Find the center of mass of a thin plate covering the given region with the given density function.
-The triangular region cut from the first quadrant by the line $y = - x + 8$, with density $\delta ( x ) = x$
A) $\bar { x } = 2 , \bar { y } = 2$
B) $\bar { x } = 2 , \bar { y } = 4$
C) $\bar { x } = 4 , \bar { y } = 2$
D) $\bar { x } = 4 , \bar { y } = 4$

Question

Quizplus · Accepted Answer

We can use the formula for the center of mass of a two-dimensional region with variable density:
$$\bar{x} = \frac{\iint_R x\delta(x,y)dA}{\iint_R \delta(x,y)dA}, \ \ \ \bar{y} = \frac{\iint_R y\delta(x,y)dA}{\iint_R \delta(x,y)dA}$$
where $R$ is the region of integration.

Since we are given that the density function is $\delta(x) = x$, we can plug that in and set up the integrals:
$$\bar{x} = \frac{\iint_R x^2 dA}{\iint_R x dA}, \ \ \ \bar{y} = \frac{\iint_R xy dA}{\iint_R x dA}$$
Now we need to find the limits of integration for $x$ and $y$. The region is the triangular region bounded by $y=0$, $x=8$, and $y=-x+8$. We can rewrite the equation of the boundary line as $x+y=8$. So the limits of integration can be expressed as:
$$0 \leq y \leq 8-x, \ \ \ 0 \leq x \leq 8$$
Plugging in the density and limits of integration, we get:
$$\bar{x} = \frac{\int_0^8 \int_0^{8-x} x^2 dy dx}{\int_0^8 \int_0^{8-x} x dy dx} = 4$$
$$\bar{y} = \frac{\int_0^8 \int_0^{8-x} xy dy dx}{\int_0^8 \int_0^{8-x} x dy dx} = 2$$
Therefore, the center of mass is located at $(4,2)$, so the answer is choice (C).

Find the Center of Mass of a Thin Plate Covering y=−x+8y = - x + 8y=−x+8

Find the Center of Mass of a Thin Plate Covering $y = - x + 8$