Answer
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} xyz{\rm{d}}V = \frac{7}{{72}}$
Work Step by Step
We have the region defined by ${\cal W} = \left\{ {0 \le x \le 1,x \le y \le 1,x \le z \le x + y} \right\}$.
Evaluate:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} xyz{\rm{d}}V = \mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{y = x}^1 \mathop \smallint \limits_{z = x}^{x + y} xyz{\rm{d}}z{\rm{d}}y{\rm{d}}x$
$ = \frac{1}{2}\mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{y = x}^1 xy\left( {{z^2}|_x^{x + y}} \right){\rm{d}}y{\rm{d}}x$
$ = \frac{1}{2}\mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{y = x}^1 xy\left( {2xy + {y^2}} \right){\rm{d}}y{\rm{d}}x$
$ = \frac{1}{2}\mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{y = x}^1 \left( {2{x^2}{y^2} + x{y^3}} \right){\rm{d}}y{\rm{d}}x$
$ = \frac{1}{2}\mathop \smallint \limits_{x = 0}^1 \left( {\left( {\frac{2}{3}{x^2}{y^3} + \frac{1}{4}x{y^4}} \right)|_x^1} \right){\rm{d}}x$
$ = \frac{1}{2}\mathop \smallint \limits_{x = 0}^1 \left( {\frac{2}{3}{x^2} + \frac{1}{4}x - \frac{2}{3}{x^5} - \frac{1}{4}{x^5}} \right){\rm{d}}x$
$ = \frac{1}{2}\mathop \smallint \limits_{x = 0}^1 \left( {\frac{2}{3}{x^2} + \frac{1}{4}x - \frac{{11}}{{12}}{x^5}} \right){\rm{d}}x$
$ = \frac{1}{2}\left( {\frac{2}{9}{x^3} + \frac{1}{8}{x^2} - \frac{{11}}{{72}}{x^6}} \right)|_0^1$
$ = \frac{1}{2}\left( {\frac{2}{9} + \frac{1}{8} - \frac{{11}}{{72}}} \right) = \frac{7}{{72}}$
So, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} xyz{\rm{d}}V = \frac{7}{{72}}$.
