Find the value generated by rotating $\mathscr{R}_3$ about $OC$

If you rotate $\mathscr{R}_3$ about $OC$, by using horizontal strip, you will form a circular washer with outer radius $\sqrt[3]{y}$ and inner radius $y^2$. Thus, the cross sectional area can be computed by substracting the area of the outer circle to the inner circle. $A_{\text{washer}} = A_{\text{outer}} - A_{\text{inner}} = \pi (\sqrt[3]{y})^2 - \pi (y^2)^2$. Therefore, the value is...

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

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