I'm having an issue finding two of the bounds in a spherical coordinates problem. I'm given the following integral:
$$\int_{-{\sqrt{2}}}^{\sqrt{2}}\int_{-{\sqrt{2-y^2}}}^\sqrt{2-y^2}\int_{\sqrt{x^2+y^2}}^{\sqrt{4-x^2-y^2}}z^2 \mathrm dz \mathrm dx \mathrm dy$$It wants me to sketch the region of integration which I've done: 
(an inverted cone inside a hemisphere of radius 2, both centered at the origin) and to write down the integral after making the change of coordinates (but not to actually evaluate the integral).
What I've done so far:
I've identified that we have $$\sqrt{x^2+y^2} \le z \le \sqrt{4-x^2-y^2}$$ $$-\sqrt{2-y^2} \le x \le \sqrt{2-y^2}$$ $$ -\sqrt{2} \le y \le\sqrt{2}$$
I've also found the intersection:
I set $$\sqrt{x^2+y^2} = \sqrt{4-x^2-y^2}$$ $$ x^2 + y^2 = 4-x^2-y^2$$ $$2x^2 + 2y^2 = 4$$ $$x^2+y^2=2$$ so the curves $z = \sqrt{4-x^2-y^2}$ and $z = \sqrt{x^2+y^2}$ intersect on the curve $x^2+y^2=2$ (circle radius 2).
With $x^2+y^2 = 2$ it can be found that the $z$-value of the intersection is at $\sqrt{4-(x^2+y^2)} = \sqrt{4-2} = \sqrt{2}$.
Now with $z=\sqrt{2}$, $ \theta$ (angle from z-axis to the curve of intersection) can be found by using $\theta = \arccos\big(\frac{z}{r}\big)$ where I believe the $r$ involved is the radius of the curve of intersection, so $$\theta = \arccos\Big(\frac{\sqrt{2}}{\sqrt{2}}\Big) = 0 \ \ \text{or} \ \ 2\pi$$ which is incorrect. I know that $0 \le \phi \le 2\pi$ because the the curve of intersection is a full circle (where $\phi$ is the azimuthal angle the curve spans in the $x$-$y$ plane) which is correct but otherwise I'm not sure what I'm doing wrong. From my sketch of the two surfaces I can see the curve of intersection occurs about half way down from the z-axis which makes the $\theta$ angle $\frac{\pi}{4}$ logical but I can't seem to be able to figure out the correct radius, as the correct bounds are $0 \le r \le 2$ and $0 \le \theta \le \frac{\pi}{4}$. Any help would be appreciated :)
In spherical coordinates,
$z = \rho cos\theta$
$x = \rho sin\theta cos\phi$
$y = \rho sin\theta sin\phi$
Equate, $z^2 = 4 - x^2-y^2$ and $z^2 = x^2+y^2$
$x^2+y^2 = 2$
Substituting back you get $z = \sqrt{2}$
$\rho^2 = x^2+y^2 + z^2 = 4 \implies \rho =2$
$\rho$ runs from 0 to 2
$\rho cos\theta =\sqrt{2}$
$cos\theta = \frac{1}{\sqrt{2}}$
$\theta = \frac{\pi}{4}$
Thus $\theta$ runs from 0 to $\frac{\pi}{4}$
$\phi$ runs from 0 to $2\pi$