# Trig substitution step (I think)

by Blastrix91
Tags: step, substitution, trig
 P: 25 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) $$\int_0^{\pi} \frac{Sin \theta}{[R^2 + Z^2 - 2 R Z Cos \theta]^{1/2}} d \theta$$ $$u = cos \theta$$ $$-du = sin \theta$$ $$\int_{-1}^1 \frac{1}{[R^2 + Z^2 - 2 R Z u]^{1/2}} d u$$ I flipped the boundaries to remove the minus sign. And here I'm stuck. I could use some help :b
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P: 25,228
 Quote by Blastrix91 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) $$\int_0^{\pi} \frac{Sin \theta}{[R^2 + Z^2 - 2 R Z Cos \theta]^{1/2}} d \theta$$ $$u = cos \theta$$ $$-du = sin \theta$$ $$\int_{-1}^1 \frac{1}{[R^2 + Z^2 - 2 R Z u]^{1/2}} d u$$ I flipped the boundaries to remove the minus sign. And here I'm stuck. I could use some help :b
Try ## u=R^2 + Z^2 - 2 R Z Cos \theta ## Everything except for theta is a constant.
Mentor
P: 21,284
 Quote by Blastrix91 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) $$\int_0^{\pi} \frac{Sin \theta}{[R^2 + Z^2 - 2 R Z Cos \theta]^{1/2}} d \theta$$ $$u = cos \theta$$ $$-du = sin \theta$$ $$\int_{-1}^1 \frac{1}{[R^2 + Z^2 - 2 R Z u]^{1/2}} d u$$ I flipped the boundaries to remove the minus sign. And here I'm stuck. I could use some help :b
An ordinary substitution will work. Let u = R2 + Z2 - 2RZ cos(θ).

 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## $$\int_{R^2 + Z^2 - 2 R Z}^{R^2 + Z^2 + 2 R Z} \frac{1}{u^{1/2}} \frac{1}{2 R Z} d u$$ $$\frac{1}{2 R Z} \int_{R^2 + Z^2 - 2 R Z}^{R^2 + Z^2 + 2 R Z} \frac{1}{u^{1/2}} d u$$ $$\frac{1}{2 R Z} [2 (R^2 + Z^2 + 2 R Z)^{2/3} - 2 (R^2 + Z^2 - 2 R Z)^{2/3}]$$ $$\frac{1}{R Z} [(R^2 + Z^2 + 2 R Z)^{2/3} - (R^2 + Z^2 - 2 R Z)^{2/3}]$$ Don't know where to go from here
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P: 25,228
 Quote by Blastrix91 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## $$\int_{R^2 + Z^2 - 2 R Z}^{R^2 + Z^2 + 2 R Z} \frac{1}{u^{1/2}} \frac{1}{2 R Z} d u$$ $$\frac{1}{2 R Z} \int_{R^2 + Z^2 - 2 R Z}^{R^2 + Z^2 + 2 R Z} \frac{1}{u^{1/2}} d u$$ $$\frac{1}{2 R Z} [2 (R^2 + Z^2 + 2 R Z)^{2/3} - 2 (R^2 + Z^2 - 2 R Z)^{2/3}]$$ $$\frac{1}{R Z} [(R^2 + Z^2 + 2 R Z)^{2/3} - (R^2 + Z^2 - 2 R Z)^{2/3}]$$ Don't know where to go from here
Factor the quadratics inside the (2/3) powers. And check that (2/3). Where did that come from?
 P: 25 Oh they were supposed to be 3/2 lol (From the integration of (random)^1/2). Anyways will try that
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P: 25,228
 Quote by Blastrix91 Oh they was supposed to be 3/2 lol. Anyways will try that
3/2 doesn't sound right either. You are integrating u^(-1/2).
 P: 25 Your last comment did the job. Thank you ^^
 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: $$\phi = \frac{Q}{8 \pi \epsilon_0} \frac{|R + Z| - |R - Z|}{R Z}$$ 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: $$\phi = \frac{Q}{8 \pi \epsilon_0} \frac{R + Z - (R - Z)}{R Z}$$ and when Z > R the equation becomes: $$\phi = \frac{Q}{8 \pi \epsilon_0} \frac{R + Z + (R - Z)}{R Z}$$ Do you know why that is? (Why we can just remove the absolute value?)
 Quote by Blastrix91 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: $$\phi = \frac{Q}{8 \pi \epsilon_0} \frac{|R + Z| - |R - Z|}{R Z}$$ 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: $$\phi = \frac{Q}{8 \pi \epsilon_0} \frac{R + Z - (R - Z)}{R Z}$$ and when Z > R the equation becomes: $$\phi = \frac{Q}{8 \pi \epsilon_0} \frac{R + Z + (R - Z)}{R Z}$$ Do you know why that is? (Why we can just remove the absolute value?)