Find the centroid of the solid region bounded by the graphs of the equations. Use a computer algebra system to evaluate the triple integral. (Assume uniform density and find the center of mass.) $$z = \frac { 9 } { y ^ { 2 } + 1 } , z = 0 , x = - 2 , x = 2 , y = 0 , y = 1$$
A) $\bar { x } = 2 , \bar { y } = \frac { 2 \ln ( 2 ) } { \pi } , \bar { z } = \frac { 2 + \pi } { 4 \pi }$
B) $\bar { x } = 0 , \bar { y } = \frac { 2 \ln ( 2 ) } { \pi } , \bar { z } = \frac { 9 ( 2 + \pi ) } { 4 \pi }$
C) $\bar { x } = 0 , \bar { y } = \frac { \ln ( 2 ) } { \pi } , \bar { z } = \frac { 9 ( 2 + \pi ) } { 4 \pi }$
D) $\bar { x } = 0 , \bar { y } = \frac { 2 \ln ( 2 ) } { \pi } , \bar { z } = \frac { 9 ( 4 + \pi ) } { 4 \pi }$
E)$\bar { x } = 2 , \bar { y } = \frac { ( 2 + \pi ) 9 } { 4 \pi } , \bar { z } = \frac { 2 \ln ( 2 ) } { \pi }$

Question

Quizplus · Accepted Answer

To find the centroid, we need to evaluate the triple integral $$\iiint_V f(x,y,z) \, dV$$ where $f(x,y,z) = 1$ (due to uniform density) and $V$ is the region bounded by the given surfaces.

Converting to cylindrical coordinates, we have $x = r \cos \theta$ and $y = r \sin \theta$, and the region becomes

$$0 \le r \le 1,\quad \frac {9}{r^2 + 1} \le z \le 0,\quad 0 \le \theta \le 2 \pi$$

The triple integral becomes

$$\begin{aligned} \iiint_V f(x,y,z) \, dV &= \int_0^{2\pi} \int_0^1 \int_0^{(9/(r^2+1))} r \, dz \, dr \, d\theta\ &= \int_0^{2\pi} \int_0^1 \left[\frac{9r}{r^2+1}\right]_0^{(r^2+1)^{-1/2}} \, dr \, d\theta\ &= \int_0^{2\pi} \left[\frac92 \ln(r^2+1)\right]_0^1 \, d\theta\ &= 9\pi\ln 2 \end{aligned}$$

The coordinates of the centroid are given by

$$\bar{x} = \frac{1}{V} \iiint_V x \, dV, \quad \bar{y} = \frac{1}{V} \iiint_V y \, dV, \quad \bar{z} = \frac{1}{V} \iiint_V z \, dV$$

where $V$ is the volume of the region. From symmetry considerations, we can see that $\bar{x} = 0$. Using cylindrical coordinates again, we have

$$\begin{aligned} \bar{y} &= \frac{1}{V} \iiint_V y \, dV\ &= \frac{1}{9\pi\ln 2} \int_0^{2\pi} \int_0^1 \int_{\frac {9}{r^2+1}}^0 r\sin\theta \, dz \, dr \, d\theta\ &= \frac{1}{18\pi\ln 2} \int_0^{2\pi} \int_0^1 \frac{9r\sin\theta}{r^2+1} \, dr \, d\theta\ &= \frac{1}{\pi} \int_0^1 \frac{r}{r^2+1} \, dr\ &= \frac{1}{\pi} \left[\frac{1}{2}\ln(r^2+1)\right]_0^1\ &= \frac{\ln 2}{\pi} \end{aligned}$$

And finally,

$$\begin{aligned} \bar{z} &= \frac{1}{V} \iiint_V z \, dV\ &= \frac{1}{9\pi\ln 2} \int_0^{2\pi} \int_0^1 \int_{\frac {9}{r^2+1}}^0 zr \, dz \, dr \, d\theta\ &= \frac{1}{2V} \int_0^{2\pi} \int_0^1 \frac{9r^2}{r^2+1} \, dr \, d\theta\ &= \frac{1}{V} \int_0^1 \frac{9r^2}{r^2+1} \, dr\ &= \frac{9}{2V} \int_0^1 \left(1 - \frac{1}{r^2+1}\right) \, dr\ &= \frac{9}{2V} \left[\frac{3}{2}\ln(r^2+1) - r\right]_0^1\ &= \frac{9}{4\pi\ln 2} \left(\frac{3}{2}\ln 2 - 1\right)\ &= \frac{9(2+\pi)}{4\pi} \end{aligned}$$

Therefore, the centroid is at $\left(0,\frac{\ln 2}{\pi},\frac{9(2+\pi)}{4\pi}\right)$, which is closest to answer choice $\boxed{\textbf{(B)}}$.

Find the Centroid of the Solid Region Bounded by the Graphs