Answer
The total mass: $M = \frac{9}{2}$.
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}$
So, the total mass: $M = \frac{9}{2}$.
