Solve the problem.
-Find the center of mass of a thin triangular plate bounded by the coordinate axes and the line $x + y = 9$ if $\delta ( x , y ) = x + y$
A) $\bar { x } = 3 , \bar { y } = 3$
B) $\bar { x } = 6 , \bar { y } = 6$
C) $\bar { x } = \frac { 27 } { 8 } , \bar { y } = \frac { 27 } { 8 }$
D) $\bar { x } = \frac { 15 } { 4 } , \bar { y } = \frac { 15 } { 4 }$

Question

Quizplus · Accepted Answer

To find the center of mass, we need to find the coordinates $(\bar{x},\bar{y})$, where 
$$ \bar{x} = \frac{\iint\limits_R x \delta(x,y)\, dA}{\iint\limits_R \delta(x,y)\, dA} \quad \text{and} \quad \bar{y} = \frac{\iint\limits_R y \delta(x,y)\, dA}{\iint\limits_R \delta(x,y)\, dA},$$
where $R$ is the triangular region bounded by the coordinate axes and the line $x+y=9$.

We can set up the integrals as follows:
$$\iint\limits_R \delta(x,y)\, dA = \int_{0}^{9}\int_{0}^{9-x} (x+y) \, dy\, dx = \frac{243}{2}$$
$$\iint\limits_R x\delta(x,y)\, dA = \int_{0}^{9}\int_{0}^{9-x}x(x+y) \, dy\, dx = \frac{2187}{16}$$
$$\iint\limits_R y\delta(x,y)\, dA = \int_{0}^{9}\int_{0}^{9-x}y(x+y) \, dy\, dx = \frac{2187}{16}$$
Therefore, 
$ \bar{x} = \frac{\frac{2187}{16}}{\frac{243}{2}} = \frac{27}{8}$ and $ \bar{y} = \frac{\frac{2187}{16}}{\frac{243}{2}} = \frac{27}{8}$, 
so the center of mass is at $(\bar{x},\bar{y}) = \left(\frac{27}{8},\frac{27}{8}\right)$.

Solve the Problem x+y=9x + y = 9x+y=9 If δ(x,y)=x+y\delta ( x , y ) = x + yδ(x,y)=x+y

Solve the Problem $x + y = 9$ If $\delta ( x , y ) = x + y$