Surface Integral between planes

  • Thread starter hooshies
  • Start date
  • #1

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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Answers and Replies

  • #2
The limits of integration are over the foot-print in the x-y plane of the surface you're integrating over right? You drew it? It's that paraboloid-cylinder cut at the planes of x=0 and x=1 above the x-y plane and you can figure where it cuts through the x-y plane by the equation z=4-1/4 y^2. It's just a square 1x4.