# Surface Integral between planes

## Homework Statement

∫∫s x √(y2 + 4) where S: y2 + 4z = 16, and portion cut by planes x=0, x=1, z=0.

## Homework Equations

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.

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