Trig substitution step (I think)

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)

[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

Dick

Try ## u=R^2 + Z^2 - 2 R Z Cos \theta ## Everything except for theta is a constant.

Mark44

An ordinary substitution will work. Let u = R² + Z² - 2RZ cos(θ).

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##

[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

Dick

Factor the quadratics inside the (2/3) powers. And check that (2/3). Where did that come from?

Blastrix91

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

Dick

3/2 doesn't sound right either. You are integrating u^(-1/2).

Blastrix91

Your last comment did the job. Thank you ^^

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:
[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?)

Dick

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.