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Trig substitution step (I think)

  • Thread starter Blastrix91
  • Start date
  • #1
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Homework Statement


I'm stuck at an attempt to solve an integration step. I think I'm supposed to trig substitute?

Homework Equations


http://img685.imageshack.us/img685/9158/unavngivetn.png [Broken]

It is the second to third equation I'm having a hard time with

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
 
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Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,258
618

Homework Statement


I'm stuck at an attempt to solve an integration step. I think I'm supposed to trig substitute?

Homework Equations


http://img685.imageshack.us/img685/9158/unavngivetn.png [Broken]

It is the second to third equation I'm having a hard time with

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
Try ## u=R^2 + Z^2 - 2 R Z Cos \theta ## Everything except for theta is a constant.
 
Last edited by a moderator:
  • #3
33,075
4,779

Homework Statement


I'm stuck at an attempt to solve an integration step. I think I'm supposed to trig substitute?

Homework Equations


http://img685.imageshack.us/img685/9158/unavngivetn.png [Broken]

It is the second to third equation I'm having a hard time with

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
An ordinary substitution will work. Let u = R2 + Z2 - 2RZ cos(θ).
 
Last edited by a moderator:
  • #4
25
0
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
Dick
Science Advisor
Homework Helper
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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
Factor the quadratics inside the (2/3) powers. And check that (2/3). Where did that come from?
 
  • #6
25
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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
Dick
Science Advisor
Homework Helper
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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).
 
  • #8
25
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Your last comment did the job. Thank you ^^
 
  • #9
25
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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
Dick
Science Advisor
Homework Helper
26,258
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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?)
It's just because |x|=x if x>=0 and |x|=(-x) if x<0. If you know the sign of x you can replace |x| with a simpler expression.
 

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