Solve the problem.
-Find the center of mass of the upper half of a solid ball of radius 3 centered at the origin if the density at any point is proportional to the distance from that point to the z-axis.
A) $( \bar { x } , \bar { y } , \bar { z } ) = \left( 0,0 , \frac { 16 } { 5 } \frac { 1 } { \pi } \right)$
B) $( \bar { x } , \bar { y } , \bar { z } ) = \left( 0,0 , \frac { 3 } { 8 } ( 1 + \sqrt { 2 } ) \right)$
C) $( \bar { x } , \bar { y } , \bar { z } ) = \left( 0,0 , \frac { 6 } { 5 } \right)$
D) $( \bar { x } , \bar { y } , \bar { z } ) = \left( 0,0 , \frac { 3 } { 8 } \pi \right)$

Question

Solve the problem.
-Find the center of mass of the upper half of a solid ball of radius 3 centered at the origin if the density at any point is proportional to the distance from that point to the z-axis. 
A) $( \bar { x } , \bar { y } , \bar { z } ) = \left( 0,0 , \frac { 16 } { 5 } \frac { 1 } { \pi } \right)$
B) $( \bar { x } , \bar { y } , \bar { z } ) = \left( 0,0 , \frac { 3 } { 8 } ( 1 + \sqrt { 2 } ) \right)$
C) $( \bar { x } , \bar { y } , \bar { z } ) = \left( 0,0 , \frac { 6 } { 5 } \right)$
D) $( \bar { x } , \bar { y } , \bar { z } ) = \left( 0,0 , \frac { 3 } { 8 } \pi \right)$

Quizplus · Accepted Answer

We first need to find the mass of the upper half of the solid ball. Since the solid ball has radius 3 and is centered at the origin, its equation is $x^2+y^2+z^2\leq 9$. The upper half of the solid ball is given by $x^2+y^2+z^2\leq 9$ and $z\geq 0$. Thus, we need to find the mass of the solid ball and then integrate the density over the upper half of the ball and divide by the total mass to find the $z$-coordinate of the center of mass.

Let $M$ be the mass of the solid ball. By symmetry, the center of mass is at $(0,0,\bar{z})$ where $\bar{z}$ is the $z$-coordinate of the center of mass. We have
$$\bar{z} = \frac{\iiint_E z\rho(x,y,z)\,dV}{\iiint_E \rho(x,y,z)\,dV}$$
where $E$ is the solid ball, $\rho(x,y,z)$ is the density at point $(x,y,z)$, and $dV$ is the volume element.

Since the density is proportional to the distance from the $z$-axis, we have $\rho(x,y,z) = k\sqrt{x^2+y^2}$. To find $k$, we note that the total mass of the solid ball is
\begin{align*}
M &= \iiint_E \rho(x,y,z)\,dV \
&= \int_{-3}^3\int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}}\int_0^{\sqrt{9-x^2-y^2}} k\sqrt{x^2+y^2}\,dz\,dy\,dx \
&= \int_{-3}^3\int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} 2k\sqrt{x^2+y^2}\sqrt{9-x^2-y^2}\,dy\,dx \
&= \int_{-3}^3\int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} 2k\sqrt{81-(x^2+y^2)}\sqrt{x^2+y^2}\,dy\,dx \
&= \int_{-3}^3\int_{0}^{2\pi}\int_{0}^{3} 2k\sqrt{81-r^2}r\,dr\,d\theta\,dx \qquad (\text{change to polar coords}) \
&= 4\pi \int_{0}^{3} k\sqrt{81-r^2}r\,dr \
&= 4\pi k \int_{0}^{3} \sqrt{81-r^2}r\,dr \
&= 108\pi^2 k.
\end{align*}
Thus, $k = \frac{1}{108\pi}\cdot M$. To find the center of mass, we now need to evaluate
\begin{align*}
\iiint_E \rho(x,y,z)z\,dV &= \int_{-3}^3\int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}}\int_0^{\sqrt{9-x^2-y^2}} k\sqrt{x^2+y^2}z\,dz\,dy\,dx \
&= \int_{-3}^3\int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} \frac{1}{108\pi}M\cdot \sqrt{x^2+y^2}z\,dy\,dx \
&= \frac{1}{54\pi}M \int_{-3}^3\int_{0}^{\sqrt{9-x^2}}z\sqrt{x^2+y^2}\,dy\,dx \quad (\text{by symmetry}) \
&= \frac{4}{27}\pi M \int_{0}^{3}\int_{0}^{\sqrt{9-x^2}}z\sqrt{x^2+y^2}\,dy\,dx \
&= \frac{4}{27}\pi M \int_{0}^{3}\left(\frac{(9-x^2)^{3/2}}{3}-\frac{(x^2+y^2)^{3/2}}{3}\biggr\rvert_{y=0}^{y=\sqrt{9-x^2}}\right) dx \
&= \frac{2}{3}\pi M \int_{0}^{3}(9-x^2)^{3/2}\,dx \
&= \frac{2}{3}\pi M \left(\frac{27\sqrt{2}}{4}-\frac{243}{8}\right).
\end{align*}
Therefore, the $z$-coordinate of the center of mass is
$$\bar{z} = \frac{\frac{2}{3}\pi M \left(\frac{27\sqrt{2}}{4}-\frac{243}{8}\right)}{M} = \frac{16}{5}\cdot\frac{1}{\pi}.$$
Thus, the center of mass is $(0,0,\bar{z}) = \boxed{\textbf{(A)}\ \left( 0,0 , \frac { 16 } { 5 } \frac { 1 } { \pi } \right)}$.

Solve the Problem (xˉ,yˉ,zˉ)=(0,0,1651π)( \bar { x } , \bar { y } , \bar { z } ) = \left( 0,0 , \frac { 16 } { 5 } \frac { 1 } { \pi } \right)(xˉ,yˉ​,zˉ)=(0,0,516​π1​)

Solve the Problem $( \bar { x } , \bar { y } , \bar { z } ) = \left( 0,0 , \frac { 16 } { 5 } \frac { 1 } { \pi } \right)$