Answer
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} xz{\rm{d}}V = \frac{{126}}{5}$
Work Step by Step
We have a solid region ${\cal W}$ bounded by the elliptic cylinder $\frac{{{x^2}}}{4} + \frac{{{y^2}}}{9} = 1$ and the sphere ${x^2} + {y^2} + {z^2} = 16$ in the first octant $x \ge 0$, $y \ge 0$, $z \ge 0$.
From Figure 14 and the figure attached, we see that ${\cal W}$ can be described as a $z$-simple region bounded below by the plane $z=0$ and bounded above by the sphere ${x^2} + {y^2} + {z^2} = 16$.
The projection of ${\cal W}$ onto the $xy$-plane is the domain ${\cal D}$, which is both vertically and horizontally simple region. Since the boundary curve of ${\cal D}$ involves square root of variable and the integrand has $x$ variable, for convenience, we choose to describe ${\cal D}$ as a horizontally simple region.
Step 1. Find the boundary of ${\cal D}$
Using $\frac{{{x^2}}}{4} + \frac{{{y^2}}}{9} = 1$, we obtain the boundary curve:
$\frac{{{x^2}}}{4} = 1 - \frac{{{y^2}}}{9}$
${x^2} = 4 - \frac{{4{y^2}}}{9} = \frac{{36 - 4{y^2}}}{9}$
$x = \frac{1}{3}\sqrt {36 - 4{y^2}} $
Step 2. Express ${\cal D}$ as a simple domain
From the results above, we obtain the description:
${\cal D} = \left\{ {\left( {x,y} \right)|0 \le y \le 3,0 \le x \le \frac{1}{3}\sqrt {36 - 4{y^2}} } \right\}$
Step 3. Evaluate the triple integral as an iterated integral
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} xz{\rm{d}}V = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \left( {\mathop \smallint \limits_{z = 0}^{\sqrt {16 - {x^2} - {y^2}} } xz{\rm{d}}z} \right){\rm{d}}A$
$ = \mathop \smallint \limits_{y = 0}^3 \mathop \smallint \limits_{x = 0}^{\frac{1}{3}\sqrt {36 - 4{y^2}} } \left( {\mathop \smallint \limits_{z = 0}^{\sqrt {16 - {x^2} - {y^2}} } xz{\rm{d}}z} \right){\rm{d}}x{\rm{d}}y$
$ = \frac{1}{2}\mathop \smallint \limits_{y = 0}^3 \mathop \smallint \limits_{x = 0}^{\frac{1}{3}\sqrt {36 - 4{y^2}} } x\left( {{z^2}|_0^{\sqrt {16 - {x^2} - {y^2}} }} \right){\rm{d}}x{\rm{d}}y$
$ = \frac{1}{2}\mathop \smallint \limits_{y = 0}^3 \mathop \smallint \limits_{x = 0}^{\frac{1}{3}\sqrt {36 - 4{y^2}} } x\left( {16 - {x^2} - {y^2}} \right){\rm{d}}x{\rm{d}}y$
$ = \frac{1}{2}\mathop \smallint \limits_{y = 0}^3 \mathop \smallint \limits_{x = 0}^{\frac{1}{3}\sqrt {36 - 4{y^2}} } \left( {16x - {x^3} - x{y^2}} \right){\rm{d}}x{\rm{d}}y$
$ = \frac{1}{2}\mathop \smallint \limits_{y = 0}^3 \left( {\left( {8{x^2} - \frac{1}{4}{x^4} - \frac{1}{2}{x^2}{y^2}} \right)|_0^{\frac{1}{3}\sqrt {36 - 4{y^2}} }} \right){\rm{d}}y$
$ = \frac{1}{2}\mathop \smallint \limits_{y = 0}^3 \left( {\frac{8}{9}\left( {36 - 4{y^2}} \right) - \frac{1}{{324}}{{\left( {36 - 4{y^2}} \right)}^2} - \frac{1}{{18}}\left( {36 - 4{y^2}} \right){y^2}} \right){\rm{d}}y$
$ = \frac{1}{2}\mathop \smallint \limits_{y = 0}^3 \left( {\frac{{14}}{{81}}{y^4} - \frac{{14}}{3}{y^2} + 28} \right){\rm{d}}y$
$ = \frac{1}{2}\left( {\frac{{14}}{{405}}{y^5} - \frac{{14}}{9}{y^3} + 28y} \right)|_0^3$
$ = \frac{1}{2}\left( {\frac{{14}}{{405}}\cdot243 - \frac{{14}}{9}\cdot27 + 28\cdot3} \right) = \frac{{126}}{5}$
Thus, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} xz{\rm{d}}V = \frac{{126}}{5}$.
