Mathematical Tools for Physics

13—Vector Calculus 2 406

I need only theˆrcomponent of the curl because the surface integral uses only the normal (rˆ) component. The
surface integral of this has the area elementdA=r^2 sinθ dθ dφ.

∫

curlF~.dA~=

∫θ 0

0

R^2 sinθ dθ

∫ 2 π

0

dφ

1

Rsinθ

(

CRcos^2 φ2 sinθcosθ+BRθ^2 sinφ

)

=R^2

∫θ 0

0

dθ

∫ 2 π

0

dφ 2 Ccos^2 φsinθcosθ

=R^22 Cπsin^2 θ 0 /2 =CR^2 πsin^2 θ 0

The other side of Stokes’ theorem is the line integral around the circle at angleθ 0.

∮

F~.d~`=

∫ 2 π

0

rsinθ 0 dφCrsinθcos^2 φ

=

∫ 2 π

0

dφCR^2 sin^2 θ 0 cos^2 φ

=CR^2 sin^2 θ 0 π (22)

and the two sides of the theorem agree. Check! Did I get the overall signs right? The direction of integration
around the loop matters. A further check: Ifθ 0 =π, the length of the loop is zero and both integrals give zero
as they should.

Conservative Fields
An immediate corollary of Stokes’ theorem is that if the curl of a vector field is zero throughout a region then line
integrals are independent of path in that region. To state it a bit more precisely, in a volume for which any closed

path can be shrunk to a point without leaving the region, if the curl of~vequals zero, then

∫b
a
F~.d~rdepends on
the endpoints of the path, and not on how you get there.
To see why this follows, take two integrals from pointato pointb.

∫

1

~v.d~r and

∫

2

~v.d~r

a b

1

2