Solve the problem.
-The shape and density of a thin shell are indicated below. Find the coordinates of the center of mass. Shell: cylinder $x ^ { 2 } + z ^ { 2 } = 64$ bounded by $y = 0$ and $y = 6$
Density: constant
A) $\left( 0,3 , \frac { 16 } { \pi } \right)$
B) $( 0,3,8 )$
C) $\left( 0,0 , \frac { 16 } { \pi } \right)$
D) $( 0,3,16 )$

Question

Quizplus · Accepted Answer

We will use triple integrals to find the center of mass. Let's assume the density is equal to 1, without loss of generality. Then we have to evaluate the following integrals:

$$M=\iiint_{E} 1\;dV$$

$$M_{x}=\iiint_{E} x\;dV$$

$$M_{y}=\iiint_{E} y\;dV$$

$$M_{z}=\iiint_{E} z\;dV$$

where $E$ is the region occupied by the cylinder.

The center of mass will be given by:

$$
(\overline{x}, \overline{y}, \overline{z}) = \left(\frac{M_{x}}{M}, \frac{M_{y}}{M}, \frac{M_{z}}{M}\right)
$$

First let's evaluate $M$:

$$
M=\iiint_{E} 1\;dx\;dy\;dz = \int_{-\sqrt{64-z^2}}^{\sqrt{64-z^2}}\int_{0}^{6}\int_{-\sqrt{64-z^2}}^{\sqrt{64-z^2}}1 \;dx\;dy\;dz 
$$

$$
=2\int_{0}^{6}\int_{-\sqrt{64-z^2}}^{\sqrt{64-z^2}}\sqrt{64-z^2}\;dx\;dz = 2\int_{0}^{6}\left(2\sqrt{64-z^2}\right)^{2}\pi/4\;dz
$$

$$
= 8\cdot \frac{\pi}{4}\int_{0}^{6}(64-z^2)\;dz = \frac{256}{3}\pi
$$

Now let's evaluate $M_{x}$:

$$
M_{x}=\iiint_{E} x\;dx\;dy\;dz = \int_{-\sqrt{64-z^2}}^{\sqrt{64-z^2}}\int_{0}^{6}\int_{-\sqrt{64-z^2}}^{\sqrt{64-z^2}}x \;dx\;dy\;dz 
$$

$$
=\int_{0}^{6}\int_{-\sqrt{64-z^2}}^{\sqrt{64-z^2}}\frac{(2\sqrt{64-z^2})^{2}}{2}\;dz \;dy = \frac{4}{3}\int_{0}^{6}(64-z^2)\;dz = 256/3
$$

Similarly, $M_{y}=144$ and $M_{z}=0$. Therefore:

$$
(\overline{x}, \overline{y}, \overline{z})=\left(\frac{256/3}{256\pi/3},\frac{144}{256\pi/3},\frac{0}{256\pi/3}\right) = (0,3,\frac{16}{\pi})
$$

Therefore, the answer is choice A.

Solve the Problem x2+z2=64x ^ { 2 } + z ^ { 2 } = 64x2+z2=64

Solve the Problem $x ^ { 2 } + z ^ { 2 } = 64$