Single Variable Calculus, Chapter 6, 6.2, Section 6.2, Problem 22

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


If you rotate $\mathscr{R}_1$ about $BC$, its cross section forms a circular washer with outer radius 1 and inner radius $1 - x^3$. Thus, the cross section area can be computed by substracting the area of the out circle to the inner circle. Hence, $A_{\text{washer}} = A_{\text{outer}} - A_{\text{inner}} = \pi (1)^2 - \pi (1 - x^3)^2$. Therefore, the value is

$\begin{equation} \begin{aligned} V &= \int^1_0 \left[ \pi (1)^2 - \pi (1 - x^3)^2 \right] dx\\ \\ V &= \pi \int^1_0 \left( 1 - (1 - x^2)^2 \right) dx\\ \\ V &= \pi \int^1_0 \left( 1 - 1 + 2x^3 - x^6 \right) dx\\ \\ V &= \pi \left[ \frac{2x^4}{4} - \frac{x^7}{7} \right]\\ \\ V &= \frac{5\pi}{14} \text{ cubic units} \end{aligned} \end{equation}$