Answer
${I_0} = \frac{{81}}{5}$
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:
${\cal D} = \left\{ {\left( {x,y} \right)|0 \le x \le 3,0 \le y \le 3 - x} \right\}$
1. Evaluate the moment of inertia relative to the $x$-axis:
${I_x} = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D} {y^2}\delta \left( {x,y} \right){\rm{d}}y{\rm{d}}x = \mathop \smallint \limits_{x = 0}^3 \mathop \smallint \limits_{y = 0}^{3 - x} {y^3}{\rm{d}}y{\rm{d}}x$
$ = \frac{1}{4}\mathop \smallint \limits_{x = 0}^3 \left( {{y^4}|_0^{3 - x}} \right){\rm{d}}x$
$ = \frac{1}{4}\mathop \smallint \limits_{x = 0}^3 {\left( {3 - x} \right)^4}{\rm{d}}x$
$ = - \frac{1}{{20}}\left( {{{\left( {3 - x} \right)}^5}|_0^3} \right) = \frac{{243}}{{20}}$
2. Evaluate the moment of inertia relative to the $y$-axis:
${I_y} = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D} {x^2}\delta \left( {x,y} \right){\rm{d}}y{\rm{d}}x = \mathop \smallint \limits_{x = 0}^3 \mathop \smallint \limits_{y = 0}^{3 - x} {x^2}y{\rm{d}}y{\rm{d}}x$
$ = \frac{1}{2}\mathop \smallint \limits_{x = 0}^3 {x^2}\left( {{y^2}|_0^{3 - x}} \right){\rm{d}}x$
$ = \frac{1}{2}\mathop \smallint \limits_{x = 0}^3 {x^2}\left( {9 - 6x + {x^2}} \right){\rm{d}}x$
$ = \frac{1}{2}\mathop \smallint \limits_{x = 0}^3 \left( {9{x^2} - 6{x^3} + {x^4}} \right){\rm{d}}x$
$ = \frac{1}{2}\left( {3{x^3} - \frac{3}{2}{x^4} + \frac{1}{5}{x^5}} \right)|_0^3$
$ = \frac{1}{2}\left( {81 - \frac{{243}}{2} + \frac{{243}}{5}} \right) = \frac{{81}}{{20}}$
Since ${I_0} = {I_x} + {I_y}$, so ${I_0} = \frac{{243}}{{20}} + \frac{{81}}{{20}} = \frac{{81}}{5}$.
