1) Eliminate the parameters to obtain an equation in rectangular coordinates, and describe the surface.

x=ucos(v), y=(u^2), z=usin(v) for 0<=u<=2 and 0<=v<(2pi)2) Find the area of the given surface.

The portion of the surface z=2x + (y^2) that is above the triangular region with vertices (0,0), (0,1), (1,1).For this one I know to use a double integral I just need help setting up the region R to use on the integrals.

Thanks.