Answer
$\displaystyle{V=\frac{13\pi}{45}}$
Work Step by Step
$\displaystyle{A(y)=\pi\left(1-0\right)^2-\pi\left(1-y^4\right)^2}\\
\displaystyle{A(y)=\pi\left(2y^4-y^8\right)}$
$\begin{aligned} V &=\int_{0}^{1} A(y) \ d y \\ V &=\int_{0}^{1} \pi\left(2y^4-y^8\right) \ d y \\ V &=\pi \int_{0}^{1} 2y^4-y^8\ dy \\ V &=\pi\left[\frac{2}{5} y^5-\frac{1}{9} y^9\right]_{0}^{1} \\ V &=\pi\left(\left(\frac{2}{5} (1)^5-\frac{1}{9} (1)^9\right)-(0)\right) \\ V &=\frac{13\pi}{45} \end{aligned}$
