# Homework Help: Surface area of a function

1. Jul 27, 2008

### fk378

1. The problem statement, all variables and given/known data
Find the area of the surface with parametric equations x=uv, y=u+v, z=u-v.
u^2 + v^2 <= 1

2. Relevant equations
A(S)= double integral over the domain D of the norm of the cross product (r_u X r_v) DA

3. The attempt at a solution
I started off with finding the norm of the cross product and got sqrt [4+2(v^2) + 2(u^2)]. I found it almost impossible to solve that integral. So I'm trying to use polar coordinates, but I don't know how to set the equations in terms of theta and radius.

2. Jul 27, 2008

### HallsofIvy

What you are thinking is exactly right. Since the region of the surface in which you are interested is given by $u^2+ v^2\le 1$ in the uv-plane, you use $u= r cos(\theta)$, $v= rsin(\theta)$. Then $dudv= r dr d\theta$ and the integral is just
[tex]\int_{r= 0}^1\int_{\theta= 0}^{2\pi}\sqrt{4+ 2r^2}rdrd\theta[/itex]

3. Jul 27, 2008

### fk378

When I compute the cross product, does r_theta X r_radius = r_u X r_v? In which case I could just compute r_u X r_v since it's easier right?

4. Jul 27, 2008

### HallsofIvy

Yes! Notice that I specifically said "the uv-plane". The reason you want to change to "polar coordinates" is because you have "$u^2+ v^2\le 1$" and $u^2+ v^2$ in your integral. This have nothing to do with the xy-plane.