Answer
The domain of integration is described by:
${\cal W} = \left\{ {\left( {x,y.z} \right)|0 \le x \le 3,0 \le y \le \sqrt {9 - {x^2}} ,0 \le z \le \sqrt {9 - {x^2} - {y^2}} } \right\}$
The projection of ${\cal W}$ onto the $xy$-plane is the domain ${\cal D}$:
${\cal D} = \left\{ {\left( {x,y} \right)|0 \le x \le 3,0 \le y \le \sqrt {9 - {x^2}} } \right\}$
$\mathop \smallint \limits_{x = 0}^3 \mathop \smallint \limits_{y = 0}^{\sqrt {9 - {x^2}} } \mathop \smallint \limits_{z = 0}^{\sqrt {9 - {x^2} - {y^2}} } xy{\rm{d}}z{\rm{d}}y{\rm{d}}x = \frac{{81}}{5}$
Work Step by Step
We have the triple integral:
$\mathop \smallint \limits_{x = 0}^3 \mathop \smallint \limits_{y = 0}^{\sqrt {9 - {x^2}} } \mathop \smallint \limits_{z = 0}^{\sqrt {9 - {x^2} - {y^2}} } xy{\rm{d}}z{\rm{d}}y{\rm{d}}x$
From the order of the integral, we see that the solid region ${\cal W}$ is a $z$-simple region bounded below by the plane $z=0$ and bounded above by the sphere ${x^2} + {y^2} + {z^2} = 9$.
Since $x \ge 0$, $y \ge 0$, $z \ge 0$, ${\cal W}$ is in the first octant. The projection of ${\cal W}$ onto the $xy$-plane is the domain ${\cal D}$ (a part of a disk of radius $3$ in the first quadrant), which is a vertically simple region defined by
${\cal D} = \left\{ {\left( {x,y} \right)|0 \le x \le 3,0 \le y \le \sqrt {9 - {x^2}} } \right\}$
So, the domain of integration:
${\cal W} = \left\{ {\left( {x,y.z} \right)|0 \le x \le 3,0 \le y \le \sqrt {9 - {x^2}} ,0 \le z \le \sqrt {9 - {x^2} - {y^2}} } \right\}$
Now, we evaluate the triple integral:
$\mathop \smallint \limits_{x = 0}^3 \mathop \smallint \limits_{y = 0}^{\sqrt {9 - {x^2}} } \mathop \smallint \limits_{z = 0}^{\sqrt {9 - {x^2} - {y^2}} } xy{\rm{d}}z{\rm{d}}y{\rm{d}}x$
$ = \mathop \smallint \limits_{x = 0}^3 \mathop \smallint \limits_{y = 0}^{\sqrt {9 - {x^2}} } xy\left( {z|_0^{\sqrt {9 - {x^2} - {y^2}} }} \right){\rm{d}}y{\rm{d}}x$
$ = \mathop \smallint \limits_{x = 0}^3 \mathop \smallint \limits_{y = 0}^{\sqrt {9 - {x^2}} } xy\sqrt {9 - {x^2} - {y^2}} {\rm{d}}y{\rm{d}}x$
$ = \mathop \smallint \limits_{x = 0}^3 x\left( { - \frac{1}{3}{{\left( {9 - {x^2} - {y^2}} \right)}^{3/2}}|_0^{\sqrt {9 - {x^2}} }} \right){\rm{d}}x$
$ = - \frac{1}{3}\mathop \smallint \limits_{x = 0}^3 x\left( {{{\left( {9 - {x^2} - 9 + {x^2}} \right)}^{3/2}} - {{\left( {9 - {x^2}} \right)}^{3/2}}} \right){\rm{d}}x$
$ = \frac{1}{3}\mathop \smallint \limits_{x = 0}^3 x{\left( {9 - {x^2}} \right)^{3/2}}{\rm{d}}x$
$ = \frac{1}{3}\left( { - \frac{1}{5}{{\left( {9 - {x^2}} \right)}^{5/2}}|_0^3} \right)$
$ = - \frac{1}{{15}}\left( { - {9^{5/2}}} \right) = \frac{{81}}{5}$
Thus, $\mathop \smallint \limits_{x = 0}^3 \mathop \smallint \limits_{y = 0}^{\sqrt {9 - {x^2}} } \mathop \smallint \limits_{z = 0}^{\sqrt {9 - {x^2} - {y^2}} } xy{\rm{d}}z{\rm{d}}y{\rm{d}}x = \frac{{81}}{5}$.
