How did they integrate the charge for Voltage question

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The discussion centers on the integration of charge potential for a disc, specifically addressing the integral involving the expression ##\sqrt{z^2 + R'^2}##. Participants clarify that the integral is not merely of the form R'dR', and emphasize the importance of applying u-substitution with ##u = z^2 + R'^2## to simplify the problem. The conversation also highlights the challenges faced when using Mathematica for solving integrals, particularly the need for clear assumptions to obtain accurate results.

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kiwibird4
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I have a question on finding potential for a disc of charge when it comes to the simple integration
upload_2016-2-2_0-18-39.png

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
 
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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}##.
 
kiwibird4 said:
I have a question on finding potential for a disc of charge when it comes to the simple integration
View attachment 95177
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
u-substitution ##u = z^2+R'^2## is applied.
 
gneill said:
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}##.

I'm SORRYYYY :cry::cry::cry::cry::cry::cry::cry::cry::H:H:H:H:H:H:H:H
I even had to check the lil button that said -I have followed the homework format
:cry:
 
kiwibird4 said:
I'm SORRYYYY :cry::cry::cry::cry::cry::cry::cry::cry::H:H:H:H:H:H:H:H
I even had to check the lil button that said -I have followed the homework format
:cry:
Somehow we must all struggle through this great tragedy and believe in the promise of a brighter tomorrow :smile:

Cheers.
 
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.
 

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