Use polar coordinates to find the volume of the solid inside the cylinder and the ellipsoid .

Question

Use polar coordinates to find the volume of the solid inside the cylinder  and the ellipsoid  .

Quizplus · Accepted Answer

The given cylinder has equation $r=11$, $0\leq 	heta \leq 2\pi$, and $0\leq z\leq 11$. The ellipsoid is given by 
$$
\frac{x^2}{11^2}+\frac{y^2}{3^2}+\frac{z^2}{7^2}\leq 1\implies r^2\cos^2	heta+\frac{r^2\sin^2	heta\sin^2\phi}{9}+\frac{r^2\sin^2	heta\cos^2\phi}{49}\leq 1.
$$
Solving for $r$, we get 
$$
r=\frac{1}{\sqrt{\frac{\cos^2	heta}{11^2}+\frac{\sin^2	heta\sin^2\phi}{9}+\frac{\sin^2	heta\cos^2\phi}{49}}}.
$$
The solid of intersection is given by $0\leq r\leq 11$, $0\leq 	heta \leq 2\pi,$ and 
$$
0\leq z\leq \frac{1}{\sqrt{\frac{\cos^2	heta}{11^2}+\frac{\sin^2	heta\sin^2\phi}{9}+\frac{\sin^2	heta\cos^2\phi}{49}}}\sqrt{49-\frac{49\cos^2	heta}{11^2}-\frac{49\sin^2	heta\sin^2\phi}{9}}.
$$
Thus, the volume is 
\begin{align*}
&\iiint_{	ext{Solid}} 1 dV= \int_{0}^{2\pi}\int_{0}^{\frac{\pi}{2}} \int_{0}^{\frac{1}{\sqrt{\frac{\cos^2	heta}{11^2}+\frac{\sin^2	heta\sin^2\phi}{9}+\frac{\sin^2	heta\cos^2\phi}{49}}}\sqrt{49-\frac{49\cos^2	heta}{11^2}-\frac{49\sin^2	heta\sin^2\phi}{9}}} r^2 \sin\phi dr d\phi d	heta\
&= \int_{0}^{2\pi}\int_{0}^{\frac{\pi}{2}} \int_{0}^{11} r^2 \sin\phi\left(\frac{1}{\sqrt{\frac{\cos^2	heta}{11^2}+\frac{\sin^2	heta\sin^2\phi}{9}+\frac{\sin^2	heta\cos^2\phi}{49}}}\sqrt{49-\frac{49\cos^2	heta}{11^2}-\frac{49\sin^2	heta\sin^2\phi}{9}}}ight)dr d\phi d	heta\
&= \frac{2}{3}\int_{0}^{2\pi}\int_{0}^{\frac{\pi}{2}} \left(49\cos	heta\sin^2\phi-11^2\sin\phi\sqrt{49-\frac{49\cos^2	heta}{11^2}-\frac{49\sin^2	heta\sin^2\phi}{9}}ight)d\phi d	heta= \frac{308\pi}{3}.
\end{align*}

Use Polar Coordinates to Find the Volume of the Solid