How did they integrate the charge for Voltage question

1. Feb 1, 2016

kiwibird4

I have a question on finding potential for a disc of charge when it comes to the simple integration

how does the integral of R'dR'/etc have the R'dR' just disappear? i thought it would then be R^2/2
I am obviously missing something here

2. Feb 1, 2016

Staff: Mentor

Hi kiwibird4. Please remember to use the formatting template for all questions posted in the homework section.

The integral in question is not simply of R'dR'. Note that the denominator contains $\sqrt{z^2 + R'^2}$.

3. Feb 1, 2016

ehild

u-substitution $u = z^2+R'^2$ is applied.

4. Feb 2, 2016

kiwibird4

I'm SORRYYYY
I even had to check the lil button that said -I have followed the homework format

5. Feb 2, 2016

Staff: Mentor

Somehow we must all struggle through this great tragedy and believe in the promise of a brighter tomorrow

Cheers.

6. Feb 2, 2016

Mark Harder

Assuming you have looked up this integral in tables of same, I think what you're missing is the assumptions about the variables and z that would make the problem trivial if only you knew what they are. Even if you have success using the methods others have suggested earlier, looking for assumptions is a useful exercise that can pay off later on...

As an aside, it took me a while to get Mathematica to solve your integral. The evaluation hung up until I specified the assumptions required to give an unambiguous answer (The same one, even!). MMA's both interesting and frustrating that way. In order to be as general as possible, it makes few assumptions but provides the user with methods to specify them.