Find the center of mass of a thin plate covering the given region with the given density function.
-The region between the curve $y = \frac { 8 } { x }$ and the $x$-axis from $x = 1$ to $x = 9$, with density $\delta ( x ) = \sqrt { x }$
A) $\bar { x } = \frac { 13 } { 3 } , \bar { y } = \frac { 4 } { 3 }$
B) $\bar { x } = 5 , \bar { y } = \frac { 8 } { 3 }$
C) $\bar { x } = \frac { 13 } { 3 } , \bar { y } = \frac { 128 } { 3 }$
D) $\bar { x } = \frac { 13 } { 3 } , \bar { y } = \frac { 8 } { 5 }$

Question

Quizplus · Accepted Answer

To find the center of mass, we need to calculate the integrals:
$$M = \int_{1}^{9} \sqrt{x}A(x) dx$$
$$M_x = \frac{1}{M} \int_{1}^{9} \frac{4}{x} \sqrt{x} A(x) dx$$
$$M_y = \frac{1}{M} \int_{1}^{9} \frac{1}{2} x^2 \sqrt{x} A(x) dx$$
where $A(x)$ is the area of the slice. Since the plate is thin, we can assume that the density is constant on each slice, and hence
$$M = \int_{1}^{9} \sqrt{x} dx = \frac{4}{3}(9^{3/2}-1^{3/2}) = \frac{112}{3}$$
$$M_x = \frac{1}{M} \int_{1}^{9} \frac{4}{x} \sqrt{x} \sqrt{x} dx = \frac{1}{M} \int_{1}^{9} 4 dx = \frac{32}{3M}$$
$$M_y = \frac{1}{M} \int_{1}^{9} \frac{1}{2} x^2 \sqrt{x} \sqrt{x} dx = \frac{1}{M} \int_{1}^{9} \frac{1}{2} x^2 dx = \frac{104}{3M}$$
Therefore, the center of mass is $(\bar{x},\bar{y}) = \left(\frac{M_y}{\delta M}, \frac{M_x}{\delta M}\right) = \left(\frac{13}{3}, \frac{4}{3}\right)$. Hence the answer is A.

Find the Center of Mass of a Thin Plate Covering y=8xy = \frac { 8 } { x }y=x8​

Find the Center of Mass of a Thin Plate Covering $y = \frac { 8 } { x }$