Find the area of the surface defined by x y z=1, x^2 2y^2<=1.

Question

Okay, so the first equation given kinda looks like a 3D 3sided pyramid when I tried graphing it myself and the second is an ellipse on the xy plane... so I'm not really sure what 'surface area' I'm even supposed to find? Can someone please give me an equation that will help me solve this problem?

This is what I know:

SA = double int.(over R)[ |(div F)|/|(div F)dot(p)|]dA

where p is the normal vector to plane R, since my ellipse is in the xy plane I would set p = k...

or SA can be explained by

SA = double int(over D)[||Tu x Tv||dudv where ||Tu x Tv|| is the norm of Tu x Tv. If I try to do this using the following parameterization:

x=u, y=v, z=g(u,v)=1-u-v

then I get ||Tu x Tv|| = sqrt.(3)

I'm still not sure what my boundaries are or if I'm solving this the correct way, please help me!

Answer 1

something is wrong with the question and it is in the wrong section....please make sure you enter this question in mathematics