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13—Vector Calculus 2 410

B~is a function of~rtoo, but I won’t write it. The first two terms have the same surface, so they combine to give
∫

S(t+∆t)

∆B~.dA~

and when you divide by∆tand let it approach zero, you get

∫

S(t)

∂B~

∂t

.dA~

Now for the next two terms, which require some manipulation. Add and subtract the surface that forms the edge
between the boundariesC(t)andC(t+ ∆t).

= _

∫

S(t+∆t)

B~(t).dA~−

∫

S(t)

B~(t).dA~=

∮

B~(t).dA~−

∫

edge

B~.dA~ (30)

The strip around the edge between the two surfaces make the surface integral closed, but I then have to subtract
it as a separate term.
You can convert the surface integral to a volume integral with Gauss’s theorem, but it’s still necessary to
figure out how to write the volume element. [Yes,∇.B~= 0, but this result can be applied in other cases too,

so I won’t use that fact here.] The surface is moving at velocity~v, so an area element∆A~will in time∆tsweep
out a volume∆A~.~v∆t. Note:~visn’t necessarily a constant in space and these surfaces aren’t necessarily flat.

∆V = ∆A~.~v∆t =⇒

∮

B~(t).dA~=

∫

d^3 r∇.B~=

∫

S(t)

∇.B d~ A~.~v∆t (31)

To do the surface integral around the edge, use the same method as in deriving Stokes’ theorem, Eq. ( 16 ).

∆A~= ∆~`×~v∆t