Answer
$\frac{\pi}{6}$
Work Step by Step
$$ V = \pi \int_{0}^{1}{({1}-{\sqrt{y}})}^2dy$$
$$ V = \pi \int_{0}^{1}[1-2\sqrt{y}+y]dy$$
$$ V = \pi [y-\frac{4}{3}y^{\frac{3}{2}}+\frac{1}{2}y^{2}]_{0}^{1}dx$$
$$ V = \pi[1-\frac{4}{3}+\frac{1}{2}]=\frac{\pi}{6}$$
