Double Integral $\int\limits_0^a\int\limits_0^a\frac{dx\,dy}{(x^2+y^2+a^2)^\frac32}$

Question

How to solve this integral?

$$\int_0^a\!\!\!\int_0^a\frac{dx\,dy}{(x^2+y^2+a^2)^\frac32}$$

my attempt

$$ \int_0^a\!\!\!\int_0^a\frac{dx \, dy}{(x^2+y^2+a^2)^\frac{3}{2}}= \int_0^a\!\!\!\int_0^a\frac{dx}{(x^2+\rho^2)^\frac{3}{2}}dy\\ \rho^2=y^2+a^2\\ x=\rho\tan\theta\\ dx=\rho\sec^2\theta \, d\theta\\ x^2+\rho^2=\rho^2\sec^2\theta\\ \int_0^a\!\!\!\int_0^{\arctan\frac{a}{\rho}}\frac{\rho\sec\theta}{\rho^3\sec^3\theta}d\theta \, dy= \int_0^a\!\!\!\frac{1}{\rho^2}\!\!\!\int_0^{\arctan\frac{a}{\rho}}\cos\theta \, d\theta \, dy=\\ \int_0^a\frac{1}{\rho^2}\sin\theta\bigg|_0^{\arctan\frac{a}{\rho}} d\theta \, dy= \int_0^a\frac{1}{\rho^2}\frac{x}{\sqrt{x^2+\rho^2}}\bigg|_0^ady=\\ \int_0^a\frac{a}{(y^2+a^2)\sqrt{y^2+2a^2}}dy$$

Update:

$$\int_0^a\frac{a}{(y^2+a^2)\sqrt{y^2+2a^2}}dy=\frac{\pi}{6a}$$

Answer 1

Let us consider your last integral $$I=\int_0^a\frac{a}{(y^2+a^2)\sqrt{y^2+2a^2}}dy$$ Changing variable $y=a z$, it becomes $$I=\frac 1 a \int_0^1 \frac{dz}{\left(z^2+1\right) \sqrt{z^2+2}}$$ Now, make a change of variable such that $$z=\frac{\sqrt{2} t}{\sqrt{1-t^2}}$$ (it did not come immediately to my mind, I must confess) $$dz=\frac{\sqrt{2}}{\left(1-t^2\right)^{3/2}}$$ and so $$\int \frac{dz}{\left(z^2+1\right) \sqrt{z^2+2}}=\int\frac{dt}{1+t^2}=\tan ^{-1}(t)=\tan ^{-1}\left(\frac{z}{\sqrt{z^2+2}}\right)$$ Now, use the bounds for the integral.

Answer 2

The natural course of action whenever you see $x^2 + y^2$ in a multiple integration problem is to convert to polar coordinates.

Set $x = r\cos\theta$, $y = r\sin\theta$, $0 \leq \theta <2\pi$. Then $dx~dy = r~dr~d\theta$, and the integrand becomes $$ \frac{r}{(r^2 + a^2)^{3/2}}~dr~d\theta. $$ This integral doesn't depend on $\theta$, so in fact when you integrate in $r$ it is a one-variable integral, and from single-variable calculus we know that the substitution $u = r^2 + a^2$ will do the trick. The only thing left to do is to find the limits of integration.

The region $0 \leq x \leq a$, $0 \leq y \leq a$ is a square of side length $a$ in the first quadrant of the plane. This can be parametrized by $0 \leq \theta \leq \frac{\pi}{4}$, $0 \leq r \leq \sqrt{a^2 + (a\sin\theta)^2} = a\sqrt{1 + \sin^2 \theta}$. (Draw a picture! This is the meat of the problem.) Therefore the integral becomes $$ \int_{\theta = 0}^{\pi/4}\int_{r=0}^{a\sqrt{1+\sin^2\theta}} \frac{r}{(r^2 + a^2)^{3/2}}~dr~d\theta. $$

Can you take it from here?