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?
 A: I can't understand why this solid is getting chopped up into two pieces. We know the $\theta$ goes all the way around, from $0$ to $2\pi$, then the boundary conditions can be solved as $z=r^2$, $r^2+z^2=r^2+r^4=2$, so $r^4+r^2-2=(r^2+2)(r^2-1)=(r^2+2)(r+1)(r-1)=0$ leads to $r=1$ as the only positive solution. So we know the limits on $r$ are from $0$ to $1$, and the limits on z are due to the paraboloid on the bottom as $z=r^2$ and the sphere on top as $z=\sqrt{2-r^2}$ so the volume is
$$\begin{align}V&=\int_0^{2\pi}\int_0^1\int_{r^2}^{\sqrt{2-r^2}}dz\,r\,dr\,d
\theta=2\pi\int_0^1\left[\sqrt{2-r^2}-r^2\right]r\,dr\\
&=2\pi\left[-\frac13(2-r^2)^{3/2}-\frac14r^4\right]_0^1=2\pi\left[-\frac13+\frac{2\sqrt2}3-\frac14\right]=\frac{\pi}6\left(8\sqrt2-7\right)\end{align}$$
Which is the same answer as everyone else got, but with fewer integrals.
A: Actually, the problem as stated is ambiguous.  What you have is a rotated parabola that divides a sphere into two regions; the problem does not state   which of the two regions is to evaluated.  But the sum of the two volumes is the volume of the sphere $(\frac43)\pi2^2$.
I cannot offer a detailed answer, but I suggest that you will have better luck in cylindrical coordinates.  I would break one region into two lenticular regions and a central figure of rotation, or into a central figure of rotation with an annular region around the rim.  The totality of these two being the complete sphere.
Hope this helps.
A: 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,
$\displaystyle V=\int_0^1\pi(r(z))^2dz+\int_1^{\sqrt{2}}2\pi(r(z))^2dz=\int_0^1\pi z\;dz+\int_1^{\sqrt{2}}\pi(2-z^2)dz$
$\displaystyle\hspace{.16 in}=\frac{\pi}{2}+\pi\left(\frac{4}{3}\sqrt{2}-\frac{5}{3}\right)=\frac{\pi}{6}\left(8\sqrt{2}-7\right)$
