3 dy
= π
[y5/3
53
]80
=96π
5.
(3) Find the volume of the solid formed by rotating the region bounded by y =1
x, x = 1,
x = 2 and the x-axis about the x-axis.
A typical slice perpendicular to the x-axis has the shape of a disk with radius y =1
x.
3
The area of this disk is
A(x) = πy2 =π
x2.
0.5 1 1.5 2 2.5 3
−2
2
Thus the volume of the solid is∫ 2
1
A(x) dx =
∫ 2
1
π
x2dx
= π
[−1
x
]21
=π
2.
(4) Find the volume of the solid formed by rotating the region bounded by y = x2, x = 0,y = 1 and the x-axis about the y-axis.
A typical slice perpendicular to the y-axis has the shape of a disk with radius x =√y.
The area of this disk is
A(y) = πx2 = πy.
−2 −1 1 2
1
2
3
4
4
Thus the volume of this paraboloid is∫ 1
0
A(y) dy =
∫ 1
0
πy dy
= π
[y2
2
]10
=π
2.
(5) Find the volume of the solid formed by rotating the region bounded by y = x3 andy = x in the upper-half plane y ≥ 0 rotated about the x-axis.
First note that the graphs of y = x3 and y = x intersect at x = 0 and x = 1when we impose the requirement that y ≥ 0. Graphing the bounded region showsthat the typical slice perpendicular to the x-axis is a washer whose outer radius is xand inner radius is x3. The area of this washer is
A(x) = πx2 − π(x3)2 = π(x2 − x6).
0.5 1 1.5 2
−2
−1
1
2
Thus the volume of the solid formed is∫ 1
0
A(x) dx =
∫ 1
0
π(x2 − x6) dx
= π
[x3
3− x7
7
]10
=4π
21.
(6) Find the volume of the solid formed by rotating the region bounded by y = secx,
x = 0, x =π
4and the x-axis, rotated about the x-axis.
This solid is made of slices perpendicular to the x-axis in the shape of disks with5
radius y = secx and hence area
A(x) = π sec2 x.
0.5 1 1.5 2
−2
−1
1
2
Thus the required volume is∫ π/4
0
A(x) dx =
∫ π/4
0
π sec2 x dx
= π [tanx]π/40
= π.
(7) Find the volume of the solid formed by rotating the region bounded by y =√x and
y = x about y = 1.
This solid is made of slices perpendicular to the line y = 1 in the shape of wash-ers with outer radius 1 − x and inner radius 1 −
√x. The area of the washer at x
between (x = 0 and x = 1) is
A(x) = π((1− x)2 − (1−√x)2).
0.2 0.4 0.6 0.8 1
0.5
1
1.5
2
2.5
6
Thus the volume of the solid formed is∫ 1
0
A(x) dx =
∫ 1
0
π((1− x)2 − (1−√x)2) dx
=
∫ 1
0
π(x4 − 2x2 − x+ 2√x) dx
= π
[x5
4− 2x3
3− x2
2+
4x3/2
3
]10
=5π
12.
(8) Set up the integral to find the volume of the region bounded by y = 4x2, x = 4 andy = 0 rotated about the y-axis.
A typical slice perpendicular to the y-axis has the shape of a washer with outer
radius 4 and inner radius x =√y
2. The area of the washer at y (for 0 ≤ y ≤ 64) is
A(y) = π
(42 −
(√y
2
)2)
= π(
16− y
4
).
−4 −2 2 4
20
40
60
80
100
Thus the volume of this solid is∫ 64
0
A(y) dy =
∫ 64
0
π(
16− y
4
)dy.
If we were also asked to evaluate this integral, it would be∫ 64
0
π(
16− y
4
)dy = π
[16y − y2
8
]640
= 512π.
7
One can also work with cross-sections of more general shapes instead of disks and washers.
Examples:
(1) Find the volume of the solid whose base is a circle with radius 1 and whose crosssections are squares perpendicular to the x-axis.
Recall that the equation of the circle in question is x2 + y2 = 1. Hence the semicirclein the upper-half plane y ≥ 0 has equation
y =√
1− x2.
The typical square slice perpendicular to the x-axis has area
A(x) = (2y)2 = 4(1− x2).
−1.5 −1 −0.5 0.5 1 1.5
−1
1
Thus the volume of the solid formed is∫ 1
−1A(x) dx =
∫ 1
−14(1− x2) dx
= 8
[x− x3
3
]10
=16
3.
(2) Find the volume of the solid whose base is the region between the parabola y = 1−x2and the x-axis and whose cross sections are squares perpendicular to the x-axis.
This is similar to the above question, except that the area of the square slice at−1 ≤ x ≤ 1 now has area
A(x) = y2 = (1− x2)2.8
−1.5 −1 −0.5 0.5 1 1.5
0.5
1
1.5
Thus the volume of the solid formed is∫ 1
−1A(x) dx =
∫ 1
−1(1− x2)2 dx
=
[x+
x5
5− 2x3
3
]10
=8
15.
(3) Find the volume of the solid whose base is the region bounded by the parabola y = x2,x = 0 and x = 1 and whose cross sections are equilateral triangles perpendicular tothe x-axis.
The area of a typical slice at 0 ≤ x ≤ 1 has side length y = x2. Since the area
of an equilateral triangle of side length a is√34a2, the area of the slice at x is
A(x) =
√3
4y2 =
√3
4x4.
0.2 0.4 0.6 0.8 1 1.2
0.5
1
1.5
9
Thus the volume of the solid formed is∫ 1
0
A(x) dx =
∫ 1
0
√3
4x4 dx
=
√3
4
[x5
5
]10
=
√3
20.
(4) Find the volume of the solid whose base is the region betweeen y = 8 − 2x, x = 0and y = 0 and whose cross sections are squares perpendicular to the y-axis.
The area of a square slice at 0 ≤ y ≤ 8 has side length x = 4− y
2and area
A(y) =(
4− y
2
)2= 16 +
y2
4− 4y.
−1 1 2 3 4
2
4
6
8
Thus the volume of the solid formed is∫ 8
0
A(y) dx =
∫ 8
0
16 +y2
4− 4y dy
=
[16y +
y3
12− 2y2
]80
=128
3.
(5) Find the volume of the solid whose base is the region betweeen y = 2− x, x = 0 andy = 0 and whose cross sections are isoceles triangles perpendicular to the x-axis withheight equal to the base (y-value).
The area of a triangular slice at 0 ≤ x ≤ 2 has base and height equal to y = 2 − x10
and hence has area
A(x) =1
2(2− x)2.
−1 −0.5 0.5 1 1.5 2
1
2
3
Thus the volume of the solid formed is∫ 2
0
A(x) dx =
∫ 2
0
1
2(2− x)2 dx
=1
2
[4x+
x3
3− 2x2
]20
=4
3.
11
6.3 Volumes by Cylindrical Shells
Another way to compute volumes of revolution is to break up the solid into cylindricalshells. For example, consider the following solid formed by about around the y-axis.
We have already seen how to compute its volume using the slice method by integration withrespect to y. But we can also integrate along x using the shell method. This avoids havingto convert y = f(x) into a function of y.
Consider a vertical strip of width dx in the graph of f . If we revolve this strip about they-axis we obtain the shell of a cylinder. This shell has height f(x), radius x and thicknessdx, and hence its volume is
dV = 2πxf(x) dx.
Since we can break up the total volume V into such shells as x goes from a to b, we have:
V =
∫ b
a
2πxf(x) dx.
Examples:
(1) Find the volume of the solid obtained by revolving the region bounded by y =1
x, y =
0, x = 1, and x = 4 about the y-axis.
Since the solid is cylindrically symmetric about the y-axis, we can use cylindricalshells and integrate along the x-axis.
12
−4 −2 2 4
1
2
3
The cylindrical shell at 1 ≤ x ≤ 4 has radius x and height y =1
x. Hence the volume
element is
dV = 2πx · 1
x· dx = 2π dx.
Thus the volume of the solid is
V =
∫ 4
1
2π dx = 6π.
(2) Find the volume of the solid obtained by revolving the region bounded by y = x−x2and y = 0 about the y-axis.
The typical cylindrical shell at x has radius x and height y = x − x2. Thus thevolume element is
dV = 2πx · (x− x2) dx.
−1.5 −1 −0.5 0.5 1 1.5
0.1
0.2
0.3
13
The volume of the solid is hence∫ 1
0
dV =
∫ 1
0
2πx · (x− x2) dx
= 2π
[x3
3− x4
4
]10
=π
6.
(3) Find the volume of the solid obtained by revolving the region bounded by y = ex2,
y = 0, x = 0, and x = 1 about the y-axis.
Since the solid is rotationally symmetric about the y-axis, we integrate cylindricalshells along the x-axis. The typical cylindrical shell at 0 ≤ x ≤ 1 has radius x andheight y = ex
2. Thus the volume element is
dV = 2πx · ex2 dx.
The volume of the solid is hence∫ 1
0
dV =
∫ 1
0
2πxex2
dx
= π
∫ 1
0
2xex2
dx.
Now let u = x2, so that du = 2xdx. Note that x = 0 =⇒ u = 0 and x = 1 =⇒u = 1. Thus the integral becomes∫ 1
0
eu du = [eu]10
= e− 1.
(4) Find the volume of the solid obtained by revolving the region bounded by x = y2,x = 0, y = 2, and y = 5 (y > 0) about the x-axis.
Since the solid is rotationally symmetric about the x-axis, we integrate cylindri-cal shells along the y-axis. The typical cylindrical shell at 2 ≤ y ≤ 5 has radius yand height x = y2. Thus the volume element is
dV = 2πy · y2 dy = 2πy3 dy.14
The volume of the solid is then∫ 5
2
dV =
∫ 5
2
2πy3 dy
= 2π
[y4
4
]52
dx
=609π
2.
(5) Find the volume of the solid obtained by revolving the region bounded by the partof y = x2 for positive x and the lines y = 0, x = 1, and x = 2, about the line x = 3.
The typical cylindrical shell at 1 ≤ x ≤ 2 has radius 3 − x and height y = x2.Thus the volume element is
dV = 2π · (3− x) · x2 dx.
The volume of the solid is then∫ 2
1
dV =
∫ 2
1
2πx2(3− x) dx
= 2π
[x3 − x4
4
]21
=13π
2.
(6) Find the volume of the solid obtained by revolving the region bounded by y =√x
and y = x2 about the line x = −1.
Here the typical cylindrical shell at 0 ≤ x ≤ 1 has radius x − (−1) = x + 1 andheight
√x− x. Thus the volume element is
dV = 2π(x+ 1)(√x− x) dx.
The volume of the solid is then∫ 1
0
dV =
∫ 1
0
2π(x+ 1)(√x− x) dx
=14π
30.
15
Many problems can be solved using more than one method, and it is sometimes easier to useone or another depending on the symmetries in the problem.
Examples:
(1) Consider the region R bounded by y = x2 and y = 2x. Set up and compare theintegrals to find the volume of the solid obtained by revolving R about the x-axisusing the washer method and the shell method.
Washer method:
The typical washer perpendicular to the y-axis has outer radius x =√y and in-
ner radius x = y2. Thus the integral we get using this method is:∫ 4
0
π
((√y)2 −
(y2
)2)dy = π
∫ 4
0
y − y2
4dy.
Cylindrical shell method:
The typical cylindrical shell at x has radius x and height 2x − x2. Thus the vol-ume element is dV = 2πx(2x− x2), and the integral we get is:∫ 2
0
2πx(2x− x2) dx.
Thus the two methods require integration along different axes. We can compute thetwo integrals to find that the answers agree:
V =8π
3.
(2) Consider the region R bounded by y = cosx, y = 0, x = 0, and x = π2. Set up and
compare the integrals to find the volume of the solid obtained by revolving R aboutthe y-axis using the disk method and the shell method. (You do not need to evaluatethe integrals.)
Disk method:
The typical disk perpendicular to the y-axis has radius x = arccos y. The integral weget using this method is: ∫ 1
0
π (arccos y)2 dy.
Cylindrical shell method:16
The typical cylindrical shell at x has radius x and height y = cos x. Thus thevolume element is dV = 2πx · cosx dx, and the integral we get is:∫ π/2
0
2πx cosx dx.
Again we see that the two methods require integration along different axes.
17
Bonus questions:
(1)
∫18 + 17x
1 + x2dx.
Use the substitution u = 1 + x2.
(2)
∫ π/4
0
1
cos2(x)√
1 + tan xdx.
Use the substitution u = 1 + tanx.
(3)
∫ 3
1
e1/x6
x7dx.
Use the substitution u = 1x6
.
(4) Find the area of the region enclosed by xy = 12, 3y = x, 3y = 4x.
(5) Find the area bounded by y = ex/2, the tangent to this curve at x = 3, the x-axisand the y-axis.
18
