# A Mysterious Calculus Operation

1. Aug 9, 2006

### Saketh

I was doing a physics problem which leads me to this equation:
$$V = \frac{\sigma}{4\pi \epsilon_0}\int_{0}^{R}\int_{0}^{2\pi} \frac{r}{\sqrt{R^2 + r^2 - 2Rr\cos{\theta}}} \,d\theta \,dr$$
Stumped by the math, I turned to the solution. However, without a word of explanation, the solution jumps from the above equation to the following:
$$V = \frac{\sigma}{4\pi \epsilon_0}R\int_{0}^{2\pi} \cos{\theta} \ln{\left ( 1 + \frac{\sqrt{2}}{\sqrt{1-\cos{\alpha}} \right ) }\,dA +8R - 2\pi R$$
I had no idea how that happened. I would appreciate it if someone could explain what I should to do arrive at this latter equation. I don't need a full explanation, just a push in the right direction.

EDIT: I understand that it's an integral, I just don't understand how the integral was done. Specifically, this one:
$$\int_{0}^{R} \frac{r}{\sqrt{R^2 + r^2 - 2Rr\cos{\theta}}} \,dr$$
Thanks.

Last edited: Aug 10, 2006
2. Aug 10, 2006

### WigneRacah

To find the solution you first have to perform the integration with respect $$r$$.

By the way, you have to observe that:

$$\frac{r}{\sqrt{r^2 +R^2 - 2 R r \cos \theta}} = \frac{r - R \cos \theta}{\sqrt{(r-R \cos \theta)^2 + R^2 \sin^2 \theta}} + \frac{R \cos \theta}{\sqrt{(r-R \cos \theta)^2 + R^2 \sin^2 \theta}}}$$.

Last edited: Aug 10, 2006