Calc 3 - outward flux question

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The discussion centers on calculating the outward flux of the velocity field F = yz(i) + xz(j) + xy(k) over a specified region bounded by a cylinder, a parabola, and planes at z = H, y = 0, and z = 0. Two methods were attempted: the double integral of the divergence of F and the double integral of F dot k dA. The divergence method yielded zero flux, leading to confusion, as the divergence theorem is applicable only to closed volumes. The correct approach is to use the second method, which accounts for the flux through a portion of the surface area.

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The region in question is bounded by:
the cylinder (x^2)+(y^2)=(R^2)
the parabola x = y-((y^2)/R)
the planes z = H, y = 0, and z = 0

and the velocity field is:

F = yz(i)+xz(j)+xy(k)

and we need to calculate the outward flux of the field of the region at z = H (the top of the region).

Ive tried doing this 2 ways:
Doing the double integral of div(F)
Doing the double integral of F (dot) k dA
the solution to this problem uses the 2nd of the two; my question is this:
why are the 2 methods in disagreement with each other? (when i take div(F) i get zero, so flux is zero)
 
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The divergence theorem applies to a closed volume. You are calculated the flux through a piece of the surface area and not the whole surface area of the whole volume. So you can't use the divergence theorem.
 

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