Single Variable Calculus, Chapter 6, 6.2, Section 6.2, Problem 20

Find the value generated by rotating $\mathscr{R}_1$ about $OC$



If you rotate $\mathscr{R}_1$ about $OC$, its cross section form a circular washer with outer radius 1 and inner radius $\sqrt[3]{y}$. Thus, the cross sectional area can be computed by subtracting the area of the outer circle to the inner circle. Hence, $A_{\text{outer}} = \pi (1)^2$ and $A_{\text{inner}} = \pi \left( \sqrt[3]{y}\right)^2$


Therefore, the value is...

$
\begin{equation}
\begin{aligned}
V &= \int^1_0 \left[ \pi (1)^2 - \pi (\sqrt[3]{y})^2 \right] dy\\
\\
V &= \pi \left[ y - \frac{y^{\frac{5}{3}}}{\frac{5}{3}}\right]^1_0\\
\\
V &= \pi \left( \left[ 1 - \frac{(1)^{\frac{5}{3}}}{\frac{5}{3}}\right] - \left[ 0 -\frac{(0)^{\frac{5}{3}}}{\frac{5}{3}} \right] \right)\\
\\
V &= \frac{2\pi}{5} \text{ cubic units}
\end{aligned}
\end{equation}
$