Find the value generated by rotating $\mathscr{R}_2$ about $BC$

If you rotate $\mathscr{R}_2$ about $BC$, by using vertical strip, you will form a circular disk with radius $1 - \sqrt{x}$. Thus, the cross section area can be computed as $A = \pi (1 - \sqrt{x})^2$. Therefore, the value is...

$ \begin{equation} \begin{aligned} V &= \int^1_0 \pi (1-\sqrt{x})^2 dx\\ \\ V &= \int^1_0 \pi (1 - 2 \sqrt{x} + x) dx\\ \\ V &= \pi \left[ x - \frac{2x^{\frac{3}{2}}}{\frac{3}{2}} + \frac{x^2}{2} \right]^1_0\\ \\ V &= 0.5236 \text{ cubic units} \end{aligned} \end{equation} $