Volumes of Revolution
D. P. Morstad, University of North Dakota
Objectives of Assignment
1.
2.
I.
To practice setting up integrals to find the volume of revolution.
To learn when to use the disk method and when to use the shell method.
Volumes of Revolution
A function y = f(x) can be revolved around either the x-axis or the y-axis to create a
volume. Consider f(x) = 2sin(x) for 0 ≤ x ≤ π/2.
y
x
Figure 1. Graph of f(x) = 2sin(x) for 0 ≤ x ≤ π /2.
-2
1
2
0
-1 0
1 2
-1
2
1
1.5
0
1
-1
0.5
-2
0
-1
0
0
1
0.5
1
1.5
Figure 2a. f(x) revolved around y-axis
.
Figure 2b. f(x) revolved around xaxis.
(X(PLORE) does not graph the volume of revolution, but it will help you integrate.)
46
Similarly, a function x = g(y) can also be revolved around either the x-axis or the y-axis
to create a volume. Consider g(y) = y2 for 0 ≤ y ≤ 2.
y
x
Figure 3. Graph of g(y) = y2 for 0 ≤ y ≤ 2.
-2
-1012
2
4
2
0
-2
-4
1
2
1.5
0
1
-1
0.5
0
-4
-2
0
2
Figure 4a. g(y) revolved around y-axis.
4
-2
0
1
2
3
4
Figure 4b. g(y) revolved around x-axis.
Each one of these three-dimensional revolutions generates two volumes of revolution.
Figure 4a. for example, has a volume “inside” the surface - that is, the volume which is between
the surface and the top of the rectangular solid in which it is graphed; and it has an “outside”
volume - the volume from the surface down to the floor of the rectangular solid.
For simple volumes of revolution, the “inside” volume contains the axis which it has
been revolved around. The “outside” volume contains the other axis. Imagine, for example, a
whirlpool being formed as water is being drained from a bathtub. The surface of the whirlpool
is similar in shape to the surface in Figure 4a. The “outside” volume of revolution corresponds
to water in the whirlpool whereas the “inside” volume corresponds to air going down into the
whirlpool.
Since each of the four revolutions above has an “inside” and an “outside” volume of
revolution, there is a total of eight volumes of revolution.
47
II. The Disk Method and “Inside” Volumes
The disk method will work very well with the “inside” volumes of Figures 2b and 4a.
The disk method will also work nicely with the “inside” volumes of Figure 2a and 4b, but only if
you can find the inverses of f(x) and g(y).
Recall the formula for the volume of a disk:
Vdisk = πr2· h
The disk method uses this formula in a Riemann sum:
Vrevolution
≈ ∑ πr2· h
≈ ∑ πf(x)2· ∆x, then take the limit as ∆x → 0 .
Vrevolution
=
∫ π f (x )
b
2
a
dx
⇐ Disk Method
The four “inside” volumes of revolution are:
III.
V2a “inside”
=
V2b “inside”
=
V4a “inside”
=
V4b “inside”
=
∫
∫
∫
∫
y=d
y =c
πf
−1
x=b
x =a
π f ( x ) 2 dx =
y=d
y= c
∫
π g( y )2 dy =
x=b
x =a
∫
( y ) 2 dy =
π g −1 ( x )2 dx =
y =0
x = π /2
x =0
∫
∫
π sin −1
x =0
dy
2
π y2
x =4
)
2
y
2
π (2sin x ) dx
( )
y =2
y =0
(
y =2
π
2
dy
( x)
2
dx .
The Shell Method and “Outside” Volumes
The shell method will work very well with the “outside” volumes of Figures 2a and 4b.
The shell method will also work nicely with the “outside” volumes of Figure 2b and 4a, but only
if you can find the inverses of f(x) and g(y).
Recall the formula for the volume of a shell is:
Vshell = circumference· height· thickness
48
The shell method uses this formula in a Riemann sum:
Vrevolution
≈ ∑ 2πr· h· t
≈ ∑ 2πx· f(x)· ∆x, then take the limit as ∆x → 0 .
Vrevolution
=
∫ 2π x f ( x ) dx .
b
a
⇐ Shell Method
The four “outside” volumes of revolution are:
IV.
x=b
V2a “outside”
=
∫
x =a
V2b “outside”
=
∫
y=c
V4a “outside”
=
V4b “outside”
=
∫
∫
2π x f ( x ) dx =
y =d
∫
x =π / 2
x =0
2π y f −1 ( y ) dy =
x=b
x=a
y= c
∫
2π x g −1 ( x ) dx =
y=d
2π y g( y ) dy =
∫
∫
2π x 2sin x dx
y =2
y= 0
2π y sin− 1 2y dx
x =4
x =0
y =2
y =0
2π x x dx
2
2π y y dy .
The Washer Method vs. The Shell Method
Suppose an area between two curves, such as the shaded area below, is revolved
around an axis to generate a volume.
Figure 5. A region between f(x) = x3 – 3x 2 + 3x – ½ and g(x) = x – ½.
49
The points of intersection, found by setting f(x) equal to g(x), are (1, ½) and (2, 1½).
Revolving this region around the x-axis results in thin washers instead of disks. The hole
in the washer comes from revolving f(x) around the x-axis. The outer circumference of the
washer comes from revolving g(x) around the x-axis.
The Washer Method is simply subtracting the volume of the hole (based on f(x)) away
from the large disk (based on g(x)). That is,
Vwasher = Vbig disk – Vsmall disk
= πR2· h – πr2· h
= π(g(x))2· ∆x – π(f(x))2· ∆x
The Washer Method uses this formula in a Riemann sum:
Vrevolution
≈ ∑ πR2· h – πr2· h
≈ ∑ π(g(x))2· ∆x – π(f(x))2· ∆x,
then taking the limit as ∆x → 0 ,
Varound x-axis =
∫ π ( g( x )
Varound x-axis =
∫
b
a
2
)
− f ( x )2 dx ⇐ Washer Method
For this example,
1
(
(
π ( x − 12 ) − x 3 − 3x 2 + 3x − 12
2
2
π
.
) ) dx = 37
84
2
Revolving the region around the y-axis lends itself to the Shell Method. The height of
each shell is now just g(x) – f(x), so the formula for this volume is just
Varound y-axis =
∫ 2π x ( g (x ) −
Varound y-axis =
∫ 2π x ( (x −
b
a
f ( x )) dx . ⇐ Shell Method
So,
2
1
1
2
50
)
) − ( x 3 − 3x 2 + 3x − 12 ) dx =
23π
.
30
BIG HINT - How to memorize all these formulas:
Don’t!
1.
2.
3.
4.
Instead, figure them out each time you work a problem:
Sketch a little graph, add disks, shells, washers - whatever is necessary.
Label radii, thickness, and height (height only for shells).
Put these into either πr2· h, 2πr· h· t, or πR2· h – πr2· h.
Integrate it over the desired interval (on the correct axis).
V.
Practice Problems
Use the “In(INTEGRAND, x = a to b)” command in X(PLORE) to quickly evaluate
the integrals to determine if you have set them up correctly. You will have to integrate by hand
on the hand-in homework. Solutions are on page 52.
In problems 1 - 4, first have X(PLORE) graph the region that will be revolved, then find the
volume of revolution using the disk method.
1.
Find the volume generated by revolving the region between f(x) = 2x +1 and the
x-axis, from x = 0 to x = 2, around the x-axis. Ans: 64.92625
2.
Find the volume generated by revolving the region between f(x) = 2x +1 and the
y-axis, from x = 0 to x = 2, around the y-axis. On this one you must first find the
inverse of f(x), and then you must integrate over the appropriate interval of the
y-axis. Ans: 16.75516
Find the volume generated by revolving the region between g(y) = sin-1y and the
y-axis, from y = 0 to y = 1, around the y-axis. ***DANGER!!!*** X(PLORE)
always plots the independent variable on the horizontal axis. Using
graph(asin(y),
y), X(PLORE) will plot the arcsine function along the x-axis. You
will need to figure out
some other way of plotting the arcsine function along the
y-axis. Ans: 1.468384
3.
4.
Find the volume generated by revolving the region between g(y) = sin-1y and the
x-axis, from y = 0 to y = 1, around the x-axis. This one is similar to #2.
Ans: 2.467401
In problems 5 - 8, first have X(PLORE) graph the region that will be revolved, then find the
volume of revolution using the shell method.
5.
Find the volume generated by revolving the region between f(x) = 2x +1 and the
x-axis, from x = 0 to x = 3, around the y-axis. Ans: 141.37167
51
6.
Find the volume generated by revolving the region between f(x) = 2x +1 and the
y-axis, from x = 0 to x = 3, around the x-axis. On this one you must first find the
inverse of f(x), and then you must integrate over the appropriate interval of the
y-axis. Ans: 282.74334
7.
Find the volume generated by revolving the region between g(y) = sin-1y and the
y-axis, from y = 0 to y = 1, around the x-axis. Ans: 2.467401
8.
Find the volume generated by revolving the region between g(y) = sin-1y and the
x-axis, from y = 0 to y = 1, around the y-axis. This one is similar to #2.
Ans: 2π
9.
The area of the plane which is bounded between f(x) = x 2 – 4x + 5 and
g(x) = 2x –3 is revolved around the x-axis. Find the volume of revolution.
Ans: 21.78171
10.
The area of the plane which is bounded between f(x) = x 2 – 4x + 5 and
g(x) = 2x –3 is revolved around the y-axis. Find the volume of revolution.
Ans: 25.13274
11.
For this problem, consider only the region of the x-y plane where –π ≤ x ≤ π.
The area of the plane which is bounded between f(x) = csc(x) and g(x) = 2 is
revolved around the x-axis. Find the volume of revolution. Ans: 15.43615
12.
For this problem, consider only the region of the x-y plane where –π ≤ x ≤ π.
The area of the plane which is bounded between f(x) = csc(x) and g(x) = 2 is
revolved around the y-axis. Find the volume of revolution. Ans: 15.346
52
Solutions to practice problems:
Region for practice problem 1.
(Region for # 5 is almost the same.)
Region for practice problem 2.
(Region for # 6 is almost the same.)
Region for practice problems 3 & 7.
Region for practice problems 4 & 8.
Volumes of revolution:
1. 64.93;
8. 2π;
2. 16.76;
9. 21.78;
3. 1.468;
10. 25.13;
4. 2.467;
11. 15.43;
53
5. 179.1;
12. 15.35.
6. 56.55;
7. 2.467;