# A double integral

1. Feb 27, 2012

### asmani

Hi all.

Suppose that we want to compute the following indefinite integral:

The correct solution by Mathematica:

Now here is the (apparently) incorrect solution by using polar coordinates:
$$\iint\frac{1}{\sqrt{x^2+y^2}}dxdy=\iint\frac{1}{r}rdrd\theta=(r+c_1)(\theta+c_2)$$
If c1=c2=0, then one solution is:
$$r\theta=\sqrt{x^2+y^2}\tan^{-1}\left ( \frac{y}{x} \right )$$
But it isn't:

What's wrong with this solution?

#### Attached Files:

File size:
992 bytes
Views:
187
File size:
966 bytes
Views:
180
• ###### 3.png
File size:
4.1 KB
Views:
196
2. Feb 28, 2012

$\iint\frac{1}{r}rdrd\theta$ looks good to me but that would just be $\iint drd\theta$ with $0<r<\infty$ and $0<\theta<2\pi$ which is $\infty$. I was already worried about that when I saw the pole at the origin but apparently you can fix it by chopping the radius of interest.