# Projecting a surface integral in 3-D onto a plane (2-D)

1. Feb 25, 2013

### Fractal20

1. The problem statement, all variables and given/known data
So the context is this arises in the method of descent, for finding a solution for the 2D heat equation from the 3d heat equation. Anyway, in one step, we must change the surface integral over a ball in 3d, to the surface integral over it's projection into a plane. In this case it is to the xy plane. So I understand that for the surface of a ball B(x1,x2,x3,t) can be represented as (y1, y2, y3) such that

(x1-y1)2 + (x2-y2)2 + (x3 - y3)2 = t2

Moreover, with the projection there is no x3 dependence so we can set it to zero and get that the surface of the projection is:

(x1-y1)2 + (x2-y2)2 + (y3)2 = t2

So my book used this and then says that dy1dy2dy3 can be written as:

[($\sigma$y3/$\sigma$y1)2 + ($\sigma$y3/$\sigma$y2)2 + 1]1/2 dy1dy2

I don't understand this step. I want to take the derivative of the expression for y3 but this certainly won't result in this expression for dy3. Can anybody explain how this step is made, or even just suggest what to look up in a calculus book? I was consulting my old calculus book but only found things describing how to integrate over a 3d surface by considering the 2d domain, so kind of the reverse of what I wanted here. Thanks!

PS is there a command so I can just type equations using the formatting from latex?

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution