Undergraduate Analysis
Handout 2
October 12, 2011
Definition. Let U ⊂ Rn be open and bounded, k ∈ {1, 2, . . .}. We say ∂U is C k if for each p ∈ ∂U
there exist r > 0 and a C k function g : Rn−1 → R such that, upon relabeling and reorienting the
coordinates axes if necessary, we have
U ∩ Br (p) = {x = (x1 , . . . , xn ) ∈ Br (p) | xn > g(x1 , . . . , xn−1 )}
and
∂U ∩ Br (p) = {x = (x1 , . . . , xn ) ∈ Br (p) | xn = g(x1 , . . . , xn−1 )}
Likewise, ∂U is C ∞ if ∂U is C k for k = 1, 2, . . . , and ∂U is analytic if the mapping g is analytic.
Definitions. (1) If ∂U is C 1 , then along ∂U is defined the outward pointing unit normal vector field
ν = (ν 1 , . . . , ν n ).
The unit normal at any point p ∈ ∂U is ν(p) = ν = (ν 1 , . . . , ν n ).
(2) Let u ∈ C 1 (Ū ). We call
∂u
:= ν · Du
∂ν
the (outward) normal derivative of u.
We will frequently need to change coordinates near a point of ∂U so as to “flatten out” the boundary.
To be more specific, fix p ∈ ∂U, and choose r, g, etc. as above. Define then
(
yi = xi =: Φi (x)
(i = 1, . . . , n − 1)
yn = xn − g(x1 , . . . , xn−1 ) =: Φn (x),
and write
y = Φ(x).
Similarly, we set
(
xi = yi =: Ψi (y)
(i = 1, . . . , n − 1)
xn = yn + g(y1 , . . . , yn−1 ) =: Ψn (x),
and write
x = Ψ(y).
Then Φ = Ψ−1 , and the mapping x → Φ(x) = y “straightens out ∂U ” near p. Observe also that
det dΦ = det dΨ = 1.
Divergence Theorem. Let R be a regular region in R3 with piecewise smooth boundary ∂R.
Suppose that F = (F1 , F2 , F3 ) is a vector field of class C 1 (R) ∩ C 0 (R̄). Then
Z
Z
div F dV =
F · n , dA,
R
∂R
where n = n(x) denotes the unit outward normal vector to ∂R at x ∈ ∂R.
Let W = {(s, t, u) ∈ R3 | a ≤ s ≤ b, c ≤ t ≤ d, e ≤ u ≤ f }, i.e. a closed cell in R3 . Partition the
region R into nonoverlapping subregions {Ri }m
i=1 , and, for each i = 1, . . . , m, let Φi : W → Ri be
1
1 − 1, onto, and belong to the class C (W ) with nonsingular dΦi |W . Thus, we have
∂Ri = ∂Φi (W ) = Φi (∂W ) for each i = 1, . . . , m,
Undergraduate Analysis
Handout 2 (Continued)
October 12, 2011
and
Z
Z
div F dV
=
R
div F dV =
Sm
i=1
=
m Z
X
i=1
m
X
¡ ¢
div F Φi dV (Φi ) =
i=1
Z
F · n dA =
Sm
∂R
i=1
F · n dA =
∂Ri
m Z
X
=
i=1
div F dV =
Ri
W
i=1
Z
Ri
m Z
X
m Z
X
i=1
m Z
X
i=1
Z
W
div F dV
Φi (W )
¡ ¢
div F Φi |JΦi | ds dt du
F · n dA =
∂Ri
m Z
X
i=1
F · n dA
Φi (∂W )
¡ ¢
F · n Φi dA(Φi )
∂W
¡ ¢
Note: In the last integral, the formula of n Φi dA(Φi ) varies along the faces of ∂W. In fact, we have

¡
¢
∂Φi ∂Φi


∓
×
dt
du
along
the
faces
∂W
∩
{s
=
a}
∪
{s
=
b}
= S1 ∪ S2 ,


∂t
∂u






¡ ¢
¡
¢
∂Φi ∂Φi
n Φi dA(Φi ) =
∓
×
du ds along the faces ∂W ∩ {t = c} ∪ {t = d} = S3 ∪ S4 ,

∂u
∂s







¡
¢

 ∓ ∂Φi × ∂Φi ds dt along the faces ∂W ∩ {u = e} ∪ {u = f } = S5 ∪ S6 ,
∂s
∂t
Furthermore, if we let
i = (1, 0, 0), j = (0, 1, 0), k = (0, 0, 1)
~ also varies along the faces of ∂W. In fact, we have
then the surface area vector dS


 ∓ i dt du along S1 ∪ S2 ,
~=
dS
∓ j du ds along S3 ∪ S4 ,


∓ k ds dt along S5 ∪ S6 .
Also, we define the vector field G = (G1 , G2 , G3 ) by
G1 = F ·
Then we have
∂Φi ∂Φi
∂Φi ∂Φi
∂Φi ∂Φi
×
, G2 = F ·
×
, G3 = F ·
×
.
∂t
∂u
∂u
∂s
∂s
∂t
Z
Z
¡ ¢
F · n Φi dA(Φi ) =
F · n dA =
∂Ri
Z
~
G · dS.
∂W
∂W
Next, if we let dW = ds dt du denote the volume element of W, then we need to show that
¡ ¢
div G dW = div F Φi |JΦi | ds dt du
If we let
r0 = Φi (s0 , t0 , u0 ), and a =
∂Φi
∂Φi
∂Φi
|r0 , b =
|r0 , c =
|r ,
∂s
∂t
∂u 0
Page 2
Undergraduate Analysis
Handout 2 (Continued)
October 12, 2011
then
∂G1 ∂G2 ∂G3
+
+
∂s
∂t
∂u
∂F ∂r ∂r
∂F ∂r ∂r
∂F ∂r ∂r
=
·
×
|r 0 +
·
× |r0 +
·
×
|r
∂s ∂t ∂u
∂t ∂u ∂s
∂u ∂s ∂u 0
¡
¢ ∂F
¡
¢ ∂F
¡
¢
∂F
|r0 · b × c +
|r0 · c × a +
|r0 · a × b
=
∂t
¡∂s
¢ ¡
¢ ∂u¡
¢ ¡
¢
= ∇F1 · a, ∇F2 · a, ∇F3 · a · b × c + ∇F1 · b, ∇F2 · b, ∇F3 · b · c × a
¡
¢ ¡
¢
+ ∇F1 · c, ∇F2 · c, ∇F3 · c · a × b
div G|r0 =
3
X
£¡
¢
¡
¢
¡
¢ ¤
∂Fi
Fi,j b × c i aj + c × a i bj + a × b i cj where Fi,j =
∂xj
i,j=1
X
£¡
¢
¡
¢
¡
¢ ¤ X
£¡
¢
¡
¢
¡
¢ ¤
Fi,j b × c i aj + c × a i bj + a × b i cj +
Fi,j b × c i aj + c × a i bj + a × b i cj
=
=
i=j
=
3
X
i6=j
£¡
¢
¡
¢
¡
¢ ¤
Fi,i b × c i ai + c × a i bi + a × b i ci
i=1
¯
¯
¯a1 a2 a3 ¯
¯
¯
=
Fi,i ¯¯ b1 b2 b3 ¯¯
¯ c1 c2 c3 ¯
i=1
¡ ¢
= div F Φi |JΦi |
3
X
Note: in the last ¯three equalities,
we have used that if {i, j, k} = {1, 2, 3}, i.e. i, j, k are all distinct,
¯
¯ bj bk ¯
¡
¢
¯ . Thus, we can conclude that
then b × c i = ± ¯¯
cj ck ¯
¯
¯
¯aj aj ak ¯
¯
¯
¡
¢
¡
¢
¡
¢
(a) if i 6= j, then b × c i aj + c × a i bj + a × b i cj = ± ¯¯ bj bj bk ¯¯ = 0.
¯ cj cj ck ¯
¯
¯
¯a1 a2 a3 ¯
¯
¯
¡
¢
¡
¢
¡
¢
(b) if i = j, then b × c i ai + c × a i bi + a × b i ci = ¯¯ b1 b2 b3 ¯¯ .
¯ c1 c2 c3 ¯
n
Remark: Let n ≥ 2, R be a regular region in R , an x1 x2 · · · xn −space, with piecewise smooth
boundary. Suppose that F is a continuously differentiable vector field of class C 1 on R̄, and Φ :
W → R is a local coordinates of R defined on a closed cell W, in ¡the s1 s2 · · · sn −space.
On the face
¢
{si = constant } ∩ ∂W, i = 1, . . . , n, we define the vector ν(x) = ν1 (x), . . . , νn (x) by
νj (x) = (−1)j+i Mij ,
for j = 1, . . . , n,
where Mij denotes the determinant of the minor matrix of dΦt with ith row, j th column removed. It
is easy to check that
∂Φ
= 0 for each j 6= i,
ν ·
∂sj
¡ ¢
i.e. ν is a normal vector to ∂W ∩ {si = constant }. Thus we find the area element ν Φi dA(Φi )
along each face of ∂W. Furthermore, if we set
¢
¡
Gi = F · (−1)1+i Mi1 , . . . , (−1)n+i Min , for each i = 1, . . . , n,
one can prove the following
Page 3
Undergraduate Analysis
Handout 2 (Continued)
October 12, 2011
Divergence Theorem. Let R be a regular region in Rn with piecewise smooth boundary ∂R.
Suppose that F = (F1 , . . . , Fn ) is a vector field of class C 1 (R) ∩ C 0 (R̄). Then
Z
Z
div F dV =
F · ν dA
R
∂R
where ν = ν(x) denotes the unit outward normal vector to ∂R at x ∈ ∂R.
Corollary (Green’s Theorem). Let R be a regular region in R2 = x1 x2 − plane with piecewise
smooth boundary ∂R, and let F = (F1 , F2 ) : R̄ → R2 be a vector field of class C 1 on R̄. Then
Z
Z
¡ ∂F2 ∂F1 ¢
F · dx =
−
dA.
∂x2
∂R
R ∂x1
Proof. Note that if ~r(t) : [a, b] → ∂R is a parametrization (or coordinate functions) of ∂R, then
Z
Z
Z b
¡
¢ ¡
¢
¡
¢ ¡ dx1 dx2 ¢
F · dx =
F1 , F2 · dx1 , dx2 =
F1 , F2 ·
,
dt.
dt dt
∂R
~
r
a
¡ dx1 dx2 ¢
¡ dx2 dx1 ¢
Since ~r 0 (t) =
,
is the tangent vector to ∂R at p = ~r(t), then ν(p) =
,−
is a
dt dt
dt
dt
normal vector there.
¡
¢ ¡
¢
Let G = G1 , G2 = F2 , −F1 . Then G is a vector field of class C 1 on R̄, and
Z
Z
Z
Z
¡
¢ ¡
¢
¡
¢ ¡
¢
F · dx =
F1 , F2 · dx1 , dx2 =
F2 , −F1 · dx2 , −dx1 =
∂R
∂R
∂R
b
¡
¢
G1 , G2 · ν dt,
a
and, by the divergence theorem, we have
Z
b
¡
¢
Z
Z
G1 , G2 · ν dt =
a
Z
G·ν=
∂R
div G dA =
R
¡ ∂F2 ∂F1 ¢
−
dA.
∂x2
R ∂x1
¥
Stokes’ Theorem. Let R ⊂ R2 be abounded, open region in s1 s2 −plane with smooth boundary
∂R, and let M ⊂ R3 be an oriented surface in x1 x2 x3 −space with boundary ∂M. Suppose that
F : M̄ → R3 is a continuously differentiable vector field defined on M., and ~r : R̄ → M ⊂ R3 is a
smooth parametrization that maps R to M, and ∂R to ∂M. Then
Z
Z
Z
Z
¡
¢
~
curl F · dA =
∇ × F · n dA =
F · T ds =
F · d~r,
M
M
∂M
∂M
where n = n(x) denotes the unit outward normal vector to the surface M ⊂ R3 at x ∈ M, ds is the
d~r
arc-length element of ∂M, and T =
is the tangent vector to ∂M at x ∈ ∂M.
ds
Proof. Let ~r(s1 , s2 ) = (x1 (s1 , s2 ), x2 (s1 , s2 ), x3 (s1 , s2 )) : R → M. Then


∂x1 ∂x1
 ∂s1 ∂s2  µ ¶
 ∂x ∂x 
ds1
 2
2
d~r|∂R = 

 ∂s1 ∂s2  ds2
 ∂x3 ∂x3 
∂s1 ∂s2
2
2
2
X
X
¡X
∂x1
∂x2
∂x3 ¢t
=
dsi ,
dsi ,
dsi
∂s
∂s
∂s
i
i
i
i=1
i=1
i=1
¡ ∂x1 ∂x2 ∂x3 ¢t
¡ ∂x1 ∂x2 ∂x3 ¢t
=
,
,
ds1 +
,
,
ds2
∂s1 ∂s1 ∂s1
∂s2 ∂s2 ∂s2
∂~r
∂~r
=
ds1 +
ds2
∂s1
∂s2
Page 4
Undergraduate Analysis
Handout 2 (Continued)
October 12, 2011
and
Z
Z
¡
F · d~r =
∂M
∂R
F ·
¢
∂~r
∂~r
ds1 + F ·
ds2 .
∂s1
∂s2
¡
If we define a 2−dimensional vector field G = G1 , G2 ) on the s1 s2 −plane by
G1 = F ·
then
Z
Z
¡
¢
G1 ds1 + G2 ds2 =
F · d~r =
∂M
∂~r
∂~r
and G2 = F ·
,
∂s1
∂s2
∂R
Z
¡ ∂G2 ∂G1 ¢
−
ds1 ds2 ,
∂s2
R ∂s1
where we have used the Green’s Theorem in the last equality. On the other hand, since
Z
Z
∂~r
∂~r
~
×
ds1 ds2
curl F · dA =
curl F |~r ·
∂s1 ∂s2
M
R
Z
Z
¡ ∂G2 ∂G1 ¢
and this integral is equal to
−
ds1 ds2 =
F · d~r since
∂s2
∂M
R ∂s1
curl F ·
∂~r
∂~r
×
∂s1 ∂s2
¯
¯
¯ ∂F3 ∂F2 ∂F1 ∂F3 ∂F2 ∂F1 ¯
¯
¯
¯ ∂x2 − ∂x3 ∂x3 − ∂x1 ∂x1 − ∂x2 ¯
¯
¯
∂x1
∂x2
∂x3
¯
¯
= ¯
¯
¯
¯
∂s1
∂s1
∂s1
¯
¯
∂x
∂x
∂x
1
2
3
¯
¯
¯
¯
∂s2
∂s2
∂s2
X ¡ ∂Fj
∂Fi ¢ ∂xi ∂xj
−
=
∂xi
∂xj ∂s1 ∂s2
i,j
X ∂Fj ∂xi ∂xj X ∂Fi ∂xj ∂xi
=
−
∂xi ∂s1 ∂s2
∂xj ∂s2 ∂s1
i,j
i,j
∂F
∂~r
∂F
∂~r
·
−
·
∂s1 ∂s2 ∂s2 ∂s1
∂G2 ∂G1
−
=
∂s1
∂s2
=
Page 5
¥
68
Theory Supplement Section M
M
PROOF OF THE DIVERGENCE THEOREM AND STOKES’ THEOREM
In this section we give proofs of the Divergence Theorem and Stokes’ Theorem using the definitions
in Cartesian coordinates.
Proof of the Divergence Theorem
Let F~ be a smooth vector field defined on a solid region V with boundary surface A oriented
outward. We wish to show that
Z
Z
~ =
div F~ dV.
F~ · dA
V
A
For the Divergence Theorem, we use the same approach as we used for Green’s Theorem; first
prove the theorem for rectangular regions, then use the change of variables formula to prove it for
regions parameterized by rectangular regions, and finally paste such regions together to form general
regions.
Proof for Rectangular Solids with Sides Parallel to the Axes
Consider a smooth vector field F~ defined on the rectangular solid V : a ≤ x ≤ b, c ≤ y ≤ d,
e ≤ z ≤ f . (See Figure M.50). We start by computing the flux of F~ through the two faces of V
perpendicular to the x-axis, A1 and A2 , both oriented outward:
Z
A1
~ +
F~ · dA
Z
A2
~ =−
F~ · dA
=
Z fZ
e
Z fZ
e
d
F1 (a, y, z) dy dz +
e
c
d
c
Z fZ
d
F1 (b, y, z) dy dz
c
(F1 (b, y, z) − F1 (a, y, z)) dy dz.
By the Fundamental Theorem of Calculus,
F1 (b, y, z) − F1 (a, y, z) =
so
Z
~ +
F~ · dA
A1
Z
A2
~ =
F~ · dA
Z fZ dZ
c
e
b
a
Z
b
a
∂F1
dx,
∂x
∂F1
dx dy dz =
∂x
Z
V
∂F1
dV.
∂x
By a similar argument, we can show
Z
A3
~ +
F~ · dA
Z
A4
~ =
F~ · dA
Z
V
∂F2
dV
∂y
and
Z
A5
~ +
F~ · dA
Z
A6
~ =
F~ · dA
Adding these, we get
Z
A
~ =
F~ · dA
Z V
∂F2
∂F3
∂F1
+
+
∂x
∂y
∂z
This is the Divergence Theorem for the region V .
dV =
Z
div F~ dV.
V
Z
V
∂F3
dV.
∂z
Theory Supplement Section M
69
A1 (back)
z
V
A6
A3
(back
left)
-
A4
A2
6
S3
(back
left)
-
S4
S2
?
V
A6
S1
(back
right) A3 (left)
A4
-
A2
y
x
t
6
s
A5 (bottom)
Figure M.50: Rectangular solid V in
xyz-space
W
S6
A1
(back
right)
y
x
z
u
6
S5 (bottom)
A5 (bottom)
Figure M.51: A rectangular solid W in stu-space and the corresponding
curved solid V in xyz-space
Proof for Regions Parameterized by Rectangular Solids
Now suppose we have a smooth change of coordinates
x = x(s, t, u),
y = y(s, t, u),
z = z(s, t, u).
Consider a curved solid V in xyz-space corresponding to a rectangular solid W in stu-space. See
Figure M.51. We suppose that the change of coordinates is one-to-one on the interior of W , and
that its Jacobian determinant is positive on W . We prove the Divergence Theorem for V using the
Divergence Theorem for W .
Let A be the boundary of V . To prove the Divergence Theorem for V , we must show that
Z
Z
~ =
F~ · dA
div F~ dV.
A
V
First we express the flux through A as a flux integral in stu-space over S, the boundary of the
rectangular region W . In vector notation the change of coordinates is
~r = ~r (s, t, u) = x(s, t, u)~i + y(s, t, u)~j + z(s, t, u)~k .
The face A1 of V is parameterized by
c ≤ t ≤ d,
~r = ~r (a, t, u),
so on this face
~ =±
dA
e ≤ u ≤ f,
∂~r
∂~r
×
dt du.
∂t
∂u
~ point outward, we must choose the negative sign. (Problem 3 on
In fact, in order to make dA
page 73 shows how this follows from the fact that the Jacobian determinant is positive.) Thus, if S 1
is the face s = a of W ,
Z
Z
∂~r
∂~r
~ =−
×
dt du,
F~ ·
F~ · dA
∂t
∂u
S1
A1
~ = −~i dt du. Therefore, if we choose a vector field
The outward pointing area element on S1 is dS
~
G on stu-space whose component in the s-direction is
G1 = F~ ·
we have
Z
A1
∂~r
∂~r
×
,
∂t
∂u
~ =
F~ · dA
Z
S1
~ · dS
~.
G
70
Theory Supplement Section M
~ by
Similarly, if we define the t and u components of G
∂~r
∂~r
G2 = F~ ·
×
∂u
∂s
then
Z
Ai
~ =
F~ · dA
Z
∂~r
∂~r
G3 = F~ ·
×
,
∂s
∂t
and
Si
~ · dS
~,
G
i = 2, . . . , 6.
(See Problem 4.) Adding the integrals for all the faces, we find that
Z
Z
~
~
~ · dS
~.
F · dA =
G
A
S
Since we have already proved the Divergence Theorem for the rectangular region W , we have
Z
Z
~ dW,
~
~
div G
G · dS =
W
S
where
~ = ∂G1 + ∂G2 + ∂G3 .
div G
∂s
∂t
∂u
Problems 5 and 6 on page 73 show that
∂(x, y, z) ∂F1
∂G1
∂G2
∂G3
∂F2
∂F3
.
+
+
=
+
+
∂s
∂t
∂u
∂(s, t, u) ∂x
∂y
∂z
So, by the three-variable change of variables formula on page 61,
Z Z
∂F1
∂F2
∂F3
div F~ dV =
+
+
dx dy dz
∂x
∂y
∂z
V
V
Z ∂F2
∂F3 ∂(x, y, z) ∂F1
+
+
ds dt du
=
∂x
∂y
∂z ∂(s, t, u) W
Z ∂G1
∂G2
∂G3
=
ds dt du
+
+
∂s
∂t
∂u
W
Z
~ dW.
div G
=
W
In summary, we have shown that
Z
A
and
Z
~ =
F~ · dA
div F~ dV =
V
Z
Z
S
~ · dS
~
G
~ dW.
div G
W
By the Divergence Theorem for rectangular solids, the right-hand sides of these equations are equal,
so the left-hand sides are equal also. This proves the Divergence Theorem for the curved region V .
Pasting Regions Together
As in the proof of Green’s Theorem, we prove the Divergence Theorem for more general regions
by pasting smaller regions together along common faces. Suppose the solid region V is formed by
pasting together solids V1 and V2 along a common face, as in Figure M.52.
The surface A which bounds V is formed by joining the surfaces A1 and A2 which bound V1
and V2 , and then deleting the common face. The outward flux integral of a vector field F~ through
A1 includes the integral across the common face, and the outward flux integral of F~ through A2
Theory Supplement Section M
71
Common face
N
V2
V1
Figure M.52: Region V formed by
pasting together V1 and V2
includes the integral over the same face, but oriented in the opposite direction. Thus, when we add
the integrals together, the contributions from the common face cancel, and we get the flux integral
through A. Thus we have
Z
Z
Z
~ =
~ +
~.
F~ · dA
F~ · dA
F~ · dA
A
But we also have
Z
A1
div F~ dV =
V
Z
A2
div F~ dV +
V1
Z
div F~ dV.
V2
So the Divergence Theorem for V follows from the Divergence Theorem for V 1 and V2 . Hence we
have proved the Divergence Theorem for any region formed by pasting together regions that can be
smoothly parameterized by rectangular solids.
Example 1
Let V be a spherical ball of radius 2, centered at the origin, with a concentric ball of radius 1
removed. Using spherical coordinates, show that the proof of the Divergence Theorem we have
given applies to V .
Solution
We cut V into two hollowed hemispheres like the one shown in Figure M.53, W . In spherical
coordinates, W is the rectangle 1 ≤ ρ ≤ 2, 0 ≤ φ ≤ π, 0 ≤ θ ≤ π. Each face of this rectangle
becomes part of the boundary of W . The faces ρ = 1 and ρ = 2 become the inner and outer
hemispherical surfaces that form part of the boundary of W . The faces θ = 0 and θ = π become the
two halves of the flat part of the boundary of W . The faces φ = 0 and φ = π become line segments
along the z-axis. We can form V by pasting together two solid regions like W along the flat surfaces
where θ = constant.
z
φ
π
φ=π
θ=0
θ=π
-
ρ=1
ρ=2
y
-
1
-
x
I
2
ρ
π
θ
φ=π
Figure M.53: The hollow hemisphere W and the corresponding rectangular region in
ρθφ-space
72
Theory Supplement Section M
Proof of Stokes’ Theorem
Consider an oriented surface A, bounded by the curve B. We want to prove Stokes’ Theorem:
Z
Z
~
~
F~ · d~r .
curl F · dA =
B
A
We suppose that A has a smooth parameterization ~r = ~r (s, t), so that A corresponds to a region R
in the st-plane, and B corresponds to the boundary C of R. See Figure M.54. We prove Stokes’ Theorem for the surface A and a continuously differentiable vector field F~ by expressing the integrals
on both sides of the theorem in terms of
R s and t, and using Green’s Theorem in the st-plane.
First, we convert the line integral B F~ · d~r into a line integral around C:
Z
Z
∂~r
∂~r
F~ · d~r =
F~ ·
ds + F~ ·
dt.
∂s
∂t
B
C
~ = (G1 , G2 ) on the st-plane by
So if we define a 2-dimensional vector field G
G1 = F~ ·
then
Z
B
∂~r
∂s
G2 = F~ ·
and
F~ · d~r =
Z
C
∂~r
,
∂t
~ · d~s ,
G
using ~s to denote the position vector
R of a point in the st-plane.
~ that occurs on the other side of Stokes’ Theorem?
What about the flux integral A curl F~ · dA
In terms of the parameterization,
Z
Z
∂~r
∂~r
~
~
×
ds dt.
curl F~ ·
curl F · dA =
∂s
∂t
R
A
In Problem 7 on page 74 we show that
∂~r
∂~r
∂G2
∂G1
curl F~ ·
×
=
−
.
∂s
∂t
∂s
∂t
Hence
We have already seen that
Z
A
~ =
curl F~ · dA
Z
B
Z R
F~ · d~r =
∂G2
∂G1
−
∂s
∂t
Z
C
ds dt.
~ · d~s .
G
By Green’s Theorem, the right-hand sides of the last two equations are equal. Hence the left-hand
sides are equal as well, which is what we had to prove for Stokes’ Theorem.
z
t
C
R
A
I
y
B
x
s
Figure M.54: A region R in the st-plane and the corresponding surface A in xyz-space; the curve C
corresponds to the boundary of B
Theory Supplement Section M
73
Problems for Section M
1. Let W be a solid circular cylinder along the z-axis, with
a smaller concentric cylinder removed. Parameterize W
by a rectangular solid in rθz-space, where r, θ, and z are
cylindrical coordinates.
2. In this section we proved the Divergence Theorem using
the coordinate definition of divergence. Now we use the
Divergence Theorem to show that the coordinate defini~
tion is the same as the geometric definition. Suppose F
is smooth in a neighborhood of (x0 , y0 , z0 ), and let UR
be the ball of radius R with center (x0 , y0 , z0 ). Let mR
~ on UR and let MR be
be the minimum value of div F
the maximum value.
(a) Let SR be the sphere bounding UR . Show that
mR ≤
R
SR
~ · dA
~
F
Volume of UR
≤ MR .
(b) Explain why we can conclude that
lim
R→0
R
SR
~ · dA
~
F
Volume of UR
~ (x0 , y0 , z0 ).
= div F
(c) Explain why the statement in part (b) remains true
if we replace UR with a cube of side R, centered at
(x0 , y0 , z0 ).
Problems 3–6 fill in the details of the proof of the Divergence
Theorem.
3. Figure M.51 on page 69 shows the solid region V in xyzspace parameterized by a rectangular solid W in stuspace using the continuously differentiable change of coordinates
~r = ~r (s, t, u),
Suppose that
∂~r
·
∂s
a ≤ s ≤ b, c ≤ t ≤ d, e ≤ u ≤ f.
∂~r
∂~r
×
∂t
∂u
is positive.
(a) Let A1 be the face of V corresponding to the face
∂~r
s = a of W . Show that
, if it is not zero, points
∂s
into W .
∂~r
∂~r
(b) Show that −
×
is an outward pointing nor∂t
∂u
mal on A1 .
(c) Find an outward pointing normal on A2 , the face of
V where s = b.
4. Show that for the other five faces of the solid V in the
proof of the Divergence Theorem (see page 70):
Z
Ai
~ · dA
~ =
F
Z
Si
~ · dS
~,
G
i = 2, 3, 4, 5, 6.
~ is a continuously differentiable vector
5. Suppose that F
field and that ~a , ~b , and ~c are vectors. In this problem we
prove the formula
~ · ~b × ~c ) · ~a + grad(F
~ · ~c × ~a ) · ~b
grad(F
~ · ~a × ~b ) · ~c = (~a · ~b × ~c ) div F
~.
+ grad(F
(a) Interpreting the divergence as flux density, explain
why the formula makes sense. [Hint: Consider the
flux out of a small parallelepiped with edges parallel
to ~a , ~b , ~c .]
(b) Say how many terms there are in the expansion of
the left-hand side of the formula in Cartesian coordinates, without actually doing the expansion.
(c) Write down all the terms on the left-hand side that
contain ∂F1 /∂x. Show that these terms add up to
∂F1
~a · ~b × ~c
.
∂x
(d) Write down all the terms that contain ∂F1 /∂y. Show
that these add to zero.
(e) Explain how the expressions involving the other
seven partial derivatives will work out, and how this
verifies that the formula holds.
~ be a smooth vector field in 3-space, and let
6. Let F
x = x(s, t, u),
y = y(s, t, u),
z = z(s, t, u)
be a smooth change of variables, which we will write in
vector form as
~r = ~r (s, t, u) = x(s, t, u)~i +y(s, t, u)~j +z(s, t, u)~k .
~ = (G1 , G2 , G3 ) on stu-space
Define a vector field G
by
~ · ∂~r ×
G1 = F
∂t
∂~r
~
G3 = F ·
×
∂s
∂~r
∂u
∂~r
.
∂t
~ ·
G2 = F
∂~r
∂~r
×
∂u
∂s
(a) Show that
∂G1
+
∂s
~
∂F
·
+
∂t
~ ∂~r
∂G2
∂G3
∂F
∂~r
+
=
·
×
∂t
∂u
∂s ∂t
∂u
~ ∂~r
∂~r
∂~r
∂F
∂~r
×
+
·
×
.
∂u
∂s
∂u ∂s
∂t
(b) Let ~r 0 = ~r (s0 , t0 , u0 ), and let
~a =
∂~r
(~r 0 ),
∂s
~b = ∂~r (~r 0 ),
∂t
~c =
∂~r
(~r 0 ).
∂u
Use the chain rule to show that
∂G1
∂G2
∂G3
+
+
∂s
∂t
∂u
=
~
r =~
r0
~ · ~b × ~c ) · ~a + grad(F
~ · ~c × ~a ) · ~b
grad(F
~ · ~a × ~b ) · ~c .
+ grad(F
74
INDEX
(c) Use Problem 5 to show that
∂G2
∂G3
∂G1
+
+
=
∂s
∂t
∂u
∂(x, y, z) ∂F1 ∂F2 ∂F3
+
+
.
∂(s, t, u) ∂x
∂y
∂z
7. This problem completes the proof of Stokes’ Theorem.
~ be a smooth vector field in 3-space, and let S
Let F
be a surface parameterized by ~r = ~r (s, t). Let ~r 0 =
~r (s0 , t0 ) be a fixed point on S. We define a vector field
in st-space as on page 72:
~ · ∂~r
G1 = F
∂s
~ · ∂~r .
G2 = F
∂t
(a) Let ~a =
∂~r
(~r 0 ),
∂s
~b = ∂~r (~r 0 ). Show that
∂t
∂G2
∂G1
(~r 0 ) −
(~r 0 ) =
∂t
∂s
~ · ~a ) · ~b − grad(F
~ · ~b ) · ~a .
grad(F
(b) Use Problem 30 on page 961 of the textbook to show
~ ·
curl F
∂~r
∂G2
∂G1
∂~r
×
=
−
.
∂s
∂t
∂s
∂t