Flux across a cone Compute the flux of $F=(x,y,z^4)$ across the cone $z=\sqrt{x^2+y^2}, z \in [0,1]$ in the downward direction. (Answer is $\pi/3$.)
For this question I tried to use the divergence theorem: $\int\int_SF = \int\int\int_V\nabla F$
I got $\nabla F = 2+4z^3$ and used cylindrical coordinates: $\int_0^{2\pi}\int_0^1\int_r^1(2+4z^3)rdzdrd\theta$ but the answer I got was $4\pi/3$. Is this correct or did I do something wrong?
 A: Observe first that by using Gauss Theorem you in fact calculated the flux outward the surface
$$\left\{\left(x,y,\sqrt{x^2+y^2}\right)\right\}\cap\{\left(x,y,1\right)\}\subset\Bbb R^3$$
Without the "upper cap", we only have the (open) cone, and "outwards" thus clearly means downwards.
The intersection of plane $\;z=1\;$ with the cone $\;z=\sqrt{x^2+y^2}\;$ is just the surface (in fact, the "canonical" unit circle at height one)  $\;S:\;r(t)=(\cos t, \sin t, 1)\;,\;\;0\le t\le 2\pi\;$ , and since we wanted the flux down the cone to begin with, we must subtract now the flux up through the above surface , thus we get that the normal vector to that circle is clearly $\;\vec n=(0,0,1)\;$ , and thus (using cylindrical coodinates):
$$\vec F\cdot\vec n=z^4\implies \iint_S\vec F\cdot\vec n\;dS=\int_0^1\int_0^{2\pi} z^4\,r\,d\theta\,dr\stackrel{z=1\;\text{here}}=\int_0^1\int_0^{2\pi} r\,d\theta\,dr=\pi$$
and thus the final flux is (according to what you already got using Gauss Theorem)
$$\frac{4\pi}3-\pi=\frac\pi3$$
A: Same solution as "1 Answer" except:
The first display with a union, not an intersection, and the restriction $x^2+y^2\le 1$ is specified. Also what is called surface $S$, the intersection of $z=1$ and the solid cone $z \le \sqrt{x^2+y^2}$, is parameterized by
$$r(a,t)=\langle a \cos(t),a \sin(t),1\rangle, \; 0 \le a \le1,\; 0 \le t \le 1.$$
