# Trouble with an integral

1. Jul 31, 2007

### rockytriton

I have a book that has the following integral:

$$6\int_\frac{-1}{2}^\frac{1}{2}\int_\frac{-1}{2}^\frac{1}{2}\frac{1}{r^2}a_r\cdot a_z dx dy$$

This integral gets converted to:

$$3\int_\frac{-1}{2}^\frac{1}{2}\int_\frac{-1}{2}^\frac{1}{2}\frac{dx dy}{(x^2 + y^2 + 1/4)^\frac{3}{2}}$$

(z = 1/2 by the way...)

I understand how it got to that point, but I'm having trouble understanding how it gets to this integral, I guess I don't understand the integration involved:

$$3\int_\frac{-1}{2}^\frac{1}{2}(\frac{x}{(y^2 + \frac{1}{4})(x^2 + y^2 + 1/4)^\frac{1}{2}})\|_\frac{-1}{2}^\frac{1}{2}dy$$

can someone explain how it gets to there?

2. Jul 31, 2007

### HallsofIvy

Staff Emeritus
If you don't tell us what ar and az are, I don't see how anyone can help you!

3. Jul 31, 2007

### rockytriton

oh, they are vectors, but they are not part of the problem. I understand how it goes from the first integral to the second integral, I'm having a problem figuring out how it goes from the second to the third integral here. The vectors have already been used up there.

4. Jul 31, 2007

### dhris

They're just doing the x-integration first. So they solve

$$\int_{-1/2}^{1/2}\frac{dx}{(y^2+x^2+1/4)^{3/2}}$$

treating y as a parameter. This is a pretty standard integral, and its solution is in brackets in your last expression. Note that

$$f(x)||_{-1/2}^{1/2}$$ means $$f(1/2)-f(-1/2)$$

so that your quantity in brackets is actually a sum of two terms, one with x=1/2 and the other with x=-1/2.