# How to evaluate an integral from Griffiths

I'm working through Griffith's Electrodynamics book during the winter break, and I'm having trouble on evaluating this integral from problem 2.7 of Introduction to Electrodynamics 4th edition.

I have my electric field right?

$$\frac{1}{4\pi\epsilon_{0}} \sigma R^{2} 2\pi \int_{0}^{\pi} \frac{z - R\cos\theta}{(R^2+z^2 -2Rz\cos\theta)^{3/2}} \sin\theta \ d\theta$$

The integral i want to evaluate is (obviously) $$\int_{0}^{\pi} \frac{z - R\cos\theta}{(R^2+z^2 -2Rz\cos\theta)^{3/2}} \sin\theta \ d\theta$$

The solution manual says to use partial fractions, but I feel like that would take quite a bit of working out to do. I want to avoid using a lot of algebra as possible (to reduce possible hiccups). Also, looking around, a lot solution of the internet just go the route of plugging it in mathematica or maple. I'm trying to avoid using those tools.

Is there an integral table that that has something of this form or is there a much clever way of evaluating this?

Thank you.

Edit: If it helps, this is the integral evaluated from $0$ to $\pi$ (based off of the solution manual).

$$\frac{1}{4\pi\epsilon_{0}} \frac{\sigma R^{2} 2\pi}{z^2} \Bigg\{ \frac{(z-R)}{|z-R|} - \frac{(-z-R)}{|z+R|} \Bigg\}$$

Their will conditions of course, such as, when $z>R$ and $z<R$, but they can be applied after evaluating the integral.

• This seems to involve elliptic integrals of first and second kind. – Claude Leibovici Dec 23 '15 at 7:14
• Are $z$ and $R$ real numbers? – GaussTheBauss Dec 23 '15 at 7:49
• @vnd sorry, it was 1 in the morning when i typed this. I missed a trig function inside of this. I've fixed it. – iron2man Dec 23 '15 at 15:43
• @GaussTheBauss Yes. They represent real numbers. $z$ is the distance of the point charge from the center of the sphere and $R$ is the radius of the sphere. Nice username by the way. – iron2man Dec 23 '15 at 15:45
• @DarthLazar Well, I'm not sure if this will help, but I couldn't help but notice that the denominator is exactly $| R e^{i \theta} - z |^3$. So you couldd possibly use complex analysis for this. And thanks haha. – GaussTheBauss Dec 23 '15 at 15:55

Putting $$R^2+z^2-2Rz\cos\theta=t^2$$ one gets $$\int_0^\pi \frac {(z-R\cos\theta)}{(R^2+z^2-2Rz\cos\theta)^{3/2}}\,\sin\theta \,d\theta=$$$$\frac 1{2Rz^2} \int _{|R-z|}^{R+z}\left [1-\frac {R^2-z^2}{t^2} \right]\, dt=$$$$\frac 1{2Rz^2} \left(2R-\frac {R^2-z^2}{|R-z|}-|R-z| \right)=\begin {cases} 0\qquad \text{if}\quad z<R\\ \\\dfrac 2{z^2}\quad\; \text{if}\quad z>R \end {cases}$$
With the substitution $u=\sqrt{R^2+z^2-2Rz\cos\theta}$, $$\frac{du}{2Rz}=\frac{\sin\theta d\theta}{u}.$$ The integral now transforms as $$\frac{1}{2Rz}\int_{R-z}^{R+z}\Big(\frac{1}{2z}-\frac{R^2-z^2}{2zu^2}\Big)du$$This you can easily evaluate and it should comeout as $0$.
• You are correct to where the electric field is zero inside the sphere where $z$(the distance of the charge) $<R$ (The radius of the sphere). Now, how would you evaluate it to where $z>R$? In all due respect, shouldn't the evaluation of the integral be something as $\Bigg\{\frac{(z−R)}{|z−R|}−\frac{(−z−R)}{|z+R|} \Bigg\}$ to where both conditions can be tested? – iron2man Dec 23 '15 at 19:47