Find the volume of the region enclosed by $x^2+y^2+z^2=2$ and $x^2+y^2=z$. 
Find the volume of the region enclosed by $x^2+y^2+z^2=2$ and $x^2+y^2=z$.

I tried to solve the problem above by doing a change of variables to the spherical coordinate system, that is,
$x= \rho \cos \theta \sin \varphi$
$y = \rho \sin \theta \sin \varphi$
$z= \rho \sin \varphi$
But I'm struggling with the range of each variable, $\rho, \theta$ and $\varphi$. What I did is set $\theta$ from $0$ to $2 \pi$, $\varphi$ from $\pi /4$ to $3 \pi /4$ and $\rho$ from $0$ to $\sqrt{2}$, so I get
$$v= \int_0^{\sqrt{2}}\left( \int_0^{2 \pi} \left( \int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \rho ^2 \sin \varphi d\varphi \right) d\theta \right)d \rho = \frac{8 \pi}{3}$$
Is this result right? If not, can you suggest me how to proceed? Thanks in advance!
EDIT: After some thinking, I've arrived to the following:
Using Tom-Tom suggestion in the comment that I should split the figure in two parts and change to cylimdral coordinates, let's call the two volumes $S_1$ and $S_2$, where $S_1$ is the upper part (from $z=1$ to $z=\sqrt{2}$) and $S_2$ the rest of it.
For $S_1$, let's first fix the variable $\rho$ which in this case goes from $0$ to $1$. Now, $z$ goes from $0$ to the sphere defined by $x^2+y^2+z^2=2$, which in cylimdral coordinates is $z^2= 2- \rho^2$, or $z=\sqrt{2-\rho^2}$. And finall, $\theta$, which moves from $0$ to $2 \pi$. So then,
$$v(S_1)=\int_{S_1} 1 = \int_0^{2\pi} \int_0^1 \int_1^{\sqrt{2-\rho^2}} \rho \ dz \ d\rho \ d\theta = 2\pi \int_0^1 \rho (\sqrt{2-\rho^2}-1)\ d\rho = 2\pi \left(\frac{\sqrt{8}}{3}-\frac{5}{6}\right)$$
Mow, for $S_2$, we fix the variable $z$ first, that in this case goes from $0$ to $1$. $\rho$ moves from $0$ to the paraboloid defined by $z=x^2+y^2$, which in cylindral coordinates is $\rho = \sqrt{z}$ and $\theta$ goes fro $0$ to $2\pi$ as before. So
$$v(S_2)=\int_{S_2} 1= \int_0^1 \int_0^{2\pi} \int_0^{\sqrt{z}} \rho \ d\rho \ d\theta \ dz = \frac{\pi}{2}$$
And since $S_1$ and $S_2$ intersect on a circle, whic has null measure in $\mathbb{R}^3$,
$$v(S_1 \cup S_2)= v(S_1) + v(S_2)$$
Is these reasoning correct or I'm stil messing it up with the change of coordinates?

We will assume that the problem refers to the region 
above the paraboloid $z=x^2+y^2$ and inside the sphere $x^2+y^2+z^2=2$.
Since the volume can be found by integrating the area of the horizontal slices of the region,
and each of the horizontal slices is a circular disc,
$\displaystyle V=\int_0^1\pi(r(z))^2dz+\int_1^{\sqrt{2}}2\pi(r(z))^2dz=\int_0^1\pi z\;dz+\int_1^{\sqrt{2}}\pi(2-z^2)dz$
$\displaystyle\hspace{.16 in}=\frac{\pi}{2}+\pi\left(\frac{4}{3}\sqrt{2}-\frac{5}{3}\right)=\frac{\pi}{6}\left(8\sqrt{2}-7\right)$
