13—Vector Calculus 2 40913.5 Reynolds’ Transport Theorem
When an integral has limits that are functions of time, how do you differentiate it? That’s pretty easy for
one-dimensional integrals.
d
dt∫f 2 (t)f 1 (t)dxg(x,t) =∫f 2 (t)f 1 (t)dx∂g(x,t)
∂t+g(f 2 (t),t)df 2 (t)
dt−g(f 1 (t),t)df 1 (t)
dt(26)
One of Maxwell’s equations for electromagnetism is
∇×E~=−
∂B~
∂t(27)
Integrate this equation over the surfaceS.
∫S∇×E~.dA~=∫
CE~.d~`=∫
S−
∂B~
∂t.dA~ (28)This used Stokes’ theorem, and I would like to be able to pull the time derivative out of the integral, but can I?
If the surface is itself time independent then the answer is yes, but what if it isn’t? What if the surface integral
has a surface that is moving? Can this happen? That’s how generators works, and you wouldn’t be reading this
now without the power they provide. The copper wire loops are rotating at high speed, and it is this motion that
provides the EMF.
I’ll work backwards and compute the time derivative of a surface integral, allowing the surface itself to
move. To do this, I’ll return to the definition of a derivative. The time variable appears in two places, so use
the standard trick of adding and subtracting a term. It’s rather like deriving the product formula for ordinary
derivatives. CallΦthe flux integral,
∫
B~.dA~.∆Φ =
∫
S(t+∆t)B~(t+ ∆t).dA~−∫
S(t)B~(t).dA~=
∫
S(t+∆t)B~(t+ ∆t).dA~−∫
S(t+∆t)B~(t).dA~+
∫
S(t+∆t)B~(t).dA~−∫
S(t)B~(t).dA~