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