9
$\begingroup$

So I have the following integral:

$$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} {{{x^2}dxdy}\over (1 + \sqrt{x^2 + y^2})^5}$$

I know that converting the integral using polar coordinates gives:

$$\int \int {{r^2 \cos^2 \theta} \over {(1 + r)^5}} rdrd \theta$$

I'm assuming $r$ is going from $0$ to infinity.

But what about $\theta$?

$\endgroup$

2 Answers 2

Reset to default
11
$\begingroup$

Note that in order to cover $\mathbb{R}^2$, $r$ extends from $0$ to $\infty$ and $\theta$ spans an entire period of $\sin(\theta)$ and $\cos(\theta)$. (For example, the first quadrant alone is covered by $\theta \in [0,\pi/2]$, $r\in [0,\infty)$.)

Hence,

$$\begin{align} \int_{-\infty}^\infty\int_{-\infty}^\infty\frac{x^2}{(1+\sqrt{x^2+y^2})^5}\,dx\,dy&=\int_0^{2\pi}\int_0^\infty \frac{r^2\cos^2(\theta)}{(1+r)^5}r\,dr\,d\theta\\\\ &=\underbrace{\left(\int_0^{2\pi}\cos^2(\theta)\,d\theta\right)}_{=\pi}\underbrace{\left(\int_0^\infty \frac{r^3}{(1+r)^5}\,dr\right)}_{=1/4}\\\\ &=\frac{\pi}{4} \end{align}$$

$\endgroup$
0
8
$\begingroup$

By symmetry (we are free to exchange $x$ and $y$) the given integral equals: $$ \frac{1}{2}\iint_{\mathbb{R}^2}\frac{x^2+y^2}{(1+\sqrt{x^2+y^2})^5}\,dx\,dy = \pi \int_{0}^{+\infty}\frac{r^3}{(1+r)^5}\,dr = \color{red}{\frac{\pi}{4}}.$$ Your $\theta$ variable simply ranges over $(0,2\pi)$.

$\endgroup$
3
  • $\begingroup$ Isn't this assuming $x^2 = y^2$?... please explain $\endgroup$
    – Max Echendu
    Mar 14, 2017 at 2:17
  • 2
    $\begingroup$ @MaxEchendu: If we apply the map $\varphi:(x,y)\mapsto (y,x)$ to $\mathbb{R}^2$ we get that $$\iint_{\mathbb{R}^2} x^2\,f(x^2+y^2)\,dx \,dy = \iint_{\mathbb{R}^2} y^2\,f(x^2+y^2)\,dx \,dy $$ hence by averaging... $\endgroup$
    – Jack D'Aurizio
    Mar 14, 2017 at 2:19
  • 1
    $\begingroup$ we get the integral over $\mathbb{R}^2$ of a radial function. $\endgroup$
    – Jack D'Aurizio
    Mar 14, 2017 at 2:20

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .