Let W be the three dimensional region under the graph of f(x,y) =$e^{x^2+y^2}$ and over the region in the plane z = 0 defined by 1≤$x^2$+$y^2$≤2.
F= (2x-xy)i - (y)j + (yz)k
Part a) required me to find the flux of the vector field F over W and I obtained $\pi$($e^2$-2$e$).
Part b) asks me to find the flux over W without the bottom (i.e. without the z=0 plane). The answer should be the same but I am not sure how I should approach the solution. I used the divergence theorem for part a).
The surface enclosing $W$, can be split into two sets: one containing the bottom surface $z=0$ ($S_0$) and the other one the rest of the surface ($S$). So that $\partial W = S_0\cup S$
$$ \int_{\partial W}{\rm d}^2{\bf S}\cdot {\bf F} = \int_{S_0}{\rm d}^2{\bf S}\cdot {\bf F} + \int_{S}{\rm d}^2{\bf S}\cdot {\bf F} \tag{1} $$
To calculate the first term on the r.h.s note that the surface element perpendicular to $S_0$ is ${\rm d}^2{\bf S} = -{\rm d}^2S {\bf k}$
$$ \int_{S_0}{\rm d}^2{\bf S}\cdot {\bf F} = \int_{S_0} {\rm d}^2S ~ (yz) = 0 \tag{2} $$
since $z=0$ on $S_0$. So, according to (1), the first term does not contribut to the flux of the field, and if you remove it, you'd get the same result you got form the divergence theorem