Integral on space Find the integral of the following function :
$$\displaystyle f(x,y,z)=\frac{1}{\sqrt{x^2+y^2+(z-2)^2}}\tag{f}$$
over the region bounded by $z=0$ two sphere's with radii $1,2$ and outside the cone $z=\sqrt{x^2+y^2}$.
My attempt: By writing this integral in spherical coordinate and using fubini theorem I got to :
$$\displaystyle\int\limits_{\theta=0}^{2\pi}\,\int\limits_{\rho=1}^{2}\,\int\limits_{\varphi=\frac{\pi}{4}}^\frac{\pi}{2}\frac{\rho^2\sin{\varphi}}{\sqrt{\rho^2-4\rho\cos{\varphi}+4}}\,\mathrm d{\varphi}\,\mathrm d\rho \,\mathrm d\theta\tag{I}$$
WRT to $\varphi$ we can easily compute the integral which leads to :
$$\int\limits_0^{2\pi}\int\limits_{1}^{2}\rho/2\sqrt{\rho^2+4}-\sqrt{\rho^2-2\sqrt{2}\rho+4}\,\mathrm d\rho \,\mathrm d\theta$$
But WRT to $\rho$ it becomes a nightmare integral.It is computable but the result is too complicated and long to achive(A lot of substituations is needed)Is there any easier way so that I am overcomplicating things?Since It is an exam problem I don't think this will be the intended solution and there should be one with less calculations.Thanks in advance!
Edit:Spheres are centered at origin.
 A: I would set it up in cylindrical coordinates,
Cone: $C1: r = z, z \geq 0$, Spheres: $S1: r^2+z^2 = 1, S2: r^2+z^2 = 4$
At intersection of cone and spheres,
$C1, S1: r = z = \frac{1}{\sqrt2}; \  C1, S2: r = z = \sqrt2$
The region is $R: r \geq z \geq 0,\  1 \leq r^2 + z^2 \leq 4$
So integral is,
$\displaystyle I = \small 2\pi \int_0^{1 / \sqrt2} \int_{\sqrt{1-z^2}}^{\sqrt{4-z^2}} \frac{r}{\sqrt{r^2 + (z-2)^2}} \ dr \ dz \ + 2\pi \int_{1 / \sqrt2}^{\sqrt2} \int_{z}^{\sqrt{4-z^2}} \frac{r}{\sqrt{r^2 + (z-2)^2}} \ dr \ dz$
As $ \displaystyle \small \int \frac{r}{\sqrt{r^2 + (z-2)^2}} \ dr = \sqrt{r^2+(z-2)^2} + C$,
$I =  \displaystyle \small 2\pi \int_0^{1 / \sqrt2} \left[2\sqrt{2-z} - \sqrt{5-4z} \right] \ dz \ + 2\pi \int_{1 / \sqrt2}^{\sqrt2} \left[2\sqrt{2-z} - \sqrt2  \sqrt{(z-1)^2+1}\right] \ dz$
$=  \displaystyle \small 4\pi \int_0^{\sqrt2} \sqrt{2-z} \ dz - 2 \pi \int_0^{1 / \sqrt2} \sqrt{5-4z} \ dz \ - 2 \sqrt2 \pi \int_{1 / \sqrt2}^{\sqrt2} \sqrt{(z-1)^2+1} \ dz$
And they are all standard integrals. Can you take it from here?
