My textbook asks the following question:
Find the flux of $F(x,y,z) = \langle x,y,z \rangle$ across the surface of the cone $ z^2 = x^2 + y^2 $, for $ 0 \leq z \leq 1 $ (normal vectors point upward).
I tried to solve this using both Divergence Thm. and directly. Using Divergence Thm., I got:
$\iint_S F\cdot n dS = \iiint_E divFdV = \int_{0}^{2\pi}\int_{0}^{1}\int_{0}^{r} (3)rdzdrd\theta = 2\pi$
However, when performing it directly, I obtained the following:
$\iint_S F\cdot n dS = \int_{0}^{2\pi}\int_{0}^1 (\langle rcos\theta, rsin\theta, r \rangle \cdot \langle -cos\theta, -sin\theta, 1 \rangle )rdrd\theta = 0 $
There is no solution listed for the problem. Are either of these correct or are neither of them correct?
Indeed, you can use the divergence theorem. You only have to compute the volume of the cone between $z=0$ and $z=1$. If you call it $E$, you have :
$$\int_E dx dy dz = 2\pi \int_0^1 \Big(\int_0^z r dr\Big)dz = \pi \int_0^1 z^2 dz = \frac{\pi}{3}$$
Therefore as $\text{div}(F(x,y,z))=3$ everywhere, you get that the flux is equal to $\pi$.