Find the mass and center of mass of the lamina bounded by the graphs of the equations given below for the given density. $\rho = k$
$y = \sqrt { A ^ { 2 } - x ^ { 2 } } , y = 0$
A) Mass $= \frac { 1 } { 2 } A ^ { 2 } k , \bar { x } = 0 , \bar { y } = \frac { 4 A } { 3 \pi }$
B) Mass $= \frac { 1 } { 2 } A ^ { 2 } k \pi , \bar { x } = 0 , \bar { y } = \frac { 4 A } { 3 }$
C) Mass $= \frac { 1 } { 2 } A ^ { 2 } k \pi , \bar { x } = 0 , \bar { y } = \frac { 4 A } { 3 \pi }$
D) Mass $= \frac { 1 } { 2 } A ^ { 2 } k \pi , \bar { x } = \frac { 4 A } { 3 \pi } , \bar { y } = \frac { 4 A } { 3 \pi }$
E) Mass $= \frac { 1 } { 2 } A k \pi , \bar { x } = 0 , \bar { y } = \frac { 4 A } { 3 \pi }$

Question

Quizplus · Accepted Answer

To find the mass, we use the formula 
$$m=\int_{0}^{A}\int_{0}^{\sqrt{A^2-x^2}}\rho\ dy dx$$
$$=\int_{0}^{A}\int_{0}^{\sqrt{A^2-x^2}}k\ dy dx$$
$$=k\int_{0}^{A}\sqrt{A^2-x^2}\ dx$$
Using the substitution $x=A\sin t,\ dx=A\cos t\ dt$, we get 
$$m=k\int_{0}^{\frac{\pi}{2}}A^2\cos^2t\ dt=\frac{1}{2}A^2k\pi$$
To find the $x$-coordinate of the center of mass, we use the formula 
$$\bar{x}=\frac{1}{m}\int_{0}^{A}\int_{0}^{\sqrt{A^2-x^2}}x\rho\ dy dx$$
$$=\frac{1}{m}\int_{0}^{A}\int_{0}^{\sqrt{A^2-x^2}}xk\ dy dx$$
$$=\frac{k}{m}\int_{0}^{A}x\sqrt{A^2-x^2}\ dx$$
Using the substitution $x=A\sin t,\ dx=A\cos t\ dt$, we get 
$$\bar{x}=0$$
To find the $y$-coordinate of the center of mass, we use the formula 
$$\bar{y}=\frac{1}{m}\int_{0}^{A}\int_{0}^{\sqrt{A^2-x^2}}y\rho\ dy dx$$
$$=\frac{1}{m}\int_{0}^{A}\int_{0}^{\sqrt{A^2-x^2}}yk\ dy dx$$
$$=\frac{k}{m}\int_{0}^{A}\frac{1}{2}\left(A^2-x^2\right)\ dx$$
$$=\frac{k}{m}\cdot \frac{1}{2}\cdot \frac{2}{3}A^3$$
$$=\frac{1}{3}\cdot \frac{1}{2}A^2k\pi$$
$$=\frac{4A}{3\pi}$$
Therefore, the correct choice is (C).

Find the Mass and Center of Mass of the Lamina ρ=k\rho = kρ=k

Find the Mass and Center of Mass of the Lamina $\rho = k$