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Trig substitution step (I think) 
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#1
Nov1912, 08:12 AM

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1. The problem statement, all variables and given/known data
I'm stuck at an attempt to solve an integration step. I think I'm supposed to trig substitute? 2. Relevant equations It is the second to third equation I'm having a hard time with 3. The attempt at a solution From second equation: (I'm only doing the integral) [tex] \int_0^{\pi} \frac{Sin \theta}{[R^2 + Z^2  2 R Z Cos \theta]^{1/2}} d \theta[/tex] [tex] u = cos \theta[/tex] [tex] du = sin \theta[/tex] [tex] \int_{1}^1 \frac{1}{[R^2 + Z^2  2 R Z u]^{1/2}} d u [/tex] I flipped the boundaries to remove the minus sign. And here I'm stuck. I could use some help :b 


#2
Nov1912, 09:16 AM

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#3
Nov1912, 09:18 AM

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#4
Nov1912, 10:51 AM

P: 25

Trig substitution step (I think)
First, thanks. I've tried using the whole expression in a substitution, but I got stuck again:
## u=R^2 + Z^2  2 R Z Cos \theta ## ## \frac{du}{d \theta}=2 R Z Sin \theta ## ## \frac{du}{2 R Z}= Sin \theta d \theta## [tex]\int_{R^2 + Z^2  2 R Z}^{R^2 + Z^2 + 2 R Z} \frac{1}{u^{1/2}} \frac{1}{2 R Z} d u [/tex] [tex]\frac{1}{2 R Z} \int_{R^2 + Z^2  2 R Z}^{R^2 + Z^2 + 2 R Z} \frac{1}{u^{1/2}} d u [/tex] [tex]\frac{1}{2 R Z} [2 (R^2 + Z^2 + 2 R Z)^{2/3}  2 (R^2 + Z^2  2 R Z)^{2/3}][/tex] [tex]\frac{1}{R Z} [(R^2 + Z^2 + 2 R Z)^{2/3}  (R^2 + Z^2  2 R Z)^{2/3}][/tex] Don't know where to go from here 


#5
Nov1912, 11:06 AM

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#6
Nov1912, 11:11 AM

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Oh they were supposed to be 3/2 lol (From the integration of (random)^1/2). Anyways will try that



#7
Nov1912, 11:12 AM

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#8
Nov1912, 11:30 AM

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Your last comment did the job. Thank you ^^



#9
Nov1912, 12:18 PM

P: 25

Oh there seems to be one more thing I don't quite get. This whole problem was how to calculate the potential of a hollow sphere
With the equation like this: [tex]\phi = \frac{Q}{8 \pi \epsilon_0} \frac{R + Z  R  Z}{R Z}[/tex] It seems that you are able to remove the absolute values when specified when in the sphere or when outside. For example when inside Z < R and the equation becomes: [tex]\phi = \frac{Q}{8 \pi \epsilon_0} \frac{R + Z  (R  Z)}{R Z}[/tex] and when Z > R the equation becomes: [tex]\phi = \frac{Q}{8 \pi \epsilon_0} \frac{R + Z + (R  Z)}{R Z}[/tex] Do you know why that is? (Why we can just remove the absolute value?) 


#10
Nov1912, 12:20 PM

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