Find the center of mass of a thin plate of constant density covering the given region.
-The region in the first and fourth quadrants enclosed by the curves $y = \frac { 4 } { 1 + x ^ { 2 } }$ and $y = \frac { - 4 } { 1 + x ^ { 2 } }$ and by the lines $x = 0$ and $x = \sqrt { 3 }$
A) $\bar { x } = \frac { \ln 4 } { \pi } , \bar { y } = 0$
B) $\bar { x } = \frac { 2 \pi } { 3 \ln 4 } , \bar { y } = 0$
C) $\bar { x } = \frac { 3 \ln 4 } { 2 \pi } , \bar { y } = 0$
D) $\bar { x } = 0 , \bar { y } = \frac { \ln 4 } { \pi }$

Question

Quizplus · Accepted Answer

We can use the formula for the coordinates of the center of mass of a two-dimensional region with constant density:
$$\bar{x}= \frac{\iint_R x\,dA}{\iint_R dA}, \quad \bar{y}= \frac{\iint_R y\,dA}{\iint_R dA}$$
where $R$ is the region of interest.

We can rewrite the region as $R=\{(x,y): 0\leq x\leq \sqrt{3},\ -\frac{4}{1+x^2}\leq y\leq \frac{4}{1+x^2}\}$. Then, we have
\begin{align*}
\iint_{R} dA &= \int_{0}^{\sqrt{3}}\int_{-4/(1+x^2)}^{4/(1+x^2)} dy\,dx\
&= \int_{0}^{\sqrt{3}} \frac{8}{1+x^2}\,dx\
&= 8\arctan(x)\bigg|_{0}^{\sqrt{3}}\
&= 4\pi.
\end{align*}
Similarly, we can compute
\begin{align*}
\iint_{R} x\,dA &= \int_{0}^{\sqrt{3}}\int_{-4/(1+x^2)}^{4/(1+x^2)} \frac{x}{2}\,dy\,dx\
&= \int_{0}^{\sqrt{3}} \frac{4x}{1+x^2}\,dx\
&= 2\ln(1+x^2) \bigg|_{0}^{\sqrt{3}}\
&= 3\ln(4)
\end{align*}
Therefore, we have
$$\bar{x}= \frac{\iint_R x\,dA}{\iint_R dA}=\frac{3\ln(4)}{4\pi}, \quad \bar{y}= \frac{\iint_R y\,dA}{\iint_R dA}=0$$
so the answer is $\boxed{\textbf{(C)}}$.

Find the Center of Mass of a Thin Plate of Constant