Find the flux of the vector field $F$ across $\sigma$ by expressing $\sigma$ parametrically.
$\mathbf{F}(x,y,z)=\mathbf{i+j+k};$ the surface $\sigma$ is the portion of the cone $z=\sqrt{x^2 +y^2}$ between the planes $z=3$ and $z=6$ oriented by downward unit normals.
I'm not sure how to use the parameterization (I'd assume in cylindrical) to solve this problem. Any help would be great! Thank you in advance!
Your surface can be parametrized by $$S:\quad{\bf r}(\rho,\phi)=(\rho\cos\phi,\rho\sin\phi,\rho)\qquad(3\leq\rho\leq6, \ 0\leq\phi\leq2\pi)\ ,$$ whereby the orientation still has to be checked. One computes $${\bf r}_\rho=(\cos\phi,\sin\phi,1),\quad{\bf r}_\phi=(-\rho\sin\phi,\rho\cos\phi,0)\ .$$ This leads to $${\bf r}_\rho\times{\bf r}_\phi=(-\rho\cos\phi,-\rho\sin\phi,\rho)\ .$$ Since the third component of this normal direction is $\rho>0$ we have to make a sign correction in the following, and obtain $$\Phi({\bf F}, S)=-\int_3^6\int_0^{2\pi}{\bf F}\bigl({\bf r}(\rho,\phi)\bigr)\cdot({\bf r}_\rho\times{\bf r}_\phi)\>d\phi\>d\rho\ .$$ Now $${\bf F}\bigl({\bf r}(\rho,\phi)\bigr)\cdot({\bf r}_\rho\times{\bf r}_\phi)=(1,1,1)\cdot(-\rho\cos\phi,-\rho\sin\phi,\rho)=\rho(1-\cos\phi-\sin\phi)\ .$$ The integral of this over $0\leq\phi\leq2\pi$ gives $2\pi\rho$, so that we finally obtain $$\Phi({\bf F}, S)=-\int_3^6 2\pi\rho\>d\rho=-\pi\>\rho^2\biggr|_3^6=-27\pi\ .$$