Answer
The total mass is $488$ g.
Work Step by Step
We have a mass density of $\delta \left( {x,y,z} \right) = 6y$ and a solid region ${\cal W}$ defined by $x \ge 0$, $y \ge 0$, ${x^2} + {y^2} \le 4$, and $x \le z \le 32 - x$.
We can consider ${\cal W}$ as a $z$-simple region whose projection onto the $xy$-plane is ${\cal D}$, a quarter of the disk of radius $2$, ${x^2} + {y^2} \le 4$, located in the first quadrant. Notice that ${\cal D}$ is both vertically and horizontally simple region. We choose to describe ${\cal D}$ as a vertically simple region such that
${\cal D} = \left\{ {\left( {x,y} \right)|0 \le x \le 2,0 \le y \le \sqrt {4 - {x^2}} } \right\}$
Thus, the description of ${\cal W}$:
${\cal W} = \left\{ {\left( {x,y,z} \right)|0 \le x \le 2,0 \le y \le \sqrt {4 - {x^2}} ,x \le z \le 32 - x} \right\}$ }
Using Eq. (1), the total mass is given by
${\rm{total{\ }mass}} = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} \delta \left( {x,y,z} \right){\rm{d}}V$
$ = 6\mathop \smallint \limits_{x = 0}^2 \mathop \smallint \limits_{y = 0}^{\sqrt {4 - {x^2}} } \mathop \smallint \limits_{z = x}^{32 - x} y{\rm{d}}z{\rm{d}}y{\rm{d}}x$
$ = 6\mathop \smallint \limits_{x = 0}^2 \mathop \smallint \limits_{y = 0}^{\sqrt {4 - {x^2}} } y\left( {z|_x^{32 - x}} \right){\rm{d}}y{\rm{d}}x$
$ = 6\mathop \smallint \limits_{x = 0}^2 \mathop \smallint \limits_{y = 0}^{\sqrt {4 - {x^2}} } \left( {32 - 2x} \right)y{\rm{d}}y{\rm{d}}x$
$ = 3\mathop \smallint \limits_{x = 0}^2 \left( {32 - 2x} \right)\left( {{y^2}|_0^{\sqrt {4 - {x^2}} }} \right){\rm{d}}x$
$ = 3\mathop \smallint \limits_{x = 0}^2 \left( {32 - 2x} \right)\left( {4 - {x^2}} \right){\rm{d}}x$
$ = 3\mathop \smallint \limits_{x = 0}^2 \left( {2{x^3} - 32{x^2} - 8x + 128} \right){\rm{d}}x$
$ = 3\left( {\left( {\frac{1}{2}{x^4} - \frac{{32}}{3}{x^3} - 4{x^2} + 128x} \right)|_0^2} \right)$
$ = 3\left( {8 - \frac{{256}}{3} - 16 + 256} \right) = 488$
So, the total mass is $488$ g.
