Finding the Flux Problem

1. Dec 10, 2013

PsychonautQQ

1. The problem statement, all variables and given/known data
Let v = -j be the velocity field of a fluid in 3 dimensional space. Computer the flow rate through the T of a triangular region with vertices (1,0,0) (0,1,0) (0,0,1) oriented with upward pointing normal vector

2. Relevant equations

3. The attempt at a solution
So can I use stokes theorem here to find the flux?
∫∫∇xF dS where F is the velocity field?

and then since this is a vector suface integral dS will equal n du dv?
What are my limits of integration? How do I parameterize this triangle in terms of u and v?

2. Dec 10, 2013

tiny-tim

Hi PsychonautQQ!
You won't have to bother with parametrisation if you choose your surface(s) so that the integral is obvious …

so can you choose them perpendicular to the three axes?

3. Dec 10, 2013

PsychonautQQ

I'm a little confused on what you mean by choose them perpendicular to the three axes. Perpendicular to the triangle you mean?

On the other hand I was thinking, do I even need calculus for this problem? Can't I just see that the vector only has a k component, and that if I project the triangle down to the xy plane it has an area of (1*1)/2 = .5.
So could the flux from the velocity vector of -j just be -1*.5 = -.5?

4. Dec 10, 2013

tiny-tim

Hi PsychonautQQ!
I meant the x y and z axes.
Yes! (but don't you need to mention the divergence theorem?)

5. Dec 10, 2013

PsychonautQQ

Mention divergence theorem? does divergence theorem give the flux?

∫∫∫(∇ dot F) dV?

that would mean
∫∫∫(-1) dxdydz where x y and z all go from 0 to 1? which would give the answer of -1 rather than .5 ;-(

My final answer is supposed to be a "flow rate" through the triangular area with an upward pointing normal vector. I don't know the units for this, and can a flow rate be negative like the -.5 answer that you approved of?

6. Dec 10, 2013

LCKurtz

There are shortcuts as you and TinyTim are discussing, but at this stage of your work I would suggest you don't use them. Remember your parameters $u,v$ can always be chosen among the original $x,y,z$ variables if convenient. In this problem your flux is in the negative $y$ direction (not $z$) so is perpendicular to the $xz$ plane. So why not try letting $x=x,~z=z,~y =1-x-z$ be your parameterization:$$\vec R(x,z) = \langle x,1-x-z,z\rangle$$and work out the flux integral. You may still see some shortcuts arise.

7. Dec 10, 2013

tiny-tim

yes … that's basically why it's called divergence!

no, you're missing the point …

you don't need to do a ∫∫∫, because divF = … ?
yup!

that's why they specified "upward pointing normal" …

if it was downward pointing normal, the flux would be minus that, ie +.5

(and the flux is presumably volume per time: if v was given in metres per second, then that would be cubic metres per second, but v is unrealistically given without units, so the flux will have to be written without units)