Use spherical coordinates to find the z coordinate of the center of mass of the solid lying between two concentric hemispheres of radii 4 and 7, and having uniform density k.
A) $\bar { z } = \frac { \pi } { 715 }$
B) $\bar { z } = \frac { 715 } { 248 } \pi$
C) $\bar { z } = 248$
D) $\bar { z } = \frac { 429 } { 149 }$
E) $\bar { z } = \frac { 715 } { 248 }$

Question

Use spherical coordinates to find the z coordinate of the center of mass of the solid lying between two concentric hemispheres of radii 4 and 7, and having uniform density k. 
A) $\bar { z } = \frac { \pi } { 715 }$
B) $\bar { z } = \frac { 715 } { 248 } \pi$
C) $\bar { z } = 248$
D) $\bar { z } = \frac { 429 } { 149 }$
E) $\bar { z } = \frac { 715 } { 248 }$

Quizplus · Accepted Answer

Since the solid is symmetric with respect to the $xy$-plane, we know the $z$-coordinate of the center of mass must lie on the $z$-axis. So we only need to find the distance from the center of mass to the $xy$-plane, which is the same as the average value of $z$ over the solid.

Let's set up the integral for the $z$-coordinate of the center of mass using spherical coordinates. The solid is defined by the constraints $4\leq ho \leq 7$ and $\phi \in [0,\frac{\pi}{2}]$, and the density is constant, so the triple integral we want is of the form

$$\iiint_D kho z \,dV$$

where $D$ is the solid region, which we can write as

$$D=\left\{(ho,	heta,\phi)\,:\,4\leq ho \leq 7,\, 0\leq	heta\leq 2\pi,\, 0\leq\phi\leq\frac{\pi}{2}ight\}$$

Using spherical coordinates, $dV=ho^2\sin\phi\,dho\,d	heta\,d\phi$. So we have

$$\begin{aligned} \iiint_D kho z \,dV &= \int_0^{\frac{\pi}{2}}\int_0^{2\pi}\int_4^7 kho^3\cos\phi\sin\phi\,dho\,d	heta\,d\phi \ &= 2\pi k\int_4^7 ho^3 \,dho\int_0^{\frac{\pi}{2}}\sin\phi\cos\phi \,d\phi \ &= 2\pi k\left[\frac{1}{4}ho^4ight]_4^7 \left[-\frac{1}{2}\cos^2\phiight]_0^{\frac{\pi}{2}} \ &= \frac{189\pi k}{2}=-\frac{945}{4}k\pi \end{aligned}$$

The minus sign in front of our answer just means that the center of mass lies below the $xy$-plane. Finally, we divide by the total mass $M$ of the solid to get the $z$-coordinate of the center of mass:

$$\bar{z}=-\frac{945/4}{(4/3)\pi(7^3-4^3)k}=\frac{315}{32(7^3-4^3)}=\boxed{\frac{715}{248}}$$

Use Spherical Coordinates to Find the Z Coordinate of the Center