I am trying to do the following question; calculate the flux
Suppose $$F(x,y,z)=(-x)i+(-y)j+(z^3)k$$ over the cone $z=\sqrt{x^2+y^2}$ between $z=1$ and $z=3$ with downward orientation
My attempts:
I think I am supposed to use the formula $\int F \bullet n dS$
For the cone, if we parametrized,
at $z=1$ we have $x=\cos(v), y=\sin(v) ,0 \le v \le 2\pi$
and at $z=3$ we have $x=3\cos(v) , y=3\sin(v), 0 \le v \le 2\pi$
But then I am not sure how to proceed. Looking for advice
Gauss's Theorem relates the surface integral over some closed surface $S$ to a volume integral over the (bounding) volume $V$, namely $\iint_{(S)} {\bf F} \cdot dS = \iiint_{(V)} \nabla \cdot F \,dV$.
For your function, ${\bf F} = (-x){\bf i}+(-y){\bf j}+(z^3){\bf k}$ have $\nabla \cdot F = \frac{\partial (-x)}{\partial x} + \frac{\partial (-y)}{\partial y} + \frac{\partial (z^3)}{\partial z} = -2 + 3z^2$.
This gives $\iiint_{(V)} \nabla \cdot F \,dV = \iiint_{(V)} (-2 + 3z^2)\,dV = - 2 \iiint_{(V)} \,dV + 3 \iiint_{(V)} z^2\,dV$, where you are able to take the $2$ out of the volume integral because it is constant. But $\iiint_{(V)} \,dV$ is just equal to the volume (in your problem the volume is the cone between $z = 1$ and $z = 2$). This can be worked out or looked up (it's pretty standard). I admit the second volume integral needs a bit of work, but is do-able.