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} $

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