Answer
$\displaystyle{V=\frac{\pi}{15}}$
Work Step by Step
$\displaystyle{A(x)=\pi\left(1-x^{\frac{1}{4}}\right)^2}\\
\displaystyle{A(x)=\pi\left(1-2x^{\frac{1}{4}}+x^{\frac{1}{2}}\right)}$
$\begin{aligned} V &=\int_{0}^{1} A(x) \ d x \\ V &=\int_{0}^{1} \pi\left(1-2x^{\frac{1}{4}}+x^{\frac{1}{2}}\right) \ d x \\ V &=\pi \int_{0}^{1} 1-2x^{\frac{1}{4}}+x^{\frac{1}{2}}\ dx \\ V &=\pi\left[x-\frac{8}{5} x^{\frac{5}{4}}+\frac{2}{3} x^{\frac{3}{2}}\right]_{0}^{1} \\ V &=\pi\left(\left(1-\frac{8}{5} (1)^{\frac{5}{4}}+\frac{2}{3} (1)^{\frac{3}{2}}\right)-(0)\right) \\ V &=\frac{\pi}{15} \end{aligned}$
