Trig substitution step (I think)

[Blastrix91]

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

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)

$$\int_0^{\pi} \frac{Sin \theta}{[R^2 + Z^2 - 2 R Z Cos \theta]^{1/2}} d \theta$$

$$u = cos \theta$$
$$-du = sin \theta$$

$$\int_{-1}^1 \frac{1}{[R^2 + Z^2 - 2 R Z u]^{1/2}} d u$$

I flipped the boundaries to remove the minus sign.

And here I'm stuck. I could use some help :b

 

[Dick]

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

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)

$$\int_0^{\pi} \frac{Sin \theta}{[R^2 + Z^2 - 2 R Z Cos \theta]^{1/2}} d \theta$$

$$u = cos \theta$$
$$-du = sin \theta$$

$$\int_{-1}^1 \frac{1}{[R^2 + Z^2 - 2 R Z u]^{1/2}} d u$$

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.

 

[Mark44]

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

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)

$$\int_0^{\pi} \frac{Sin \theta}{[R^2 + Z^2 - 2 R Z Cos \theta]^{1/2}} d \theta$$

$$u = cos \theta$$
$$-du = sin \theta$$

$$\int_{-1}^1 \frac{1}{[R^2 + Z^2 - 2 R Z u]^{1/2}} d u$$

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).

 

[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

 

[Dick]

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?

 

[Blastrix91]

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

 

[Dick]

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).

 

[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:
$$\phi = \frac{Q}{8 \pi \epsilon_0} \frac{|R + Z| - |R - Z|}{R Z}$$

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?)

 

[Dick]

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?)

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.