Answer
$\displaystyle{V=\frac{17\pi}{45}}$
Work Step by Step
$\displaystyle{A(y)=\pi\left(1-y^4\right)^2-\pi\left(1-y\right)^2}\\ \displaystyle{A(y)=\pi\left(y^8-2y^4-y^2+2y\right)}$
$\begin{aligned} V &=\int_{0}^{1} A(y) \ d y \\ V &=\int_{0}^{1} \pi\left(y^8-2y^4-y^2+2y\right) \ d y \\ V &=\pi \int_{0}^{1} y^8-2y^4-y^2+2y\ dy \\ V &=\pi\left[\frac{1}{9} y^9-\frac{2}{5} y^5-\frac{1}{3}y^3+y^2\right]_{0}^{1} \\ V &=\pi\left(\left(\frac{1}{9} (1)^9-\frac{2}{5} (1)^5-\frac{1}{3}(1)^3+(1)^2\right)-(0)\right) \\ V &=\frac{17\pi}{45} \end{aligned}$
