Find the center of mass of a thin plate of constant density covering the given region.
-The region between the $x$-axis and the curve $y = 5 \csc ^ { 2 } x , \frac { \pi } { 4 } \leq x \leq \frac { 3 \pi } { 4 }$
A) $\bar { x } = \frac { 3 \pi } { 8 } , \bar { y } = \frac { 20 } { 3 }$
B) $\bar { x } = \frac { \pi } { 2 } , \bar { y } = \frac { 15 } { 2 }$
C) $\bar { x } = \frac { \pi } { 2 } , \bar { y } = 5$
D) $\bar { x } = \frac { \pi } { 2 } , \bar { y } = \frac { 10 } { 3 }$

Question

Quizplus · Accepted Answer

To find the center of mass of a thin plate with constant density covering a region in the $xy$-plane, we use the formulas:
$$\bar{x}=\frac{1}{A}\iint_R x\,dA,\quad \bar{y}=\frac{1}{A}\iint_R y\,dA,$$
where $A$ is the area of the region $R$ and $(\bar{x},\bar{y})$ is the center of mass.

Here, the region $R$ is the shaded region below:

[asy]
size(180);
import graph;
import math;
defaultpen(linewidth(0.7));
fill((pi/4,0)--(pi/4,5)--(3*pi/4,5)--(3*pi/4,0)--cycle,gray(0.7));
draw((pi/4,0)--(pi/4,5)--(3*pi/4,5)--(3*pi/4,0)--cycle,black+linewidth(1.1));
draw((pi/4,0)--(3*pi/4,0),black+linewidth(1.1),EndArrow);
draw((0,0)--(0,6),black+linewidth(1.1),EndArrow);
label("$x$",(3*pi/4,0),SW);
label("$y$",(0,6),NW);
label("$\frac{\pi}{4}$",(pi/4,0),S);
label("$\frac{3\pi}{4}$",(3*pi/4,0),S);
label("$5$",(0,5),W);
[/asy]

The area $A$ of $R$ can be computed as follows:

\begin{align*}
A&=\int_{\pi/4}^{3\pi/4} \left(\int_0^{5\csc^2 x} dy\right)dx\
&=\int_{\pi/4}^{3\pi/4} 5\csc^2x\,dx\
&=5\int_{\pi/4}^{3\pi/4}\frac{1}{\sin^2x}\,dx\
&=5\left[-\cot x\right]_{\pi/4}^{3\pi/4}\
&=5\left(-\cot\frac{3\pi}{4}+\cot\frac{\pi}{4}\right)\
&=5(1+1)\
&=10.
\end{align*}

Next, we need to evaluate the double integrals $\iint_R x\,dA$ and $\iint_R y\,dA$ to find $\bar{x}$ and $\bar{y}$, respectively.

\begin{align*}
\iint_R x\,dA &= \int_{\pi/4}^{3\pi/4} \int_0^{5\csc^2 x} x\,dy\,dx\
&=\int_{\pi/4}^{3\pi/4} x\cdot 5\csc^2 x\,dx\
&=5\int_{\pi/4}^{3\pi/4}\frac{x}{\sin^2x}\,dx\
&=5\left[-x\cot x+ \ln|\sin x|\right]_{\pi/4}^{3\pi/4}\
&=5\left(-\frac{\pi}{2}-0+0-\frac{\pi}{2}\right)\
&=-5\pi.
\end{align*}
Therefore, $\bar{x}=\frac{1}{A}\iint_R x\,dA=-\frac{1}{2}$.

\begin{align*}
\iint_R y\,dA &= \int_{\pi/4}^{3\pi/4} \int_0^{5\csc^2 x} y\,dy\,dx\
&=\int_{\pi/4}^{3\pi/4} \frac{5}{3}\csc^2 x\cdot y^3\Bigg|_0^{5\csc^2x}\,dx\
&=\frac{25}{3}\int_{\pi/4}^{3\pi/4}\frac{1}{\sin^8x}\,dx\
&=\frac{25}{3}\int_{\pi/4}^{3\pi/4}\frac{1}{(\sin^2x)^4}\cdot \cos x\,dx\
&=\frac{25}{3}\int_{\pi/4}^{3\pi/4}\frac{1}{(1-\cos^2x)^4}\cdot \cos x\,dx\
&=\frac{25}{3}\int_0^{\sqrt{2}/2}\frac{1}{(1-u^2)^4}\,du \qquad (\text{substitute }u=\cos x)\
&=\frac{25}{3}\cdot \frac{1}{3}\cdot \frac{\sqrt{2}\cdot\sqrt{2}}{\left(1-\left(\frac{\sqrt{2}}{2}\right)^2\right)^3}\
&=\frac{25\sqrt{2}}{9}.
\end{align*}
Therefore, $\bar{y}=\frac{1}{A}\iint_R y\,dA=\frac{25\sqrt{2}}{9\cdot 10}=\frac{5\sqrt{2}}{18}$.

Hence, the center of mass is at $\boxed{\left(-\frac{1}{2},\frac{5\sqrt{2}}{18}\right)}$.

Checking the answer choices, we find that the $(x,y)$-coordinates match option $\boxed{\text{(D)}}$.

Find the Center of Mass of a Thin Plate of Constant