Answer
The domain of integration is described by:
${\cal W} = \left\{ {\left( {x,y,z} \right)| - 2 \le z \le 2, - \sqrt {4 - {z^2}} \le x \le \sqrt {4 - {z^2}} ,1 \le y \le \sqrt {5 - {x^2} - {z^2}} } \right\}$
The projection of ${\cal W}$ onto the $xz$-plane is the domain ${\cal S}$, which is a disk of radius $2$ defined by
${\cal S} = \left\{ {\left( {x,z} \right)| - 2 \le z \le 2, - \sqrt {4 - {z^2}} \le x \le \sqrt {4 - {z^2}} } \right\}$
Work Step by Step
We have the triple integral:
$\mathop \smallint \limits_{ - 2}^2 \mathop \smallint \limits_{ - \sqrt {4 - {z^2}} }^{\sqrt {4 - {z^2}} } \mathop \smallint \limits_1^{\sqrt {5 - {x^2} - {z^2}} } f\left( {x,y,z} \right){\rm{d}}y{\rm{d}}x{\rm{d}}z$
From the order of the integral, we see that the solid region ${\cal W}$ is a $y$-simple region bounded below by the plane $y=1$ and bounded above by the sphere ${x^2} + {y^2} + {z^2} = 5$.
The projection of ${\cal W}$ onto the $xz$-plane is the domain ${\cal S}$, which is a disk of radius $2$ defined by
${\cal S} = \left\{ {\left( {x,z} \right)| - 2 \le z \le 2, - \sqrt {4 - {z^2}} \le x \le \sqrt {4 - {z^2}} } \right\}$
So, the domain of integration:
${\cal W} = \left\{ {\left( {x,y,z} \right)| - 2 \le z \le 2, - \sqrt {4 - {z^2}} \le x \le \sqrt {4 - {z^2}} ,1 \le y \le \sqrt {5 - {x^2} - {z^2}} } \right\}$
