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
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!
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
by the luck of the draw the area of the rectangle is also so the Area vector
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
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 =
Where dA is the projection of the surface into it's domain so in this case the xy plane as you wanted.
Thanks for the help! I didn't even think to do it in a non integral way. That makes it much easier.
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.
Originally Posted by HallsofIvy
If it didn't specify, how would I do the problem? Would I just assume it's a plane?