∫∫s x √(y2 + 4) where S: y2 + 4z = 16, and portion cut by planes x=0, x=1, z=0.
I attempted to solve using the surface area integral formula, whereby this double integral is transformed to ∫∫f(x,y,g(x,y)) √((∂z/∂x)2 + (∂z/∂y)2 + 1) dA
The Attempt at a Solution
Solving for z in the S region, and finding partials with respect to x and y yields A(S) of √(1+(1/4)y2) which can be rewritten as 1/2 √(4+y2)
Multiplying this by the original function, which is a function of just x and y, gives ∫∫ x/2*(4 + y2) dA.
I'm having trouble finding the limits of integration for the given planes.