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S
74 5. STOKES™ THEOREM

(6) Let S be any curve that connects the point (0, 0) to the point (3, 0). What is
ω? WHY???
S

Exercise 5.14. Calculate the volume of a ball of radius 1, {(ρ, θ, φ)|ρ ¤ 1}, by inte-
grating some 2-form over the sphere of radius 1, {(ρ, θ, φ)|ρ = 1}.

Exercise 5.15. Calculate
13
x3 dx + x + xy 2 dy
3
C

where C is the circle of radius 2, centered about the origin. Answer: 8π

Exercise 5.16. Suppose ω = x dx + x dy is a 1-form on R2 . Let C be the ellipse
y2
x2
+ = 1. Determine the value of ω by integrating some 2-form over the region
4 9
C
bounded by the ellipse. (Hint: the region bounded by the ellipse can be parameterized
by φ(r, θ) = (2r cos(θ), 3r sin(θ)), where 0 ¤ r ¤ 1 and 0 ¤ θ ¤ 2π.) Answer: 6π


4. Vector calculus and the many faces of Stokes™ Theorem

Although the language and notation may be new, you have already seen Stokes™
Theorem in many guises. For example, let f (x) be a 0-form on R. Then df = f ′ (x)dx.
Let [a, b] be a 1-cell in R. Then Stokes™ Theorem tells us

b

f ′ (x)dx = f ′ (x)dx = f (x) = f (b) ’ f (a)
f (x) =
a b’a
[a,b] ‚[a,b]

Which is, of course, the Fundamental Theorem of Calculus. If we let R be some
2-chain in R2 then Stokes™ Theorem implies

‚Q ‚P

P dx + Q dy = d(P dx + Q dy) = dx dy
‚x ‚y
‚R R R

This is what we call “Green™s Theorem” in Calculus. To proceed further, we
restrict ourselves to R3 . In this dimension there is a nice correspondence between
vector ¬elds and both 1- and 2-forms.
4. VECTOR CALCULUS AND THE MANY FACES OF STOKES™ THEOREM 75




1
” ωF = Fx dx + Fy dy + Fz dz
F = Fx , Fy , Fz
2
” ωF = Fx dy § dz ’ Fy dx § dz + Fz dx § dy

On R3 there is also a useful correspondence between 0-forms (functions) and
3-forms.
3
f (x, y, z) ” ωf = f dx § dy § dz

We can use these correspondences to de¬ne various operations involving functions
and vector ¬elds. For example, suppose f : R3 ’ R is a 0-form. Then df is the
1-form, ‚f dx + ‚f dy + ‚f dz. The vector ¬eld associated to this 1-form is then
‚x ‚y ‚z
‚f ‚f ‚f
. In calculus we call this vector ¬eld grad f , or ∇f . In other words, ∇f
,,
‚x ‚y ‚z
is the vector ¬eld associated with the 1-form, df . This can be summarized by the
equation
1
df = ω∇f

It will be useful to think of this diagrammatically as well.

grad
f ’ ’ ∇f
’’
¦
¦
f ’ ’ df
’’
d




Example 5.9. Suppose f = x2 y 3z. Then df = 2xy 3z dx+3x2 y 2z dy +x3 y 3 dz.
The associated vector ¬eld, grad f , is then ∇f = 2xy 3 z, 3x2 y 2 z, x3 y 3 .

1
Similarly, if we start with a vector ¬eld, F, form the associated 1-form, ωF ,
di¬erentiate it, and look at the corresponding vector ¬eld, then the result is called
1
curl F, or ∇—F. So, ∇—F is the vector ¬eld associated with the 2-form, dωF. This
can be summarized by the equation
1 2
dωF = ω∇—F

This can also be illustrated by the following diagram.
76 5. STOKES™ THEOREM


curl
F ’’ ∇—F
’’
¦
¦ ¦
¦
1 1
ωF ’ ’
’’ dωF
d



Example 5.10. Let F = xy, yz, x2 . The associated 1-form is then

ωF = xy dx + yz dy + x2 dz.
1


The derivative of this 1-form is the 2-form
1
dωF = ’y dy § dz + 2x dx § dz ’ x dx § dy.

The vector ¬eld associated to this 2-form is curl F, which is

∇ — F = ’y, ’2x, ’x .



Lastly, we can start with a vector ¬eld, F = Fx , Fy , Fz , and then look at the
3-form, dωF = ( ‚Fx + ‚Fy + ‚Fz )dx § dy § dz (See Exercise 4.12). The function,
2
‚x ‚y ‚z
+ ‚Fy + ‚Fz is called div F, or ∇ · F. This is summarized in the following
‚Fx
‚x ‚y ‚z
equation and diagram.
2 3
dωF = ω∇·F

div
F ’’ ∇·F
’’
¦
¦ ¦
¦
2 2
ωF ’ ’ dωF
’’
d



Example 5.11. Let F = xy, yz, x2 . The associated 2-form is then

ωF = xy dy § dz ’ yz dx § dz + x2 dx § dy.
2


The derivative is the 3-form
2
dωF = (y + z) dx § dy § dz.

So div F is the function ∇ · F = y + z.
4. VECTOR CALCULUS AND THE MANY FACES OF STOKES™ THEOREM 77

Two important vector identities follow from the fact that for a di¬erential form,
ω, calculating d(dω) always yields zero (see Exercise 4.8 of Chapter 4). For the ¬rst,
consider the following diagram.
grad curl
f ’ ’ ∇f ’ ’ ∇ — (∇f )
’’ ’’
¦ ¦
¦ ¦
f ’ ’ df ’ ’
’’ ’’ ddf
d d

This shows that if f is a 0-form then the vector ¬eld corresponding to ddf is
∇ — (∇f ). But ddf = 0, so we conclude

∇ — (∇f ) = 0

For the second identity, consider this diagram.
curl div
F ’ ’ ∇ — F ’ ’ ∇ · (∇ — F)
’’ ’’
¦ ¦
¦ ¦ ¦
¦
1 1 1
ωF ’ ’
’’ ’’
’’
dωF ddωF
d d
1
This shows that if ddωF is written as g dx § dy § dz then the function g is equal
1
to ∇ · (∇ — F). But ddωF = 0, so we conclude

∇ · (∇ — F) = 0

In vector calculus we also learn how to integrate vector ¬elds over parameterized
curves (1-chains) and surfaces (2-chains). Suppose ¬rst that σ is some parameterized
curve. Then we can integrate the component of F which points in the direction of
the tangent vectors to σ. This integral is usually denoted F · ds, and its de¬nition
σ
1
is precisely the same as the de¬nition we learned here for ωF . A special case of this
σ
1
integral arises when F = ∇f , for some function, f . In this case, ωF is just df , so the
∇f · ds is the same as
de¬nition of df .
σ σ
We also learn to integrate vector ¬elds over parameterized surfaces. In this case,
the quantity we integrate is the component of the vector ¬eld which is normal to the
surface. This integral is often denoted F · dS. Its de¬nition is precisely the same as
S
2
(see Exercises 2.20 and 2.21). A special case of this is when F = ∇—G,
that of ωF
S
78 5. STOKES™ THEOREM

2 1
(∇ — G) · dS must
for some vector ¬eld, G. Then ωG is just dωG , so we see that
S
1
be the same as dωG .
S
The most basic thing to integrate over a 3-dimensional region (i.e. a 3-chain),
„¦, in R3 is a function f (x, y, x). In calculus we denote this integral as f dV . Note
„¦
3
A special case is when f = ∇ · F, for some
that this is precisely the same as ωf .
„¦
(∇ · F)dV . But we can write this integral
vector ¬eld F. In this case f dV =
„¦ „¦
2
with di¬erential forms as dωF.
„¦
We summarize the equivalence between the integrals developed in vector calculus
and various integrals of di¬erential forms in the following table:

Vector Calculus Di¬erential Forms

1
F · ds ωF
σ σ
∇f · ds df
σ σ

2

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