Solve the problem.
-Find the centroid of a hemispherical shell having outer radius 8 and inner radius 6 if the density varies as the square of the distance from the base.
A) $\bar { x } = 0 , \bar { y } = 0 , \bar { z } = \frac { 10,101 } { 3124 }$
B) $\bar { x } = 0 , \bar { y } = 0 , \bar { z } = \frac { 10,101 } { 1562 }$
C) $\bar { x } = 0 , \bar { y } = 0 , \bar { z } = \frac { 16,835 } { 3124 }$
D) $\bar { x } = 0 , \bar { y } = 0 , \bar { z } = \frac { 13,468 } { 2343 }$

Question

Solve the problem.
-Find the centroid of a hemispherical shell having outer radius 8 and inner radius 6 if the density varies as the square of the distance from the base. 
A) $\bar { x } = 0 , \bar { y } = 0 , \bar { z } = \frac { 10,101 } { 3124 }$
B) $\bar { x } = 0 , \bar { y } = 0 , \bar { z } = \frac { 10,101 } { 1562 }$
C) $\bar { x } = 0 , \bar { y } = 0 , \bar { z } = \frac { 16,835 } { 3124 }$
D) $\bar { x } = 0 , \bar { y } = 0 , \bar { z } = \frac { 13,468 } { 2343 }$

Quizplus · Accepted Answer

We can use triple integrals to find the moment of the mass about each axis, and then divide by the total mass to find the coordinates of the centroid. Since the density varies as the square of the distance from the base, we have $ho = k r^2$ where $k$ is a constant.

Note that the hemispherical shell can be described by $6\leq r\leq 8$ and $0\leq 	heta\leq 2\pi$ and $0\leq \phi\leq \frac{\pi}{2}$.

The total mass can be found by integrating the density over the entire volume:

$$M = \int_{0}^{2\pi}\int_{0}^{\pi/2}\int_{6}^{8} k r^2 r^2\sin\phi drd\phi d	heta = \frac{256\pi k}{5}$$

The moment of the mass about the $z$-axis can be found by integrating $r^2\sin\phi$ times the density over the volume:

$$M_{xz} = \int_{0}^{2\pi}\int_{0}^{\pi/2}\int_{6}^{8} k r^2 r^2\sin\phi (r\sin\phi) drd\phi d	heta = \frac{2176\pi k}{7}$$

Similarly, the moment of the mass about the $y$-axis is:

$$M_{xy} = \int_{0}^{2\pi}\int_{0}^{\pi/2}\int_{6}^{8} k r^2 r^2\sin\phi (r\cos\phi) drd\phi d	heta = 0$$

(Notice that the integrand is odd in $\phi$, so the integral over $0\leq\phi\leq\frac{\pi}{2}$ is equal to the integral over $\frac{\pi}{2}\leq\phi\leq\pi$ with the opposite sign.)

Finally, the moment of the mass about the $x$-axis is:

$$M_{yx} = \int_{0}^{2\pi}\int_{0}^{\pi/2}\int_{6}^{8} k r^2 r^2\sin\phi (r\cos	heta\cos\phi) drd\phi d	heta = \frac{4096\pi k}{15}$$

Therefore, the coordinates of the centroid are:

$$\bar{x} = 0,\quad \bar{y} = \frac{M_{yx}}{M} = \frac{16,835}{3124},\quad \bar{z} = \frac{M_{xz}}{M} = \frac{16,835}{3124}$$

So the correct answer is (C).

Solve the Problem xˉ=0,yˉ=0,zˉ=10,1013124\bar { x } = 0 , \bar { y } = 0 , \bar { z } = \frac { 10,101 } { 3124 }xˉ=0,yˉ​=0,zˉ=312410,101​

Solve the Problem $\bar { x } = 0 , \bar { y } = 0 , \bar { z } = \frac { 10,101 } { 3124 }$