Answer
a. $\frac{\pi}{3}h^2b$
b. $\frac{\pi}{3}hb^2$
Work Step by Step
a. Draw a diagram as shown in the figure and identify the line equation also shown. With the enclosed region revolving around the x-axis, using disk approximation, we have
$V=\int_0^b \pi (h(1-\frac{x}{b}))^2 dx=\pi h^2\int_0^b (1-\frac{2x}{b}+\frac{x^2}{b^2}) dx=\pi h^2(x-\frac{x^2}{b}+\frac{x^3}{3b^2})|_0^b =\frac{\pi}{3}h^2b$
b. With the enclosed region revolving around the y-axis, using cylinder approximation, we have
$V=\int_0^b 2\pi x(h(1-\frac{x}{b}))dx= 2\pi h\int_0^b (x-\frac{x^2}{b})dx= 2\pi h(\frac{1}{2}x^2-\frac{1}{3b}x^3)|_0^b= 2\pi h(\frac{1}{2}b^2-\frac{1}{3b}b^3)=\frac{\pi}{3}hb^2$
