# Flux Integrals

• Aug 1st 2010, 09:46 AM
Raguvian
Flux Integrals
Hi, I'm having trouble with some simple flux integrals.

A surface is given with points (0,0,2), (0,1,2), (1,0,0), and (1,1,0), and the vector field is v= 2i + 3j + 5k, and I am trying to find the flux of the vector field through the surface. I have found the normal to this surface by taking the cross product, and my normal is

<2,0,1>

I'm having problems setting up my integral. What is my dA? I don't know if it's in terms of dx, dy, or dz. The back of the book says the answer is 9, and the only way I can get this is by using dxdy for my dA, but I'm not sure why this is the case.

Any help is appreciated. Thanks!
• Aug 1st 2010, 10:21 AM
TheEmptySet
You have a constant vector field over a flat surface so an integral is not needed. You area vector is given by a unit normal to your surface multiplied by it's area. You found the the a normal vector to be <2,0,1> if you normalize this you get

$\vec{n}=\frac{1}{\sqrt{5}}<2,0,1>$
by the luck of the draw the area of the rectangle is also $\sqrt{5}$ so the Area vector $\vec{S}=<2,0,1>$

The flux of a constant vector field through a flat surface is the dot product of the area vector and the V.F so we get

$\vec{F}\cdot \vec{S}=(2)(2)+(3)(0)+(5)(1)=9$

IF you really want to use an integrate the (from your equation of the normal vector) the plane has equation 2x+z=2 or z=2-2x

Then use the formula Flux = $\int \vec{F}\cdot \left( -\frac{\partial z}{\partial x}\vec{i} -\frac{\partial z}{\partial y}\vec{j}+\vec{k} \right)dA$
Where dA is the projection of the surface into it's domain so in this case the xy plane as you wanted.
• Aug 1st 2010, 12:14 PM
Raguvian
Thanks for the help! I didn't even think to do it in a non integral way. That makes it much easier.
• Aug 2nd 2010, 01:15 AM
HallsofIvy
By the way, in your first post you say only "a [b]surface is given with points ..." with the points happening to lie on a plane, and then you assume the surface is a plane. There exist an infinite number of different surfaces passing through 4 given points. Are you specifically told that the surface is a plane?
• Aug 2nd 2010, 01:31 PM
Raguvian
Quote:

Originally Posted by HallsofIvy
By the way, in your first post you say only "a [b]surface is given with points ..." with the points happening to lie on a plane, and then you assume the surface is a plane. There exist an infinite number of different surfaces passing through 4 given points. Are you specifically told that the surface is a plane?

Sorry, I didn't specify in the original post that there was a graph given, with those 4 points being the corners.

If it didn't specify, how would I do the problem? Would I just assume it's a plane?