Answer
$V$ = $\frac{\pi}{3}(2-\sqrt 2)$
Work Step by Step
$V$ = $\int_0^{2\pi}\int_0^{\frac{1}{\sqrt 2}}[\sqrt {1-r^2}-r]rdrdθ$
$V$ = $\int_0^{2\pi}\int_0^{\frac{1}{\sqrt 2}}[r\sqrt {1-r^2}-r^2]drdθ$
use integration by substitution
$u$ = $1-r^2$
$du$ = $-2rdr$
$-\frac{1}{2}du$ = $dr$
$V$ = $\int_0^{2\pi}[-\frac{1}{3}(1-r^2)^\frac{3}{2}-\frac{1}{3}r^3]_0^{\frac{1}{\sqrt 2}}dθ$
$V$ = $\int_0^{2\pi}[(-\frac{1}{6\sqrt 2}-\frac{1}{6\sqrt 2})-(-\frac{1}{3})]dθ$
$V$ = $\int_0^{2\pi}(-\frac{1}{3\sqrt 2}+\frac{1}{3})dθ$
$V$ = $(-\frac{1}{3\sqrt 2}+\frac{1}{3})[θ]_0^{2\pi}$
$V$ = $\frac{\pi}{3}(2-\sqrt 2)$
