Mathematical Tools for Physics

13—Vector Calculus 2 405

Multiply and divide each term in the sum ( 18 ) by∆Akand you have

∑

k

[

1

∆Ak

∮

Ck

~v.d~`

]

∆Ak=

∮

C

~v.d~` (19)

Now increase the number of subdivisions of the surface, finally taking the limit as all the∆Ak→ 0 , and the
quantity inside the brackets becomes the normal component of the curl of~vby Eq. ( 17 ). The limit of the sum
is the definition of an integral, so

Stokes’ Theorem:

∫

A

curl~v.dA~=

∮

C

~v.d~` (20)

What happens if the vector field~vis the gradient of a function,~v=∇f? By Eq. ( 11 ) the line integral
in ( 20 ) depends on only the endpoints of the path, but in this integral the initial and final points are the same.
That makes the integral zero: f 1 −f 1. That implies that the surface integral on the left is zero no matter
what the surface spanning the contour is, and that can happen only if the thing being integrated is itself zero.
curl gradf= 0. That’s one of the common vector identities in problem9.36. Of course this statement requires
the usual assumption that there are no singularities of~vwithin the area.

^

q 0

Example n
Verify Stokes’ theorem for that part of a spherical surfacer=R, 0 ≤θ≤θ 0 , 0 ≤φ <
2 π. Use for this example the vector field

F~=ˆrAr^2 sinθ+θBrθˆ^2 cosφ+φCrˆ sinθcos^2 φ (21)

To compute the curl ofF~, use Eq. (9.29), getting

∇×F~=ˆr

1

rsinθ

(

∂

∂θ

(

sinθ Crsinθcos^2 φ

)

−

∂

∂φ

(

Brθ^2 cosφ

)

)

+···

=ˆr

1

rsinθ

(

Crcos^2 φ2 sinθcosθ+Brθ^2 sinφ

)

+···