Changing to polar coordinates I am stuck with this problem:
$$\int_0^1\int_{x^2}^x\sqrt{x^2+y^2}\,dy\,dx$$
I would like to change to polar coordinates, but I am having a hard time parametrizing the domain. Do you guys have any suggestion? 
Thanks you :)
 A: First draw a picture. It should be clear that $\theta$ ranges from $0$ (the $+x$-axis) to $\dfrac{\pi}{4}$ (the line $y = x$). 
Now, draw the radial line from the origin which makes an angle of $\theta$ with the $+x$-axis. This line intersects the region in a segment that starts at the origin $r = 0$ and ends on the curve $y = x^2$. 
The curve $y = x^2$ in polar is $r\sin\theta = (r\cos\theta)^2$. Now simplify this to get the upper bound for $r$. 
Once you have the bounds $0 \le \theta \le \dfrac{\pi}{4}$ and $0 \le r \le \ ?$, you can write the double integral as 
$\displaystyle\int_{0}^{1}\int_{x^2}^{x}\sqrt{x^2+y^2}\,dx\,dy = \int_{0}^{\pi/4}\int_{0}^{?}r \cdot r\,dr\,d\theta$. 
This is easy to evaluate if you use the identity $\tan^2\theta = \sec^2\theta - 1$ and use a substitution. 
A: The integral in polar coordinates is
$$\int _{0}^{\frac{\pi}{4} }\!\int _{0}^{{\frac {\sin \left( \theta \right) }
{ \left( \cos \left( \theta \right)  \right) ^{2}}}}\!{r}^{2}{dr}\,{d
\theta} = {\frac {2}{45}}+{\frac {2}{45}}\,\sqrt {2}
$$
