3
$\begingroup$

I want to use polar coordinates to prove this:

$$\int_{\mathbb R^2}\frac{dxdy}{(x^2+y^2+1)^{3/2}}=2\pi$$

I'm thinking to make the following substitution:

$x^2+y^2=r^2, dx=-r\sin\theta d\theta, dy=r\cos\theta d\theta$

Am I in the right way? I need help how to proceed.

  • 3
    You have to use the Jacobian instead. Namely, $\int_{\Bbb R^2}f(x,y) dx dy$ becomes $\int_0^{2\pi} \int_0^{\infty} f(r,\theta) r dr d\theta$ – 2017-01-29
  • 1
    I'm pretty sure I have answered this in the past. I can't find it. I suspect the question was deleted. Nevermind, [different integral](http://math.stackexchange.com/questions/515948/hard-integral-frac11x2y2z22). It should still be useful. – 2017-01-29

2 Answers 2

4

Your way to reason is correct.

Apply a polar coordinates transformation as you wrote, to obtain:

$$\text{d}x\text{d}y \to \text{Jacobian} = r\ \text{d}r\ \text{d}\theta$$

$$x^2 + y^2 = r^2$$

Hence the integral becomes

$$\int_0^{+\infty}\int_0^{2\pi} \frac{r\ \text{d}r\ \text{d}\theta}{(r^2+1)^{3/2}}$$

The integral over $\theta$ is trivial and you get immediately a factor of $2\pi$.

It's not difficult to show thence that

$$\int_0^{+\infty} \frac{r}{(r^2+1)^{3/2}}\ \text{d}r = 1$$

once the final result is

$$2\pi$$

Write me if you need details to solve the latter integral!

How to solve the integral

With a simple substitution:

$$k = r^2+1$$

$$\text{d}k = 2r\ \text{d}r$$

hence

$$\frac{1}{2}\int\frac{\text{d}k}{k^{3/2}}$$

You can easily proceed by yourself.

1

There is a faster way, that is integrating along "shells":

$$ \iint_{\mathbb{R}^2}\frac{dx\,dy}{(1+x^2+y^2)^{3/2}} = \int_{0}^{+\infty}\frac{2\pi\rho}{(1+\rho^2)^{3/2}}\,d\rho = \left.-\frac{2\pi}{\sqrt{1+\rho^2}}\right|_{0}^{+\infty} = \color{red}{2\pi}$$ since the length of the circle $x^2+y^2=\rho^2$ is just $2\pi\rho$.

  • 1
    I don't see what you mean by "integrating along shells". Would you please elaborate? It just looks like a polar-coordinates transformation, multiplying by $2\pi$ directly because the integral is independent of $\theta$ anyway. – 2017-01-29
  • 1
    @OpenBall: I mean dealing with the integral this (Lebesgue-style) way: $$\int_{0}^{+\infty} f(x)\,dx =\int_{-\infty}^{+\infty} \mu\left\{x: f(x)=k\right\} \,dk $$ Since the integrand function just depends on the distance from the origin, this is actually the same as performing a change of variables and switching to polar coordinates, I agree, but we do not need to introduce an angle, then remove it: we just exploit the fact that the length of a circle with radius $\rho$ is $2\pi\rho$ in the first step. – 2017-01-29
  • 1
    This is a useful technique for dealing with integrals like $$\iiint_{x^2+y^2+z^2\leq 1}(x^2+y^2+z^2)^2\,dx\,dy\,dz, $$ for instance. We may switch to polar coordinates, introduce two angles and remove them, or just notice that the surface area of a sphere with radius $\rho$ is $4\pi\rho^2$, hence the previous integral is $$ \int_{0}^{1}(4\pi\rho^2)\cdot \rho^4\,d\rho = \frac{4\pi}{7},$$ fast and clean. – 2017-01-29
  • 0
    Oh, now I see. Thanks! – 2017-01-29