1. Finding the Volume of a Solid of Revolution

2. Consider two functions 𝑓 and 𝑔 bounding a region 𝐷
For this example, 𝑓 𝑥 = 1 8 𝑥 4 and 𝑔 𝑥 = 𝑥 3 −3 𝑥 2 +3𝑥
Now imagine we rotate 𝐷 around the 𝑥 axis

3. (Using Mathematica)
We investigate the problem of finding the volume of this object

4. Let’s consider a small slice of width 𝑑𝑥 of the region 𝐷
We can find the volume when it is rotated around the x axis.
Width=𝑑𝑥
Area=𝜋 𝑔 𝑥 2 −𝑓 𝑥 2
𝑑𝑉=Width×Area
= 𝜋 𝑔 𝑥 2 −𝑓 𝑥 2 𝑑𝑥
Integrate the volume element to get volume. a b
Volume= 𝑎 𝑏 𝜋 𝑔 𝑥 2 −𝑓 𝑥 2 𝑑𝑥
Where [𝑎,𝑏] is the interval over which 𝐷 lies.

5. Example:
Let 𝐷 be the object bounded by the graphs of 𝑦= 1 8 𝑥 4 and 𝑦= 𝑥 3 −3 𝑥 2 +3𝑥.
Let 𝐸 be the object obtained by rotating 𝐷 around the 𝑥-axis.
Find the volume of 𝐸.
Solution
We first sketch 𝐷.
We see that the interval is 𝑎,𝑏 =[0,2].
The formula for the volume of 𝐸 is…
Volume= 0 2 𝜋 𝑥 3 −3 𝑥 2 +3𝑥 2 − 𝑥 𝑑𝑥
= 𝜋 − 𝑥 𝑥 7 7 − 𝑥 6 +3 𝑥 5 − 9 2 𝑥 4 +3 𝑥
= 88𝜋 63 ∎

6. Example 2:
Consider the region 𝐷 in the first quadrant bounded by 𝑦= sin 𝑥 and 𝑦= cos 𝑥 as in the picture.
Let 𝐸 be the solid obtained by rotating 𝐷 about the 𝑦-axis.
Find the volume of 𝐸.
Consider a small sliver of 𝐷 of width 𝑑𝑥.
Solution
When rotated around the 𝑦-axis, it becomes…
a cylindrical shell of radius 𝑥, height cos 𝑥 − sin 𝑥 and thickness 𝑑𝑥.
Hence 𝑑𝑉=2𝜋×radius×height 𝑑𝑥
=2𝜋𝑥 cos 𝑥 − sin 𝑥 𝑑𝑥
And the integration will go from 0 to 𝜋/4 since sin 𝑥= cos 𝑥 when 𝑥=𝜋/4
Volume= 0 𝜋/4 2𝜋𝑥 cos 𝑥 − sin 𝑥 𝑑𝑥
= 2𝜋 cos 𝑥 +𝑥 cos 𝑥 − sin 𝑥 +𝑥 sin 𝑥 0 𝜋/4
= 𝜋 2 𝜋 2 −4

7. Example 3
Let 𝐷 be the region between 𝑦=𝑥+1 and 𝑦= 𝑥−1 2 .
Find the volume of the solid obtained by revolving 𝐷 around the 𝑥-axis.
Solution
We draw a sketch.
Now we find the volume of a small slice.
𝑑𝑉=Area×𝑑𝑥
=𝜋 𝑓 𝑥 2 −𝑔 𝑥 2 𝑑𝑥
𝑉= 0 3 𝜋 𝑥+1 2 − 𝑥−1 4 𝑑𝑥
=𝜋 3 𝑥 2 − 5 3 𝑥 3 + 𝑥 4 − 1 5 𝑥
= 72𝜋 5

8. Example
Let 𝐷 be the region bounded by 𝑦= ln 𝑥 , 𝑥=1, 𝑦= ln 3 .
Let 𝐸 be the solid obtained by rotating 𝐷 about the line 𝑥=3.
Find the volume of 𝐸.
Solution
We draw a sketch.
We pick a sliver of width 𝑑𝑥.
Now we observe what happens to the sliver as it is rotated about
the line 𝑥=3.
We obtain a cylindrical shell with volume:
𝑑𝑉=2𝜋×Radius×Height×Thickness
=2𝜋
3−𝑥
ln 3 − ln 𝑥 𝑑𝑥
𝑉= 1 3 𝑑𝑉
= 1 3 2𝜋 3−𝑥 ln 3 − ln 𝑥 𝑑𝑥
= 1 3 2𝜋 𝑥−3 ln 𝑥 3 𝑑𝑥
Integration by parts
= 6𝜋𝑥− 𝜋 𝑥 2 2 −6𝜋𝑥 ln 𝑥 3 +𝜋 𝑥 2 ln 𝑥
=8𝜋−5𝜋 ln 3

9. See sections 7.2 and 7.3 in Stewart for problems on finding the volumes of solids of revolution.