# Finding surface area with a function. Double Integration.

1. Nov 21, 2012

### Unart

1. The problem statement, all variables and given/known data
Calculate ∫∫f(x,y,z)DS for the given surface and function.

Part of the plain x+y+z=0, contained in the cylinder x^2+y^2=1 f(x,y,z)=z^2.
2. Relevant equations

∫∫F(x,y,z)Ds=∫∫F(g(u,v)*||n(u,v)||

N= TuXTv
Tu= G(u,v)(du); Tv is the same only the derivative is with respect to v

3. The attempt at a solution
So, where I confused... is how to find the vector to use for G(u,v) from the cylinder and the plain. Aside from that I think I pretty much understand. Oh, and whould the upper and lower bounds be 1? Or do I use polar coordinates?

2. Nov 21, 2012

### LCKurtz

You have choices about how to parameterize the surface. You could choose $x$ and $y$ for your parameters$$x = x,\, y = y,\, z = 1-x-y\,\quad \vec R(x,y)=\langle x,y,1-x-y\rangle$$or you could choose $r$ and $\theta$ and parameterize it like this$$x=r\cos\theta,\,y=r\sin\theta,\,z = 1 - r\cos\theta-r\sin\theta$$and parameterize it accordingly. And, of course, the limits of integration depend on your variables. Even if you start with the $x,y$ parameterization you will likely change the resulting integral to polar coordinates because the domain is a circle.

3. Nov 24, 2012

### Unart

on the z=1-rcos(theta)-rsin(theta)... how did you get the 1
if x+y+z=0
So are you saying that the g path function would be (x,y,z)=(rcos(theta),rsin(theta),-rcos(theta)-rsin(theta)

4. Nov 24, 2012

### LCKurtz

That was a typo. I had read the equation as $x+y+z=1$. So put 0 instead of 1 in the appropriate places. Sorry.

That isn't the "g path". What it is is a parameterization of the portion of the plane inside the cylinder in terms of $r$ and $\theta$.

Last edited: Nov 24, 2012