In the first case make sure you are familiar with the surface area formula for a function z = f(x,y)

In your first problem use the analagous formula where x = f(y,z)

In particular:

x= (1-y^2)^(1/2)

dS = [(dx/dy)^2 + (dx/dz)^2 +1]^(1/2) dzdy

By Symmetry you can do half and double it

The region of integration is the trapezoidal region in the yz plane

where z varies from 0 to y + 1 and y varies from 0 to 1.

In the second problem you should know how to find the equation of a plane using 3 points. You probably did this much earlier in the semester so if you forgot look it up.

Once you have z = f(x,y) then it is fairly easy to see the region of integration in the xy plane is the triangular region with vertices

(-1,0), (1,2), and (3,0)