Single Variable Calculus, Chapter 6, 6.2, Section 6.2, Problem 30

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



If you rotate $\mathscr{R}_3$ about $BC$, by using vertical strip, you will form a circular washer with radius $1 - x^3$ and their radius $1 - \sqrt{x}$. Thus, the cross sectional area can be computed by subtracting the area of the outer circle to the inner circle. $A_{\text{washer}} = A_{\text{outer}} - A_{\text{inner}} = \pi ( 1 - x^3)^2 - \pi (1-\sqrt{x}^2)$
Therefore, the value is

$
\begin{equation}
\begin{aligned}
V &= \int^1_0 \left[\pi(1-x^3)^2 - \pi (1- \sqrt{x})^2 \right]dx\\
\\
V &= \pi \int^1_0 \left( 1 - 2x^3 + x^6 - 1 + 2\sqrt{x} - x \right) dx\\
\\
V &= \pi \left[ x - \frac{2x^4}{4} + \frac{x^7}{7} - x + \frac{2x^{\frac{3}{2}}}{\frac{3}{2}} - \frac{x^2}{2} \right]^1_0\\
\\
V &= \pi \left[ x - \frac{2x^4}{4} + \frac{x^7}{7} - x + \frac{2x^{\frac{3}{2}}}{\frac{3}{2}} - \frac{x^2}{2} \right]^1_0\\
\\
V &= \frac{10\pi}{21} \text{ cubic units}
\end{aligned}
\end{equation}
$