Finding the Volume of a Solid of Revolution

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

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

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.

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 − 1 8 𝑥 4 2 𝑑𝑥 = 𝜋 − 𝑥 9 576 + 𝑥 7 7 − 𝑥 6 +3 𝑥 5 − 9 2 𝑥 4 +3 𝑥 3 0 2 = 88𝜋 63 ∎

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

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 5 0 3 = 72𝜋 5

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 3 1 3 =8𝜋−5𝜋 ln 3

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