Find the center of mass of a thin plate of constant density covering the given region.
-The region bounded by the parabola $y = 16 - x ^ { 2 }$ and the $x$-axis
A) $\bar { x } = 0 , \bar { y } = \frac { 5 } { 32 }$
B) $\bar { x } = \frac { 32 } { 5 } , \bar { y } = 0$
C) $\bar { x } = 0 , \bar { y } = \frac { 32 } { 5 }$
D) $\bar { x } = 0 , \bar { y } = \frac { 8192 } { 15 }$

Question

Quizplus · Accepted Answer

We use the formula for the center of mass of a two-dimensional object:
$$\bar{x} = \frac{\iint_R x dA}{\iint_R dA}$$
$$\bar{y} = \frac{\iint_R y dA}{\iint_R dA}$$
where $R$ is the region of interest.
The density is constant, so we can pull it out of the integral:
$$\iint_R dA = \iint_R \rho dA = M$$
where $M$ is the mass of the plate.
We can compute the integral in the numerator by using a double integral:
$$\iint_R x dA = \int_{-4}^4 \int_0^{16-x^2} x dy dx$$
$$= \int_{-4}^4 x(16-x^2) dx = 0$$
where we used the fact that the integrand is odd and the limits of integration are symmetric about the $y$-axis.
Similarly, we can compute the integral for $\bar{y}$:
$$\iint_R y dA = \int_{-4}^4 \int_0^{16-x^2} y dy dx$$
$$= \int_{-4}^4 \frac{1}{2}(16-x^2)^2 dx = \frac{2048}{15}$$
Putting it all together, we get:
$$\bar{x} = 0$$
$$\bar{y} = \frac{\frac{2048}{15}}{M} = \frac{32}{5}$$
Therefore, the answer is $\boxed{\textbf{(C)}}$.

Find the Center of Mass of a Thin Plate of Constant