StokesI Theorem
+
Our text states and proves Stokes' Theorem in 12.11, but ituses
the scalar form for writing both the line integral and the surface integral involved. In the applications, it is the vector form of the theorem that is most likely to be quoted, since the notations dxAdy =d the like are not in common use (yet) in physics and engineering.
Therefore we state and prove the vector form of the theorem here. The proof is the same as in our text, but not as condensed. --.I
Definition. Let F =
PT + QT + $
be a continuously differentiable
vector field defined in an open set U of R3
.
We define another vector
field in U, by the equation
curla =
-adz)+
i +
()R/ay
7
(JP/>Z
- aR/Jx)
,
.
+ (>Q/& -ap/Jy) 7 .
We discuss later the physical meaning of this vector field. An easy way to remember this definition is to introduce the symbolic operator "delM, defined by the equation md to note that
curl ?
=
*
curl F can be evaluated by computing the symbolic determinant
7'
7
+
V ~ F=
-9
det
Theorem. (Stokes1 theorem).
Let S be a simple smooth parametrized
surface in R , parametrized by a function g : T
S
,
where T is a
region in the (u,v) plane. Assume that T is a Green's region, bounded
by a simple closed piecewise-smooth curve D, and that
has continuous
second-order partial deri~tivesin an open set containing T ard D.
Let C be the curve L(D).
If F is a continuously differentiable vector field defined in
an open set of R~
containing S and C, then
J
( F a
Jc
--t
T) ds
-
JJs ((curl T)-$] d~ .
Here the orientation of C is that derived from the counterclockwise
orientation of D; and the normal
direction as 3rJu
Y
1gJv
fi
to the surface S points in the same
.
4
Remark 1. The relation between T and
2
is often described
informally as follows: "If youralkaround C in the direction specified by
3
with your head in the direction specified by ?i
,
then the surface S
is on your left." The figure indicates the correctness of this informal
description.
/
Remark 2. We note that the equation is consistent with a change of parametrization. Suppose that we reparametrize S by taking a function 9 : W -s T carrying a region in the (s,t) plane onto T, and use
new parametrization R(s,t) = g(q(s,t))
.
the
What happens to the integrals
/
in the statement of the th?orem? If det Dg
.
0
, then the left side of
the equation is unchanged, for we know that g carries the countercloclcwise orientation of
because
2~
to the counterclockwise orientation of
-
) R /as
x&/J t
=
2 T. Furthermore,
( J dx 3rJk) det Dg , the unit normal
determined by the parametrization g is the same as that determined by g
,
so the right side of the equation is also unchanged. Orr the other hand, if det D g i 0,
then the counterclockwise orientation
of J W goes to the opposite direction on C,
so that
7
changes
sign. But in that case, the unit normal determined by g is opposite to
that determined by
r.
Thus both sides of the equation change sign.
Proof of the theorem.
---
The proof consists of verifying the following three equations: m e theorem follo&
by adding these equations together. We shall in fact verify only the first equation. The others are proved similarly. Alternatively, if one makes the substitutions >
4
-+
3
-4
1
j and j + k
m.d k -+i m d x--sy and y + z
and z - e x ,
then each equation is transformed into the next one. This corresponds to
3
ar? orientation-presedng chmge of ~riablesin R , so it leaves the
orientations of C and S unchanged.
-3
Sa let F henceforth denote the vector field Pi ; we prove Stokes'
theorem in that case.
The idea of the proof is to express the line and surface integrals of the theorem as integrals over D and TI respectively, and then to apply Green's theorem to show they are equal. Let r(uIv) =
l(t)
4
.
(X(U.V)~ Y(u,v), Z(u,v)),
as usual. Choose a counterclockwise parametrization of D
for a 5 t
< b.
;
call it Then the function is the parametrization of C that we need to compute the line integral of our theorem.
k a We compute as follows: where 2 U u and aX&v
are evaluated at d_(t), of course. We can write
this as a line integral over D. Indeed, if we let p and q be the functions
ax
p(uIv)
= P(g(uIv) )*ru(ufv)
s(u,v)
= P(g(u,v) 1 )
8vru$-
ax
d
I
then our integral is just the line integral New by Green's theorem, this line integral equals I*)
(J~/JU
-
J , / ~ v ) du dv
.
We use the chain rule to compute the integrand. We have where P and its partials are evaluated at ~(u,v), of course.
Subtracting, we see that the first and last terms cancel each other.
The double integral ( * ) then takes the form
Nclw we compute the surfac~integral of our theorem.
T->P/>~
2 , formula
3
curl F = > P / ~ Z
Since
(12.20) on p. 435 of our text
tells us we have Here
2P/> z
and 2 P/3y
-
are e~luateda t r(u,v) , of course.
,
Our theorem is thus proved. Exercises
Let
1. S
on tfie diverqence
theorem z = 9
be the portion of the surface
-
x2
-
lying y2
above t h e xy plane. Let 8 be the unit upward normal to S. Apply t h e divergence theorem'to the solid bounded by S and the
J'I P
xy-plane to evaluate
S
(a)
1 = ~ i n ( ~ + z ) ?+
2. I
D =
Let
S2
S1
2
)d.
(b)
8ln. denote t h e surface
z = 1
let
t h e unit normal to
y o d l d S and
S1
2
if:
denote t h e unit d i s c
(2wcy)f + zk;
d~
s X Z J + ( x +y
y 2 z 3 + yf + zR.
Anewers:
(a) 8 1 ~ / 2 *
(b)
il
3,
x2
+
y2
-
x
2
-
z = 0
1
Let
b e t h e unit normal t o
both with positive
S2,
Jj
1
s2 * a 2 d s .
Y 2 ; z 2 0.
d
S,
= xt
and let
component.
Let
3,
be
Evaluate
- - -
Grad 'Curl Div and all that.
-'
We study two questions about these operations: DO they have natural (i.e., coordinate-free) physical I.
or geometric interpretations? What is the relation between them? 11.
We already have a natural interpretation of the I.
gradient. For divergence. the question is answered in 12.20 of
Apostol.
The theorem of that section gives a coordinate-free
-+
1
definition of divergence F.
and the subsequent discussion +
gives a physical interpretation, in the case where F is the flux density vector of a moving fluid. Apostol treats curl rather more briefly.
i
Formula (12.62) on p. 461 gives a coordinate-free expression for +
r i -
curl F (a)
- , as follows:
where
-L
is the circle of radius r
C(r)
lying in the plane perpendicular to
the point
a,
and
C(r)
centered at
-a
and passing through
is directed in a counterclockwise
direction when viewed from the tip of
!
called the circulation of
5
at a
-
+
n.
This number is +
-
around the vector
it is clearly independent of coordinates.
n;
Then one has a -b
coordinate-free definition of
+
curl F
at
-a
curl F
as follows: points in the direction of the vector +
around which the circulation of
F
is a maximum, and its magnitude equals this maximum circulation. YOU
will note a strong analogy here with the relation between the gradient and the directional derivative. ,
-+
For a physical interpretation of
+
imagine F
curl F,
let us to be the velocity vector field of a moving fluid. I
in the fluid, Let us place a small paddle wheel of radius P
-+
with its axis along n.
Eventually, the paddle wheel settles down to rotating steadily with angular speed
w
(considered as positive if counterclockwise as biewed from the tip of if). + +
The tangential component F * T of velocity will tend to in- crease the speed
w
if it is positive and to decrease
o
if i t is n e g a t i v e .
On p h y s i c a l grounds, i t i s r e a s o n a b l e t o
1
suppose t h a t
average v a l u e of
+
-b
( F - T ) = -
,
speed of a p o i n t
on one of t h e p a d d l e s
That is,
&I
C~
+
+
T ds = r w .
I t follows t h a t
w e have ( i f
s o t h a t by formula
d
+
i s very s m a l l )
,
-
c u r l F ( a ) = 2w.
+
I n p h y s i c a l terms t h e n , t h e v e c t o r k u r l F(=)]
points i n the
d i r e c t i o n of t h e a x i s around which o u r p a d d l e wheel s p i n s
most r a p i d l y (in a c o u n t e r c l o c k w i s e d i r e c t i o n ) , and i t s magnit u d e e q u a l s t w i c e this maximum a n g u l a r speed.
What a r e t h e r e l a t i o n s between t h e o p e r a t i o n s
11.
,
grad, c u r l , and d i v ?
Grad goes
goes from v e c t o r
Here i s one way of e x p l a i n i n g them.
-
from s c a l a r f i e l d s t o v e c t o r f i e l d s , C url
-
f i e l d s t o v e c t o r f i e l d s , and Div goes from
vector f i e l d s t o s c a l a r f i e l d s .
his i s s u m a r i z e d i n t h e
diagram:
0 (5)
Scalar f i e l d s
Vector f i e l d s
curl
1
Vector f i e l d s
a
Scalar f i e l d s
Jl
I ,(
(XI
-
L e t us c o n s i d e r f i r s t t h e t o p two o p e r a t i o n s ,
curl.
grad
and
W e r e s t r i c t o u r s e l v e s t o s c a l a r and v e c t o r f i e l d s t h a t
a r e c o n t i n u o u s l y d i f f e r e n t i a b l e on a r e g i o n
U
of
R
3
.
Here i s a theorem w e have a l r e a d y proved:
Theorem 1.
r
-L
-=
F * du
0
+
F
-i s-a
-
gradient i n
f o r every closed
+
Theorem 2.
If
F = grad
~ r o o f . W e compute
curl
curl
U
if and
-
piecewise-smooth p a t h
4
--
3
by t h e formula
f o r some
4,
i n U.
-
only i f
then
+
+
c u r l F = 0.
$
w e know t h a t i f
continuous, t h e n
i s a g r a d i e n t , and t h e p a r t i a l s o f
D.F
1
+
-!-
= D.F
j J i
f o r all
i , j.
are
F
Hence
curlF=O.
If curl
= 8 i n-a- star-convex r e g i o n U ,
t h e n $ = grad $ -f o r some + d e f i n e d i n U.
The f u n c t i o n $(XI
= + ( x ) + c i s t h e most g e n e r a l f uncTheorem 3 .
_I--
t i o n such t h a t
--Proof.
j.
If
U
-b
F = grad $. If
+
+
c u r l F = 0,
D.F = D.F
I j I i
i s star-convex, t h i s f a c t i m p l i e s t h a t
by t h e Poincar6 l e m m a .
g r a d i e n t i n U,
Theorem 4 .
The c o n d i t i o n
-
c u r l $ =
does
ot
-n-
d
in
-
-
circle
F
i,
is a
17 3
-i s-a
gradient i n U.
Consider t h e v e c t d r f i e l d
It is defined i n t h e region
cept f o r the
for a l l
u
i n g e n e r a l imply t h a t
Proof.
TO show
then
z-axis.
U
c o n s i s t i n g of a l l of
I t i s e a s y t o check t h a t
is not a gradient i n
U,
we let
C
curl
R
?
3
ex-
=
8.
be t h e u n i t
~ ( t=) (cos t , s i n t , 0 ) ;
xy-plane,
i n the
rt
and compute
f o l l o w s from Theorem
Remark.
]I
A region
that
U
in
$
cannot be a g r a d i e n t i n
U.
i s c a l l e d "simply
R~
connected" i f , roughly speaking, e v e r y c l o s e d curve i n
bounds an o r i e n t a b l e s u r f a c e l y i n g i n
U.
The r e g i o n
U
R3
-
( o r i g i n ) i s simply connected, f o r example, b u t t h e r e g i o n
R
3
- (z-axis)
i s not.
rt t u r n s o u t t h a t i f
,
curl $ =
8
in
U,
U
i s simply connected and i f
is a gradient i n
then
U.
The proof
goes roughly a s follows:
Given a c l o s e d c u r v e
able surface i n
t o t h a t surface.
U
which
in
U,
bounds.
let
S
be an o r i e n t -
Apply S t o k e s t theorem
One o b t a i n s t h e e q u a t i o n
Then Theorem 1 shows t h a t
NOW
C
C
P
i s a gradient i n
U.
l e t us consider t h e next t w o operations,
curl
and
Again, w e c o n s i d e r o n l y f i e l d s t h a t a r e c o n t i n u o u s l y
3
d i f f e r e n t i a b l e i n a region U of R
There a r e analogues of
div.
.
a l l t h e e a r l i e r theorems:
I
0
Theorem 5 .
11 5 t
dS = 0
S If F i s- a- c-u r l i n U, then f o r every orientable closed surface
-
~roof. Let
in
-
S
-
be a c lased s u r f a c e t h a t l i e s i n
S
(While we assume t h a t
in
U
U,
U.
S
lies
w e do n o t assume t h a t
includes t h e
region t h a t
S
)-dimensional
bounds.)
Break
s up i n t o two s u r f a c e s S1
S2
U. and
that intersect in their
common boundary, which i s a simple smooth c l o s e d curve C. Now
+by h y p o t h e s i s , G = c u r l ? f o r some
d e f i n e d i n U. W e cornpute :
n ds =
G
11
S2
Adding, w e see t h a t
ema ark.
curl
s2
.
F m ii d~
=
-
1 Hm -
da.
C
,
The convers'e of Theorem 5 h o l d s a l s o , b u t w e
s h a l l n o t a t t e m p t t o prove it.
Theorem 6 .
P roof.
-
If
-
2
= curl
By assumption,
?
f o r some
--
?,
t h e n d i v 2 = 0.
-
Then
div
-+
G =
(D1D2F3-D1D 3F 21 - (D2D1F3-DZD3Fl)+(D3DlF2-D3D2F1)
If div
= 0 in-a- star-convex region U,
+
then 2 = curl 3 -for some F defined in U.
+
+
The function H = F + grad $ --is the most general func+
tion such that
= curl H.
--Theorem 7.
We shall not prove this theorem in full generality.
The
proof is by direct computation, as in the ~oincar6lemma.
when
A proof that holds when U is a 3-dimensional box, or
U is all of R3, is given in section 12.16 of Apostol.
This proof also shows how to construct a specific such function
in the given cases. Note that if
+ +
curl(H-F) =
for some
8.
= curl ?$ and
Hence by Theorem 3,
-+
-+
G = curl H,
H
-
5
then
= grad 4
4.
Theorem 8.
-t
The condition
-
divG=O
---
2
&
I U
- if -is-a- curl
- in
does not in general imply that
U. in U,
Proof.
Let
2
be t h e v e c t o r f i e l d
which i s d e f i n e d i n t h e r e g i o n
except f o r t h e o r i g i n .
that
If
S
U
c o n s i s t i n g of a l l of
R3
One r e a d i l y shows by d i r e c t computation
d i v G = 0.
i s t h e u n i t s p h e r e c e n t e r e d a t t h e o r i g i n , t h e n w e show
that
his w i l l imply (by Theorem 5 ) t h a t
(x,y,z) i s a p o i n t o f S t t h e n
+
+
+
+ *
G(x,y,z) = xi + yj + zk = n o T h e r e f o r e
If
so
is n o t a c u r l .
11 b*;
dA =
I (x,y,z)B = 1,
1 dA = ( a r e a of s p h e r e ) # 0.
0
erna ark.
Suppose w e s a y t h a t a r e g i o n
simply c o n n e c t e d w i f e v e r y c l o s e d s u r f a c e i n
s o l i d region l y i n g i n
U
bounds a
U = R -(origin)
simply connected", f o r example, b u t t h e r e g i o n
i s "two-
R3
3
The r e g i o n
U.*
in
U
i s nott'twoaxis)
U = R3-(z
is.
-
rt
-b
div G = 0
I1
U is"two-simply connected and i f
turns out that i f
in
U,
then
2
is a curl in
The proof goes U.
roughly a s follows: Given a c l o s e d s u r f a c e
it bounds.
Since
S
in
U,
let
V
be t h e r e g i o n
i s by h y p o t h e s i s d e f i n e d on a l l of
V,
w e can apply Gauss' theorem t o compute
Then t h e converse of Theorem 5 implies. t h a t
+
G
is a c u r l i n
U.
There i s much more one can s a y a b o u t t h e s e matters, b u t
one needs t o i n t r o d u c e a b i t of a l g e b r a i c topology i n o r d e r t o do
so.
It i s a b i t l a t e i n t h e semester f o r t h a t !
*The proper mathematical term for t h i s is ffharalogicallyt r i v i a l in dimension two."
MIT OpenCourseWare
http://ocw.mit.edu
18.024 Multivariable Calculus with Theory
Spring 2011
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