Question: Let R be the region bounded by \(y={{x}^{2}}\) and y = 4. Compute the volume of the solid formed by revolving R about the given line.

(a) y = 4 (b) the y-axis (c) y = 6
\[\begin{aligned} & A=\pi {{r}^{2}} \\ & V=\pi {{r}^{2}}dx \\ & r=4-{{x}^{2}} \\ & V=\int_{-2}^{2}{\pi (4-{{x}^{2}}}{{)}^{2}}dx \\ & =2\pi \int_{0}^{2}{(16-8{{x}^{2}}+{{x}^{4}})dx} \\ & =2\pi \left[ 16x-\frac{8}{3}{{x}^{3}}+\frac{{{x}^{5}}}{5} \right]_{0}^{2} \\ & =2\pi \left[ 16(2)-\frac{8}{3}{{(2)}^{3}}+\frac{{{2}^{5}}}{5} \right]_{-2}^{2} \\ & =2\pi (32-\frac{64}{3}+\frac{32}{5}) \\ & =2\pi (\frac{256}{15}) \\ & =\frac{512\pi }{15} \\ \end{aligned}\]

(d) y = –2 (e) x = 2 (f) x = –4