Evaluating $\int_{0}^{1}\int_{0}^{1}\sqrt{x^{2}+y^{2}}dxdy$ using polar coordinates. I want to evaluate the integral $\int_{0}^{1}\int_{0}^{1}\sqrt{x^{2}+y^{2}}dxdy$ using the substitution $x=r\cos\theta$ and $y=r\sin\theta$ where the Jacobian is given as $|J(r, \theta)|=r$ and thus $dxdy=rdrd\theta$.
I need to set the range for $r$ and $\theta$. For $\theta$, it seems obvious that it ranges from $0$ to $\frac{\pi}{2}$ to cover all parts of the rectangle. But I am confused with the range for $r$. Would it be OK to set minimum $r$ as $0$ and maximum $r$ as $\sqrt{2}$, which is the diagonal of the square domain? I think I am confused with the fundamentals of integration.
 A: With symmetry we cut the line in half at $y=x$ and claim that the integral is equal to twice its value on the bottom triangle. The $r$ bounds go from the origin to the line $x=1$, which translates to  $r =\sec\theta$. Then the integral becomes
$$2 \int_0^{\frac{\pi}{4}} \int_0^{\sec\theta} r^2 dr d\theta = \frac{2}{3} \int_0^{\frac{\pi}{4}} \sec^3\theta d\theta$$
Which you can take from here with trig identities, or use the substitution $\tan\theta = \sinh(t)$:
$$\frac{2}{3} \int_0^{\sinh^{-1}(1)} \cosh^2(t) \: dt = \frac{1}{3} \int_0^{\sinh^{-1}(1)} 1 + \cosh(2t)\:dt $$ $$ = \frac{t}{3} + \frac{1}{3}\sinh(t)\cosh(t)\Biggr|_0^{\sinh^{-1}(1)} = \frac{1}{3}\sinh^{-1}(1)  + \frac{\sqrt{2}}{3}$$
A: In the square, $r$ will go from $0$ to $\sqrt2$, as the furthest point from the
origin within the square is $(1,1)$. But the range of allowable $\theta$ will depend
on $r$. If $r<1$ we get all $\theta$ between $0$ and $\pi/2$. But for $1<r<\sqrt2$
the smallest $\theta$ we get corresponds to the point $(1,\sqrt{r^2-1})$ and
so is $\theta=\tan^{-1}\sqrt{r^2-1}$. The largest theta we get is its complement.
Therefore
\begin{align}
\int_0^1\int_0^1\sqrt{x^2+y^2}\,dx\,dy
&=\int_0^1\int_0^{\pi/2}r^2\,d\theta\,dr
+\int_1^{\sqrt2}\int_{\tan^{-1}\sqrt{r^2-1}}^{\pi/2-\tan^{-1}\sqrt{r^2-1}}
r^2\,d\theta\,dr\\
&=\frac\pi2\int_0^1r^2\,dr+\int_1^{\sqrt2}\left(\frac\pi2
-2\tan^{-1}\sqrt{r^2-1}\right)r^2\,dr.
\end{align}
Good luck with that second integral!
A: The integrand $\sqrt{x^2+y^2}$ is symmetric about the line $y=x$ yielding
$$\iint_{[0,1]\times [0,1]}\sqrt{x^2+y^2}\;dydx=2\int_0^1\int_0^x\sqrt{x^2+y^2}\;dydx=2\int_0^\frac{\pi}{4}\int_0^{\sec(\theta)}r^2\;drd\theta=\frac{2}{3}\int_0^\frac{\pi}{4}\sec^3(\theta)\;d\theta$$ $$=\frac{1}{3}(\sec({\pi\over 4})\tan({\pi\over4})+\ln(\sec({\pi\over4})+\tan({\pi\over4})))=\frac{1}{3}(\sqrt2+\ln(1+\sqrt2))$$ 
