Volumes of Revolution
You should be able to find the volume of a solid obtained by rotating a region about any horizontal line or
vertical line. The strategy is as follows. (Please follow this strategy, instead of just memorizing formulas).
1. Slice the region into n slices of equal width (either horizontally or vertically, depending on whether it
is easier to figure out horizontal or vertical distance). Each slice is approximately a thin rectangle.
2. Focus on the k-th slice. Estimate the volume that results from rotating the k-th slice.
3. Sum up over all slices to get a Riemann sum which gives an estimate for the volume of the whole solid.
4. Take the limit as n goes to infinity and express the limit as a definite integral.
5. Evaluate the definite integral.
Remark: It is not important whether you can visualize the whole solid. You just need to know what happens
to a typical slice when you rotate it. When you rotate a typical slice (the k-th slice), you will get one of
these three shapes:
• disk (volume: πr2 × thickness, where r is the radius)
• washer (volume: (πR2 − πr2 ) × thickness, where R is the outer radius and r is the inner radius)
• cylindrical shell (volume: 2πrh × thickness, where r is the radius and h is the height)
To find the volume of the k-th slice rotated, you need to figure out r (or R and r; or r and h) in terms of xk
or yk . It helps you figure out if you indicate them in the picture. Remember that
vertical distance=upper y coordinate-lower y coordinate
horizontal distance=right x coordinate-left x coordinate.
1. Here is one loop of the sine curve.
1
Π
(a) If you rotate this region about the x-axis, what shape do you get? What is its volume? You may
leave your answer as an integral.
Solution. The solid looks like a football.
1
To find its volume, we slice the region and figure out what happens to the k-th slice when we
rotate it. Let’s slice vertically. We’ll slice the interval [0, π] and label ∆x, x0 = 0, . . . , xn = π as
usual. We focus on the k-th slice.
1
xk
Π
-1
Rotating the k-th slice gives approximately a disk with radius sin xk and thickness ∆x, so
volume of k-th slice rotated ≈ π(sin xk )2 ∆x
Summing over all slices and taking the limit as the number of slices tends to ∞, the volume is
Z π
n
X
equal to lim
π(sin xk )2 ∆x, or
π(sin x)2 dx .
n→∞
k=1
0
(b) If you rotate the region about the horizontal line y = −1, what shape do you get? What is its
volume? You may leave your answer as an integral.
Solution. This looks like a bead. But even if you cannot visualize the shape, you can still
compute the volume by slicing, as long as you know what happens to a typical slice when you
rotate it.
We will slice the region the same way as in (a). Again, we want to see what happens when
we rotate the k-th slice. This time, rotating the k-th slice doesn’t give a disk; instead, it gives
something close to a washer (disk with a round hole cut out, like a CD). The outer radius of the
washer is sin xk + 1, the inner radius is 1, and the thickness is ∆x. Therefore,
volume of k-th slice rotated ≈ π(sin xk + 1)2 − π · 12 ∆x
Z
Summing over all slices and taking the limit as n → ∞, the volume is
0
2
π
π(sin x + 1)2 − π dx .
(c) If you rotate the region about the vertical line x = 4, what shape do you get? What is its volume?
You may leave your answer as an integral.
Solution. This looks like a Bundt cake.
We will slice the region the same way as in (a). Again, we want to see what happens when we
rotate the k-th slice:
1
xk
Π
4
-1
Now, we get something which is very close to being a toilet paper tube (or “cylindrical shell”) of
radius 4 − xk , height sin xk , and thickness ∆x. Therefore,
volume of k-th slice rotated ≈ 2π(4 − xk ) sin xk ∆x
Z
Summing over all slices and taking the limit as n → ∞, the volume is equal to
π
2π(4 − x) sin x dx .
0
2. Let R be the region enclosed by the circle of radius 3 centered at the origin. Using vertical slices, find
the volume generated when R is rotated about the line x = 4. (Please evaluate your integral.)
Note: It is also possible to find the volume using horizontal slices, and you might want to try that for
extra practice.
Solution. The shape of the solid is like a bagel.
We slice the interval [−3, 3] into n pieces of equal width ∆x, label x0 = −3, . . . , xn = 3 as usual, and
focus on the k-th slice.
3
3
-3
xk
3
4
-3
√
√
The equation of the circle is√x2 + y 2 = 9, or y = ± 9 − x2 . y = 9 − x2 is the equation of the top
half of the circle, and y = − 9 − x2 is the equation of the bottom half of the circle.
Rotatingpthe k-th slice gives (approximately) a paper towel tube with thickness ∆x, radius 4 − xk , and
height 2 9 − x2k . So,
q
q
volume of k-th slice rotated ≈ 2π(4 − xk ) 2 9 − x2k ∆x = 4π(4 − xk ) 9 − x2k ∆x.
Z
Summing over all slices and taking the limit as n → ∞, the volume is given by the integral
p
x) 9 − x2 dx. To evaluate this integral, let’s first multiply it out a little:
Z
3
Z
p
4π(4 − x) 9 − x2 dx = 16π
3
−3
−3
Z
p
9 − x2 dx − 4π
3
x
p
3
4π(4 −
−3
9 − x2 dx.
−3
√
Now, we have two integralsZto evaluate. Notice that y = 9 − x2 is just the graph of the top half of
3 p
the circle in our picture, so
9 − x2 is the area of the top half of the circle. We know that a circle
−3
Z 3p
9π
of radius 3 has area 9π, so
9 − x2 dx =
.
2
−3
Z 3 p
The second integral
x 9 − x2 dx is actually even simpler: the integrand is an odd function, and
−3
the interval of integration is symmetric about 0, so the integral is equal to 0. Alternatively, this integral
can be done using substitution.
So, our final answer is 16π ·
9π
2
= 72π 2 .
3. (a) What shape do you get if you rotate the following region about the x-axis?
4
1
-1
- 12€€€
1
y ‡ €€€€ Ix3 + 1M
2
1
€€€
2
1
y‡2x-1
Solution. The solid looks like a bell:
(b) Write down an integral or sum of integrals that expresses the volume of the solid from (a).
Solution. We will do this problem 2 ways, once using vertical slices and again using horizontal
slices. You will see that slicing horizontally is much easier.
• Method 1: Vertical Slices
If we slice the region vertically, there are two types of slices, which we need to deal with
separately:
1
-1
- 12€€€
1
€€€
2
1
Between x = −1 and x = 1/2, the slices look like the yellow slice: they go from the y-axis to
5
3
the curve y = x 2+1 . Between x = 1/2, the slices look like the magenta slice: they go from
3
y = 2x − 1 to y = x 2+1 . So, we’ll split the problem up further:
– The volume from x = −1 to x = 1/2
Now, we’ll slice the interval [−1, 1/2] into n pieces and label x0 = −1, . . . , xn = 1/2 as
usual. As always, we focus on the k-th slice:
1
- 12€€€
-1
1
€€€
2
xk
1
x3 +1
If we rotate this slice, we get a solid which is approximately a disk of radius k2 and
thickness ∆x. Therefore, the volume of the k-th slice of the solid is approximately
3 2
x +1
∆x.
π k2
Summing and taking the limit as n → ∞, the volume from x = −1 to x = 1/2 is
3
2
2
Z 1/2 3
n
X
xk + 1
x +1
lim
π
∆x =
π
dx.
n→∞
2
2
−1
k=1
– The volume from x = 1/2 to x = 1
Let’s slice the interval [1/2, 1] into n pieces and label x0 = 1/2, . . . , xn = 1 as usual. We’ll
again focus on a single slice, the k-th slice:
1
- 12€€€
-1
1
€€€
2
xk
1
Rotating this, we get a solid which is approximately a washer (or CD) with inner radius
x3 +1
2xk −1, outer radius k2 , and thickness ∆x. Therefore,
the volume of the k-th slice of the
3 2
xk +1
solid is approximately π
− (2xk − 1)2 ∆x. Summing and taking the limit as
2
6
n
X
"
n → ∞, the volume from x = 1/2 to x = 1 is lim
π
n→∞
k=1
"
#
3
2
Z 1
x +1
π
− (2x − 1)2 dx.
2
1/2
x3k + 1
2
2
#
− (2xk − 1)2 ∆x =
So, the total volume of the bell is
Z
1/2
π
−1
x3 + 1
2
2
Z
"
1
dx +
π
1/2
x3 + 1
2
2
#
− (2x − 1)2
dx
(This may look like quite a mess, but it is not hard to evaluate if you multiply out the
integrand; the volume is 17π
42 .)
• Method 2: Horizontal Slices
Now, there is just one type of slice, which makes things slightly easier. We’ll slice the interval
[0, 1] (on the y-axis) into n pieces and label y0 = 0, . . . , yn = 1 as always. Here is the k-th
slice:
1
yk
-1
- 12€€€
1
€€€
2
1
If we rotate this slice, we get a toilet paper tube (or “cylindrical shell”) of radius yk , height
of whatever the length of this slice is, and thickness ∆y.
We need to write everything in our Riemann sum in terms of yk and ∆y, so we want to find
3
the length of this slice in terms of yk . The slice goes from the curve y = x 2+1 to y = 2x − 1.
We can rewrite these two curves as x = (2y − 1)1/3 and x = y+1
2 . Therefore, the length of
yk +1
1/3
the k-th slice is 2 − (2yk − 1) . So, the k-th toilet paper tube has approximate volume
2πyk yk2+1 − (2yk − 1)1/3 ∆y. Summing these up and taking the limit, the volume of the
Z 1
y+1
1/3
bell is
2πy
− (2y − 1)
dy . (This integral may also look like a big mess, but
2
0
you should be able to evaluate it by substituting u = 2y − 1. Of course, the answer will be
17π
42 , just like it was when we used vertical slices.)
Homework: Do every problem step by step. Your writing should be like the writing in Problem 1 above.
You don’t need to draw the whole solid, but please draw the shape you get by rotating a typical slice: do
you get a disk? washer? cylindrical shell?
7
1. Consider the region bounded by the graph of y = x2 and the x-axis from x = 0 to x = 1. Find the
following volumes of revolution by following our strategy of slicing.
(a) Find the volume of the solid obtained by rotating the region around the x-axis.
(b) Find the volume of the solid obtained by rotating the region around the horizontal line y = 2.
(c) Find the volume of the solid obtained by rotating the region around the vertical line x = 3.
√
2. Consider the region enclosed by the curve y = x and the line y = 12 x. Find the volume obtained
by revolving the region around the vertical line x = −1. Do this two ways, once slicing vertically and
once slicing horizontally. Evaluate both integrals and show that you get the same answer.
3. Consider the region between y = 1 and y = 2 bounded by y = 2x + 2 and y = x1 . Write an integral
that gives the volume generated when the region is rotated around:(for this problem, it is much easier
to slice horizontally. You do NOT need to simplify or evaluate the integral.)
(a) the x-axis
(b) the vertical line x = 1
8