Compute the flux of the vector field $F = <x, y, 1>$ through the closed surface bounded by $z = x^2 + y^2$ and the plane $z = 1$, using the outward normals.
I computed the flux using two integrals, one of the paraboloid and one for the "cap." The flux through the cap is $\pi$ and I know that is correct. However, the flux through the paraboloid is not. This is what I have:
$\iint_{S_1} F \cdot dS = \int_{S_1} <x,y,1> \cdot <-2x,-2y,1> dA = \iint_{S_1} -2x^2 – 2y^2 + 1 dA$
and I know this should be $0$ but it is not.
Thanks for your help.
Best Answer
There are a couple other ways to find the flux through the paraboloid portion easily. The paraboloid has axial symmetry about the $ \ z-$ axis. The vector field $ \ \mathbf{F} \ = \ \langle \ x, y, 1 \ \rangle \ $ , at every "level" of $ \ z \ $ , has the same magnitude, $ \ \Vert \ \mathbf{F}(z) \ \Vert \ = \ \sqrt{x^2+y^2+1} \ = \ \sqrt{z+1} \ $ , at every point of the "boundary" circle $ \ r \ = \ \sqrt{z} \ $ , and for every point $ \ (x,y) \ $ on that circle, there is a point $ \ (-x, y ) \ $ (or a point $ \ (x, -y ) \ $ ) where the flux through the circle has the opposite direction. So there is complete cancelation of flux at every level of the paraboloid, and thus over its entire surface.
One can also apply the Divergence Theorem. The total flux through the "capped" paraboloid's volume is given by
$$ \iiint_V \ \nabla \cdot \mathbf{F} \ \ dV \ \ = \ \ \iiint_V \ 2 \ \ dV $$
$$ 2 \ \int_0^1 \ \pi \ [ \ r(z) \ ]^2 \ \ dz \ \ = \ \ 2 \pi \int_0^1 \ z \ \ dz \ \ = \ \ 2 \pi \ \cdot \ \left( \ \frac{1}{2}z^2 \ \right) \ \vert_0^1 \ = \ \pi \ \ .$$
But you have already established that the surface flux through the "cap" at $ \ z \ = \ 1 \ $ is
$$ \iint_S \ \mathbf{F} \cdot \mathbf{n} \ \ dS \ = \ \ \iint_S \ 1 \ \ dA \ = \ 1 \ \cdot \ \pi \ \cdot \ 1^2 \ = \ \pi \ \ , $$
so all the net flux is through the "cap", leaving zero net flux through the paraboloid.