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Calculate the flux of the vector field $F(x,y,z)=(x,y,xe^{y+z^2})$ through the surface of the cone $z^2=x^2+y^2$ between $z=0$ and $z=1$.

The flux is just the respective surface integral. I think we can "close" the surface with the upper cap of the cone (the circle with $T(r,\theta)=(r\cos\theta,r\sin\theta,1),\ r\in [0,1],\ \theta\in[0,2\pi]$) and then apply Gauss' divergence theorem, calculating the triple integral of the divergence of $F$ in the cone. Then, we just need to subtract the surface integral of $F$ on the circle represented from $T$ and we have the result.

Therefore, I obtain that the final flux is: $\frac{2\pi}{3}-0=\frac{2\pi}{3}$.

Can you verify my two results? What is especially troubling to me is that the surface integral of $F$ on the circle represented from $T$ is $0$...

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