$\int_{0}^{1} \int_{0}^{1} \sqrt{x^2+y^2} dxdy$

Question

I would like to solve the given integral:

$$\int_{0}^{1} \int_{0}^{1} \sqrt{x^2+y^2} dxdy$$

The integral is doable just using the regular iterated integral with $x$ and $y$. We integrate with respect to $x$ first with the substitution $x=y\tan{\theta}$, and we end up with:

$$\int_{0}^{1} \int_{0}^{\arctan{(\frac{1}{y})}}y^2\sec^3{\theta}d\theta dy$$

This... ends up to be really ugly. We turn it into the form:

$$\frac{1}{2}\int \left[\sqrt{1+y^2}+y^2\ln{(1+\sqrt{1+y^2}})-y^2\ln{y}\right]dy$$

Again, really ugly, but I've managed to wring out an answer of $\frac{1}{3} \sqrt{2}+\frac{1}{3}\ln{(1+\sqrt{2}}).$

My question: Is there another way to do this? My instict is to turn the bounds into polar, so the integrand becomes just $r$. But.... a square is the last thing to turn into polar. How would we do the integral to turn it into polar?

And my second question, is there a more elegant way to do the integral better than both of these?

Thanks.

Answer 1

\begin{align} \int_0^1 \int_0^1 \sqrt{x^2+y^2 } \, dy \, dx &= 2 \int_0^{\frac{\pi}4} \int_0^{\sec \theta} r^2 \, dr d \theta \\ &= \frac23 \int_0^{\frac{\pi}4} \sec^3 \theta \,d\theta \\ &= \frac23 \int_0^{\frac{\pi}4} \sec\theta (\sec^2 )\theta \,d\theta \end{align}

Try to complete the task.