Why is it that the surface integral of the flux of a vector field is the same as the surface integral of the vector field itself?

1.9k Views Asked by At

In other words, this:

definition

http://www.math.ucla.edu/~archristian/teaching/32b-w17/week-7.pdf

Is this just a definition because what we really care about is how much the vectors are "pushing" through the surface? Or is it an actual equality?

1

There are 1 best solutions below

0
On

This is just notation: $d\mathbf S = \mathbf n\,dS$. Both sides are computing the flux of $\mathbf F$, i.e., the net amount of $\mathbf F$ pushing outwards across the surface.