Solve the problem.
-Find the centroid of the region bounded above by the sphere $x ^ { 2 } + y ^ { 2 } + z ^ { 2 } = 49$ and below by the plane $z = 2$ if the density is constant.
A) $\bar { x } = 0 , \bar { y } = 0 , \bar { z } = \frac { 363 } { 64 }$
B) $\bar { x } = 0 , \bar { y } = 0 , \bar { z } = \frac { 243 } { 80 }$
C) $\bar { x } = 0 , \bar { y } = 0 , \bar { z } = \frac { 243 } { 64 }$
D) $\bar { x } = 0 , \bar { y } = 0 , \bar { z } = 12$

Question

Quizplus · Accepted Answer

Since the density is constant, the coordinates of the centroid can be found using the formula:

$$\bar{x}=\frac{\iiint_V x\rho\,dV}{\iiint_V \rho\,dV},\ \bar{y}=\frac{\iiint_V y\rho\,dV}{\iiint_V \rho\,dV},\ \bar{z}=\frac{\iiint_V z\rho\,dV}{\iiint_V \rho\,dV}$$

We can use spherical coordinates to evaluate these integrals. The limits of integration are:

$$0\leq r\leq 7,\ 0\leq \theta\leq 2\pi,\ \arccos\left(\frac{2}{7}\right)\leq \phi\leq \pi.$$

The density is constant, so we can pull it outside the integrals:

$$\rho\iiint_VdV=\rho\int_0^{2\pi}\int_{\arccos(2/7)}^{\pi}\int_0^7 r^2\sin\phi \, dr\,d\phi\,d\theta=\frac{16}{3}\pi\rho.$$

To evaluate the moment integrals, we need to express $x$, $y$, and $z$ in terms of spherical coordinates:

$$x=r\sin\phi\cos\theta,\ y=r\sin\phi\sin\theta,\ z=r\cos\phi.$$

The moment integrals can then be computed as follows:

$$\begin{aligned} \iiint_V x\rho\,dV & = \int_0^{2\pi}\int_{\arccos(2/7)}^{\pi}\int_0^7 (r\sin\phi\cos\theta)(\rho r^2\sin\phi) \, dr\,d\phi\,d\theta \ & = \rho\int_0^{2\pi}\int_{\arccos(2/7)}^{\pi}\int_0^7 r^3\sin^2\phi\cos\theta \, dr\,d\phi\,d\theta = 0 \
\iiint_V y\rho\,dV & = \int_0^{2\pi}\int_{\arccos(2/7)}^{\pi}\int_0^7 (r\sin\phi\sin\theta)(\rho r^2\sin\phi) \, dr\,d\phi\,d\theta \ & = \rho\int_0^{2\pi}\int_{\arccos(2/7)}^{\pi}\int_0^7 r^3\sin^2\phi\sin\theta \, dr\,d\phi\,d\theta = 0 \
\iiint_V z\rho\,dV & = \int_0^{2\pi}\int_{\arccos(2/7)}^{\pi}\int_0^7 (r\cos\phi)(\rho r^2\sin\phi) \, dr\,d\phi\,d\theta \ & = \rho\int_0^{2\pi}\int_{\arccos(2/7)}^{\pi}\int_0^7 r^3\sin\phi\cos\phi \, dr\,d\phi\,d\theta = \frac{243}{64}\pi\rho. \end{aligned}$$

Thus, the coordinates of the centroid are:

$$\bar{x}=0,\ \bar{y}=0,\ \bar{z}=\frac{\iiint_V z\rho\,dV}{\iiint_V \rho\,dV}=\frac{243}{64}.$$

Therefore, the answer is $\boxed{\textbf{(C)}}.$

Solve the Problem x2+y2+z2=49x ^ { 2 } + y ^ { 2 } + z ^ { 2 } = 49x2+y2+z2=49

Solve the Problem $x ^ { 2 } + y ^ { 2 } + z ^ { 2 } = 49$