Use a spherical coordinate integral to find the volume of the given solid.
-the solid between the spheres $\varrho = \cos \varphi$ and $\varrho = 6$
A) $\frac { 863 } { 6 } \pi$
B) $288 \pi$
C) $\frac { 1727 } { 6 } \pi$
D) $144 \pi$

Question

Quizplus · Accepted Answer

The region of integration is bounded above by $ho=6$ and below by $ho=\cos\phi$, where $0\leq\phi\leq\frac{\pi}{2}$ by symmetry. The region is also symmetric about the $xy$-plane, and is a full sphere after we reflect all points through the $xy$-plane. We will use spherical coordinates.
\begin{align*}V&=\int_{0}^{2\pi}\int_{0}^{\frac{\pi}{2}}\int_{\cos\phi}^{6}ho^2\sin\phi\,\mathrm{d}ho\,\mathrm{d}\phi\,\mathrm{d}	heta \ &=\int_{0}^{2\pi}\int_{0}^{\frac{\pi}{2}}\frac{1}{3}\left(6^3-\cos^3\phiight)\sin\phi\,\mathrm{d}\phi\,\mathrm{d}	heta \ &=\int_{0}^{2\pi}\left[54-\frac{1}{3}\int_{0}^{\frac{\pi}{2}}\cos^3\phi\,\mathrm{d}\phiight]\mathrm{d}	heta \ &= \int_{0}^{2\pi}\left[54-\frac{1}{3}\left(\sin\phi-\frac{1}{3}\sin^3\phiight)\Big|_{0}^{\frac{\pi}{2}}ight]\mathrm{d}	heta\ &=\boxed{\frac{1727}{6}\pi}\end{align*}
See the page on the volume element for more information on how this all works.

Use a Spherical Coordinate Integral to Find the Volume of the Given