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

## Expert Answers 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 $\sqrt{y}$. 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 $A_{\text{inner}} = \pi \left( \sqrt{y}\right)^2$

Therefore, the value is...

\begin{equation} \begin{aligned} V &= \int^1_0 \left[ \pi (1)^2 - \pi (\sqrt{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}

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