Surface Area Calculation: u^2+v^2≤1

[bodensee9]

Could someone help with the following?
I am asked to find the surface area of the following surface with parametric equations x = uv, y= u+v, z = u-v, and u^2+v^2≤1.

So d/du is <v,1,-1> and d/dv is <u,1,-1> And the cross product is -2i + (u+v)j + (v-u)k. So the magnitude of the vector is 4+2v^2+2u^2. If I convert this to polar coordinates, the surface area is ∫∫√(4+2r)drdθ. I am wondering, would r be between 0 and 1? And what about θ, to me that would seem to be between 0 and 2π? But this seems wrong as this would be the solution if the area were u^2+v^2 = 1. Could someone explain what the significance of the ≤ 1 is? Thanks!

 

[gammamcc]

I would verify the condition u^2+v^2≤1 is satisfied by directly substituting u=rcos \theta , v=r sin\theta