Integration Issues: How to Join Parts and Techniques

  • Thread starter Prologue
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In summary: So when they write dr*2r, they really mean d(r^2).In summary, the conversation involves a discussion about a substitution technique used in electromagnetism, specifically for finding the electric field on a disk. The substitution involves replacing 2rdr with du, where u = r^2. The conversation also touches on the perspective of replacing r^2 with u, and the notation of d(r^2) being equivalent to dr*2r.
  • #1
Prologue
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I am not sure exactly what is happening between the first part and second part. What happens to the 2?

Also, what technique do you use to integrate after that? I am a little unsure as to what slick moves are being used here. Any help is appreciated.

http://img511.imageshack.us/img511/2091/14157612rc6.jpg
http://g.imageshack.us/g.php?h=511&i=14157612rc6.jpg
 
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  • #2
I don't understand the context this is in but it looks like the 2r(dr) is being replaced by the substitution d(r^2) = 2r(dr) (this resembles usub, but I've never encountered this before). The denominator in the first step is just brought up top with a negative exponent. From step two to three, it looks like polynomial integration.

Hmm are you learning electromagnetism right now?
 
  • #3
Yep electromagnetism. It is out of the first chapter on the subject in Serway and Jewett. The example is for finding the electric field on a disk.

I am really stuck on that two, it seems like it has to be a mistake. There is probably some perspective I am missing here, that dr*2r=dr^2, but I am not seeing it now.

I can work it out fine with a substitution, but I really like to understand everything I encounter, and this doesn't mesh well.
 
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  • #4
If you let u = r^2, then you would replace 2rdr with du, wouldn't you?
Here, they have just not bothered renaming r^2.
 
  • #5
Big-T said:
If you let u = r^2, then you would replace 2rdr with du, wouldn't you?
Here, they have just not bothered renaming r^2.

Perfect, yep that is the angle I was missing. Of course I made the substitution u=x^2 +r^2 so du=2rdr. So thinking this way, keep going du=d(x^2+r^2)=d(r^2) etc.

Thank you for the advice.
 
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  • #6
Prologue said:
There is probably some perspective I am missing here, that dr*2r=dr^2, but I am not seeing it now.

It's kind of an abuse of notation. What's d(r^2)? It's the differential of r^2, which equals 2r dr.
 

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