Considering S to be a closed surface consisting of a paraboloid $z=x^2+y^2$, with ($0≤z≤1$), and capped by the disc $x^2+y^2 ≤1$ on the plane $z=1$.
If the vector field is $\vec F (x,y,z) = z\hat j − y\hat k$, is it correct to say that the flux in the direction that points out across the surface S is zero?
Best Answer
Yes, since $\operatorname{div}\vec F=0$, so the result follows from the divergence theorem.
Just to convince ourselves further, we can explicitly compute the integral over the surface. Let $S_1$ and $S_2$ denote the "faces" of the surface $S=S_1\cup S_2$, where
$$\begin{align} S_1&=\left\{\vec s_1(u,v)=(u\cos v,u\sin v,u^2)\mid0\le u\le1,0\le v\le2\pi\right\}\\[1ex] S_2&=\left\{\vec s_2(u,v)=(u\cos v,u\sin v,1)\mid0\le u\le1,0\le v\le2\pi\right\} \end{align}$$
Take the normal vectors to be
$$\begin{align} \vec n_1&=\frac{\partial\vec s}{\partial v}\times\frac{\partial\vec s}{\partial u}=2u^2\cos v\,\vec\imath+2u^2\sin v\,\vec\jmath-u\,\vec k\\[1ex] \vec n_2&=\frac{\partial\vec s}{\partial u}\times\frac{\partial\vec s}{\partial v}=u\,\vec k \end{align}$$
Then the flux of $\vec F$ across $S$ is
$$\begin{align} \iint_S\vec F(x,y,z)\cdot\mathrm d\vec S&=\int_0^{2\pi}\int_0^1\vec F(u\cos v,u\sin v,u^2)\cdot(\vec n_1+\vec n_2)\,\mathrm du\,\mathrm dv\\[1ex] &=\int_0^{2\pi}\int_0^12u^4\sin v\,\mathrm du\,\mathrm dv\\[1ex] &=0 \end{align}$$
since the integral with respect to $v$ vanishes.