Answer
a. $\frac{2\pi}{3}$
b. $\frac{8\pi}{3}$
Work Step by Step
a. Draw a diagram as shown in the figure. With the enclosed region revolving around $x=1$, using cylinder approximations, we have
$V=\int_0^1 2\pi (1-x)(2x)dx=2\pi\int_0^1 (2x-2x^2)dx=2\pi (x^2-\frac{2}{3}x^3)|_0^1=2\pi (1-\frac{2}{3})=\frac{2\pi}{3}$
b. With the enclosed region revolving around $x=2$, using cylinder approximations, we have
$V=\int_0^1 2\pi (2-x)(2x)dx=4\pi\int_0^1 (2x-x^2)dx=4\pi (x^2-\frac{1}{3}x^3)|_0^1=4\pi (1-\frac{1}{3})=\frac{8\pi}{3}$
