# Flux across surface

Find A uniform fluid that flows vertically downward is described by the vector field F (x, y, z) = (0, 0, −1).

Find the flux through the cone z = $z= \sqrt{x^2 + y^2}$, $x^2 + y^2 \le 1$.

I attempted this question with spherical coordinates and i don't know why it didn't work out. I used that $\sin(\theta) = \frac{\pi}{4}$ and evaluated the cross product T$\phi$ x Tr

This came to $(r\sin(\theta)\cos(\theta)\cos(\phi),-r\sin(\theta)\cos(\theta)\sin(\phi),r\sin^2(\theta))$ I then integrated

$\int_0^{2\pi}\int_0^1(0,0,-1).(r\sin(\theta)\cos(\theta)\cos(\phi),-r\sin(\theta)\cos(\theta)\sin(\phi),r\sin^2(\theta))~dr~d\phi$

=$-\int_0^{2\pi}\int_0^1r\sin^2(\theta)~dr~d\phi$

=$-\frac{\pi}{2}$ as $\theta = \frac{\pi}{4}$

Because I didn't get the right answer I checked my lecture notes, and I thought 'hey there is a $\hat n$ at the end of this integral and i didn't divide by magnitude. But to no avail.

I ended up getting -2$\pi$ by dividing by the magnitude and the answer is $\pi$

I now know that I can do it better with cylindrical coordinates and parametrization. But this way i don't divide by the magnitude

Do I divide by the normals magnitude to get $\hat n$ or not?

• There seems to be some ambiguity. The flux could be $\pi$ or $-\pi$ as stated. Either don't worry about the sign part, or you may find some statement regarding orientation. Can you show us your work? – Brady Trainor May 29 '13 at 4:54
• Ah it also says take the normal pointing outwards from the surface if thats what you mean? – Jesse Ross May 29 '13 at 4:58
• How can there be an outward with the surface described above? Outward and inward implicitly assume there is a bounded region I think. – Brady Trainor May 29 '13 at 5:34

Denote the cone surface as $S=\{(x,y,z):z = \sqrt{x^2+y^2},\;z\leq 1\}$, the region surrounded by this surface is $D = \{(x,y,z):\sqrt{x^2+y^2}< z< 1\}$.
This region has a "cap": $C = \{(x,y,z): x^2+y^2\leq 1, z= 1\}$ such that $\partial D = S\cup C$. Apply the divergence theorem for $F = (0,0,-1)$: $$\iiint_D \nabla\cdot F\,dxdydz = \iint_{\partial D} F\cdot \nu\,dS = \iint_{S} F\cdot \nu\,dS + \iint_{C} F\cdot \nu\,dS,$$ where $\nu$ are the unit surface normal pointing outward. The divergence of $F$ is zero, what we wanna compute is the flux across the cone surface $S$: $$\iint_{S} F\cdot \nu\,dS = - \iint_{C} F\cdot \nu\,dS.\tag{1}$$ Basically this identity from divergence theorem tells us that the amount of fluid flowing across the cone surface outward away from the cone, is the same with the amount of the fluid flowing into the cone from the cap. Flowing in/out yields the sign difference.
Now back to evaluating (1): $\nu$ on the cap is $(0,0,1)$, thus $F\cdot \nu = -1$, the right side is just the area of the cap $C$, which is $\pi$. Therefore: $$\iint_{S} F\cdot \nu\,dS = \pi,$$ and we do not have to deal with the integral on a parametrized surface which is significantly messier than this approach.