Stumped on Calculating Total Charge on Conducting Surface

[Dawei]

I'm trying to find the total charge on a conducting surface but am stuck at the last part of this integral. Maybe I'm just dumb but it looks really complex to me and I can't seem to figure it out.

http://www.flickr.com/photos/54479402@N07/5045800952/#/photos/54479402@N07/5045800952/lightbox/

(Sorry, I made this kind of fast--there should be no dφ term, just dθ.)

R and a are constants.

 

[Dawei]

Really? No one can do it?

I wrote it simpler here:
http://www.flickr.com/photos/54479402@N07/5047681891/

 

[The_Duck]

Suppose you make the substitution u = cos(theta). I think that ought to simplify it a bit more.

Edit: I had sin(theta) earlier but I think you want cos.

 

[paulfr]

Yes u = cos theta and du = - sin(theta) d(theta)
yields ... minus Integral (p + qu)^(-3/2) du
where ... p = R^2 + a^2, q = -2Ra
which becomes
- 2 [ (p+qu)^(-1/2) ] + C
backsubstituting
- 2 [ (p+qcos(theta))^(-1/2) ] + C

I think I may be off by a constant factor here.
The (p+qu)^(-3/2) term needs to be modified to
k (p/q +u)^(-3/2) before integrating.

 

[HallsofIvy]

Just because you don't get a response immediately doesn't mean "no one can do it". In fact, that's a fairly elementary integral.

Let $u= R^2+ a^2- 2aR cos(\theta)$ so that $du= 2aR sin(\theta) d\theta$.

When $\theta= 0$, $u= R^2- 2aR+ a^2= (R- a)^2$.

When $\theta= \pi$, $u= R^2+ 2aR+ a^2= (R+ a)^2$.

Your integral becomes
$$\int_{(R-a)^2}^{(R+a)^2} u^{-3/2} du$$
$$= -\frac{2}{3}u^{-1/2}\right|_{(R- a)^2}^{(R+a)^2}$$
$$= \frac{2}{3}\left(\frac{1}{R-a}- \frac{1}{R+ a}\right)$$
$$= \frac{4}{3}\frac{a}{R^2- a^2}$$