Answer
a) $\frac{2\pi}{15}$
b) $\frac{\pi}{6}$
c) $\frac{8\pi}{15}$
Work Step by Step
a) A cross section is a washer with inner radius $x^{2}$ and outer radius $x$
$V$ = $\int_0^1{\pi}[(x)^{2}-(x^{2})^{2}]dx$
$V$ = $\int_0^1{\pi}(x^{2}-x^{4})dx$
$V$ = $\pi[\frac{1}{3}x^{3}-\frac{1}{5}x^{5}]_0^1$
$V$ = $\pi\left[\frac{1}{3}-\frac{1}{5}\right]$
$V$ = $\frac{2\pi}{15}$
b) A cross section is a washer with inner radius $y$ and outer radius $\sqrt y$
$V$ = $\int_0^1{\pi}[(\sqrt y)^{2}-y^{2}]dy$
$V$ = $\int_0^1{\pi}(y-y^{2})dy$
$V$ = $\pi[\frac{1}{2}y^{2}-\frac{1}{3}y^{3}]_0^1$
$V$ = $\pi[\frac{1}{2}-\frac{1}{3}]$
$V$ = $\frac{\pi}{6}$
c) A cross section is a washer with inner radius $2-x$ and outer radius $2-x^{2}$
$V$ = $\int_0^1{\pi}[(2-x^{2})^{2}-(2-x)^{2}]dx$
$V$ = $\int_0^1{\pi}(x^{4}-5x^{2}+4x)dx$
$V$ = $\pi[\frac{1}{5}x^{5}-\frac{5}{3}x^{3}+2x^{2}]_0^1$
$V$ = $\pi[\frac{1}{5}-\frac{5}{3}+2]$
$V$ = $\frac{8\pi}{15}$
