Answer
The center of mass: $\left( {{x_{CM}},{y_{CM}}} \right) = \left( {\frac{3}{4},\frac{3}{2}} \right)$.
Work Step by Step
We have the triangular domain ${\cal D}$ bounded by the coordinate axes and the line $y=3-x$, with mass density $\delta \left( {x,y} \right) = y$.
From the figure attached, we see that ${\cal D}$ can be considered as a vertically simple region with the description:
${\cal D} = \left\{ {\left( {x,y} \right)|0 \le x \le 3,0 \le y \le 3 - x} \right\}$
Using Eq. (1), we evaluate the total mass:
$M = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \delta \left( {x,y} \right){\rm{d}}A = \mathop \smallint \limits_{x = 0}^3 \mathop \smallint \limits_{y = 0}^{3 - x} y{\rm{d}}y{\rm{d}}x$
$ = \frac{1}{2}\mathop \smallint \limits_{x = 0}^3 \left( {{y^2}|_0^{3 - x}} \right){\rm{d}}x$
$ = \frac{1}{2}\mathop \smallint \limits_{x = 0}^3 \left( {9 - 6x + {x^2}} \right){\rm{d}}x$
$ = \frac{1}{2}\left( {9x - 3{x^2} + \frac{1}{3}{x^3}} \right)|_0^3$
$ = \frac{1}{2}\left( {27 - 27 + 9} \right) = \frac{9}{2}$
1. Evaluate the $x$-coordinate of the center of mass:
${x_{CM}} = \frac{{{M_y}}}{M} = \frac{1}{M}\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} x\delta \left( {x,y} \right){\rm{d}}A = \frac{2}{9}\mathop \smallint \limits_{x = 0}^3 \mathop \smallint \limits_{y = 0}^{3 - x} xy{\rm{d}}y{\rm{d}}x$
$ = \frac{1}{9}\mathop \smallint \limits_{x = 0}^3 x\left( {{y^2}|_0^{3 - x}} \right){\rm{d}}x$
$ = \frac{1}{9}\mathop \smallint \limits_{x = 0}^3 \left( {9x - 6{x^2} + {x^3}} \right){\rm{d}}x$
$ = \frac{1}{9}\left( {\frac{9}{2}{x^2} - 2{x^3} + \frac{1}{4}{x^4}} \right)|_0^3$
$ = \frac{1}{9}\left( {\frac{{81}}{2} - 54 + \frac{{81}}{4}} \right) = \frac{3}{4}$
2. Evaluate the $y$-coordinate of the center of mass:
${y_{CM}} = \frac{{{M_x}}}{M} = \frac{1}{M}\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} y\delta \left( {x,y} \right){\rm{d}}A = \frac{2}{9}\mathop \smallint \limits_{x = 0}^3 \mathop \smallint \limits_{y = 0}^{3 - x} {y^2}{\rm{d}}y{\rm{d}}x$
$ = \frac{2}{{27}}\mathop \smallint \limits_{x = 0}^3 \left( {{y^3}|_0^{3 - x}} \right){\rm{d}}x$
$ = \frac{2}{{27}}\mathop \smallint \limits_{x = 0}^3 {\left( {3 - x} \right)^3}{\rm{d}}x$
$ = - \frac{1}{{54}}\left( {{{\left( {3 - x} \right)}^4}|_0^3} \right) = - \frac{1}{{54}}\left( { - 81} \right) = \frac{3}{2}$
Thus, the center of mass: $\left( {{x_{CM}},{y_{CM}}} \right) = \left( {\frac{3}{4},\frac{3}{2}} \right)$.
