How do I compute this integral? I'm wondering how to compute the integral $$ \int_2^3\int_0^\sqrt{3x-x^2}\frac{1}{(x^2+y^2)^{1/2}}\,\mathrm{d}y\mathrm{d}x. $$ Clearly it is too complicated to do it directly, so I'm guessing you have to do some change of variables. But what kind of change? This is not really a sphere, so I don't think that polar coordinates would be so good.
 A: You'll actually want to use polar coordinates. Set your first integral to y, square that y, and you should see you can move x to the left-hand side. You can then turn this into your r value for polar coordinates. You'll also note that it's mapping a sphere in the upper-right coordinate (which allows you to pick the theta values). From there it's a rather simple integration. I'll purposely leave the work to you, if you need clarification let me know.  
A: Starting with the integral
\begin{equation*}
\int \frac{1}{\sqrt{x^2+y^2}}dx
\end{equation*}
Use the substitution $x=y\tan(u),~dx=y\sec^2(u)du.$ The integral becomes
\begin{equation*}
y\int \frac{\sec(u)}{y}du=\int\sec(u)du=\int \frac{\sec^2(u)+\tan(u)\sec(u)}{\tan(u)+\sec(u)}du.
\end{equation*}
Use the substitution $s=\tan(u)+\sec(u):$
\begin{equation*}
\int \frac{1}{s}ds=\log(s).
\end{equation*}
Substituting back again and using $\sec(\tan^{-1}(z))\sqrt{z^2+1}$ and $\tan(\tan^{-1}(z))=z$ we get
\begin{equation*}
\log(\sqrt{x^2+y^2}+x).
\end{equation*}
Can you finish the rest of the integration? 
A: in polar coordinates, the integral is equivalent to 
$$\int _0^{\arctan \left(\left.\sqrt{2}\right/2\right)}\int _{2 \sec  \theta }^{3\cos  \theta }drd\theta =$$
$$\int_0^{\arctan \left(\left.\sqrt{2}\right/2\right)} (3\cos  \theta -2 \sec  \theta ) \, d\theta =$$
$$\sqrt{3}-\ln \left(2+\sqrt{3}\right)$$
$r$ goes from the straight line $r=2 sec  \theta $ to the circle.
