I don't think you have the correct region.
Of course to set up the integral in spherical coordinates, you need to describe the region in spherical coordinates.
So, let's do that. First, from the given integral, let's determine the solid $S$ defined by the limits of integration. Towards this end, let's call the innermost integral $g(x,y)$. Then we can write the given integral as $\tag{1} \int_0^{\sqrt 2} \int_y^{\sqrt{4-y^2}} g(x,y) \,dx\,dy. $ We will now determine the region $R$ in the $x$-$y$ plane that the limits of integration in $(1)$ determine. Since the outermost integral is with respect to $y$, the region $R$ is "generated" by line segments $l_y$ that are parallel to the $x$-axis. The line segments start at $y=0$ and end at $y=\sqrt2$ (the limits of integration of the outer integral). For a fixed $y$ in $[0,\sqrt2]$, the endpoints of the line segment $l_y$ are determined by the limits of integration of the inner integral: $l_y$ has endpoints on the graphs of $x=y$ and $x=\sqrt{4-y^2}$.
Alternatively, you can read a description of $R$ from the limits: $ R=\bigl\{\, (x,y)\mid 0\le y\le \sqrt2, y\le x\le \sqrt{4-y^2}\,\bigr\}. $
Either way, we see that $R$ is the blueish-shaded region below:

Now back to the triple integral. Considering the inner most integral, we see that the solid is bounded below by the region $R$ (as the lower limit of integration is $z=0$) and bounded above by the sphere of radius 2 (as the upper limit of integration is $z=\sqrt{4-x^2-y^2}$).
So $S$ is a spherical sector and is described by the spherical coordinates:
$ \eqalign{ 0&\le r\le 2\cr 0&\le\theta\le\pi/4\cr 0&\le \phi\le\pi/2 } $ (I use $\phi$ as the angle to the $z$-axis.)
And from this you can now set up the integral in spherical coordinates.