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
A
F~ · d ~A = Z
V
div ~F dV.
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, A1and A2, both oriented outward:
Z
A1
F~ · d ~A + Z
A2
F~ · d ~A =− Z f
e
Z d c
F1(a, y, z) dy dz + Z f
e
Z d c
F1(b, y, z) dy dz
= Z f
e
Z d 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) = Z b
a
∂F1
∂x dx,
so Z
A1
F~ · d ~A + Z
A2
F~ · d ~A = Z f
e
Z d c
Z b a
∂F1
∂x dx dy dz = Z
V
∂F1
∂x dV.
By a similar argument, we can show Z
A3
F~ · d ~A + Z
A4
F~ · d ~A = Z
V
∂F2
∂y dV and Z
A5
F~ · d ~A + Z
A6
F~ · d ~A = Z
V
∂F3
∂z dV.
Adding these, we get Z
A
F~ · d ~A = Z
V
∂F1
∂x +∂F2
∂y +∂F3
∂z
dV =
Z
V
div ~F dV.
This is the Divergence Theorem for the region V .
x
y z
A4
A6
A2
A3 - (back left)
(backA1 right)
6 A5(bottom)
V
Figure M.50:Rectangular solid V in xyz-space
s
t u
S4
S6
S2
S3 - (back left)
(backS1 right)
6 S5(bottom)
W
x
y z
V
6 A5(bottom) A3(left) -
? A1(back)
A2
A6
A4
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 coordinatesx = 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
A
F~ · d ~A = Z
V
div ~F dV.
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 A1of V is parameterized by
~r = ~r (a, t, u), c≤ t ≤ d, e ≤ u ≤ f, so on this face
d ~A =±∂~r
∂t ×∂~r
∂udt du.
In fact, in order to make d ~A point outward, we must choose the negative sign. (Problem 3 on page 73 shows how this follows from the fact that the Jacobian determinant is positive.) Thus, if S1
is the face s = a of W , Z
A1
F~ · d ~A =− Z
S1
F~ · ∂~r
∂t ×∂~r
∂u dt du,
The outward pointing area element on S1is d~S = −~i dt du. Therefore, if we choose a vector field G~ on stu-space whose component in the s-direction is
G1= ~F ·∂~r
∂t ×∂~r
∂u,
we have Z
A1
F~ · d ~A = Z
S1
G~ · d~S .
Similarly, if we define the t and u components of ~G by G2= ~F · ∂~r
∂u ×∂~r
∂s and G3= ~F ·∂~r
∂s ×∂~r
∂t,
then Z
Ai
F~ · d ~A = Z
Si
G~ · d~S , i = 2, . . . , 6.
(See Problem 4.) Adding the integrals for all the faces, we find that Z
A
F~ · d ~A = Z
S
G~ · d~S .
Since we have already proved the Divergence Theorem for the rectangular region W , we have Z
S
G~ · d~S = Z
W
div ~G dW, where
div ~G = ∂G1
∂s +∂G2
∂t +∂G3
∂u . Problems 5 and 6 on page 73 show that
∂G1
∂s +∂G2
∂t +∂G3
∂u =
∂(x, y, z)
∂(s, t, u)
∂F1
∂x +∂F2
∂y +∂F3
∂z
. So, by the three-variable change of variables formula on page 61,
Z
V
div ~F dV = Z
V
∂F1
∂x +∂F2
∂y +∂F3
∂z
dx dy dz
= Z
W
∂F1
∂x +∂F2
∂y +∂F3
∂z
∂(x, y, z)
∂(s, t, u)
ds dt du
= Z
W
∂G1
∂s +∂G2
∂t +∂G3
∂u
ds dt du
= Z
W
div ~G dW.
In summary, we have shown that Z
A
F~ · d ~A = Z
S
G~ · d~S
and Z
V
div ~F dV = Z
W
div ~G dW.
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 V1and V2along a common face, as in Figure M.52.
The surface A which bounds V is formed by joining the surfaces A1and A2which 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
N Common face
V1
V2
Figure M.52:Region V formed by pasting together V1and 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
A
F~ · d ~A = Z
A1
F~ · d ~A + Z
A2
F~ · d ~A . But we also have Z
V
div ~F dV = Z
V1
div ~F dV + Z
V2
div ~F dV.
So the Divergence Theorem for V follows from the Divergence Theorem for V1and 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.
x y z
ρ = 1 - ρ = 2 -
θ = 0 θ = π -
I φ = π
φ = π
ρ
θ φ
π
1 2
π
Figure M.53:The hollow hemisphere W and the corresponding rectangular region in ρθφ-space
Proof of Stokes’ Theorem
Consider an oriented surface A, bounded by the curve B. We want to prove Stokes’ Theorem:
Z
A
curl ~F · d ~A = Z
B
F~ · d~r .
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’ The- orem for the surface A and a continuously differentiable vector field ~F by expressing the integrals on both sides of the theorem in terms of s and t, and using Green’s Theorem in the st-plane.
First, we convert the line integralR
BF~ · d~r into a line integral around C:
Z
B
F~ · d~r = Z
C
F~ ·∂~r
∂sds + ~F · ∂~r
∂t dt.
So if we define a 2-dimensional vector field ~G = (G1, G2)on the st-plane by G1= ~F ·∂~r
∂s and G2= ~F ·∂~r
∂t,
then Z
B
F~ · d~r = Z
C
G~ · d~s , using ~s to denote the position vector of a point in the st-plane.
What about the flux integralR
Acurl ~F · d ~A that occurs on the other side of Stokes’ Theorem?
In terms of the parameterization, Z
A
curl ~F · d ~A = Z
R
curl ~F ·∂~r
∂s ×∂~r
∂t ds dt.
In Problem 7 on page 74 we show that curl ~F ·∂~r
∂s ×∂~r
∂t = ∂G2
∂s −∂G1
∂t .
Hence Z
A
curl ~F · d ~A = Z
R
∂G2
∂s −∂G1
∂t
ds dt.
We have already seen that Z
B
F~ · d~r = Z
C
G~ · d~s .
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.
R
C
s t
x
y z
I B A
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
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
be the minimum value of div ~F on UR and let MRbe the maximum value.
(a) Let SRbe the sphere bounding UR. Show that
mR≤ R
SR
F~ · d ~A
Volume of UR ≤ MR. (b) Explain why we can conclude that
Rlim→0
R
SR
F~ · d ~A Volume of UR
= div ~F (x0, y0, z0).
(c) Explain why the statement in part (b) remains true if we replace URwith 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 xyz- space parameterized by a rectangular solid W in stu- space using the continuously differentiable change of co- ordinates
~r = ~r (s, t, u), a≤ s ≤ b, c ≤ t ≤ d, e ≤ u ≤ f.
Suppose that ∂~r
∂s ·
∂~r
∂t ×∂~r
∂u
is positive.
(a) Let A1 be the face of V corresponding to the face s = aof W . Show that ∂~r
∂s, if it is not zero, points into W .
(b) Show that −∂~r
∂t ×∂~r
∂u is an outward pointing nor- 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
F~ · d ~A = Z
Si
G~ · d~S , i = 2, 3, 4, 5, 6.
5. Suppose that ~F is a continuously differentiable vector field and that ~a , ~b , and ~c are vectors. In this problem we prove the formula
grad( ~F ·~b × ~c ) · ~a + grad( ~F · ~c × ~a ) · ~b + grad( ~F · ~a ×~b ) · ~c = (~a · ~b × ~c ) div ~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 coor- dinates, 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
~a ·~b × ~c∂F1
∂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.
6. Let ~F be a smooth vector field in 3-space, and let 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 . Define a vector field ~G = (G1, G2, G3)on stu-space by
G1= ~F ·∂~r
∂t ×∂~r
∂u G2= ~F ·∂~r
∂u ×∂~r
∂s G3= ~F ·∂~r
∂s ×∂~r
∂t. (a) Show that
∂G1
∂s +∂G2
∂t +∂G3
∂u =∂ ~F
∂s ·∂~r
∂t ×∂~r
∂u +∂ ~F
∂t ·∂~r
∂u ×∂~r
∂s +∂ ~F
∂u ·∂~r
∂s ×∂~r
∂t. (b) Let ~r0= ~r (s0, t0, u0), and let
~a =∂~r
∂s(~r0), ~b = ∂~r
∂t(~r0), ~c = ∂~r
∂u(~r0).
Use the chain rule to show that
∂G1
∂s +∂G2
∂t +∂G3
∂u
~ r =~r0
=
grad( ~F ·~b × ~c ) · ~a + grad( ~F · ~c × ~a ) · ~b + grad( ~F · ~a ×~b ) · ~c .
(c) Use Problem 5 to show that
∂G1
∂s +∂G2
∂t +∂G3
∂u =
∂(x, y, z)∂(s, t, u)
∂F1
∂x +∂F2
∂y +∂F3
∂z
. 7. This problem completes the proof of Stokes’ Theorem.
Let ~F be a smooth vector field in 3-space, and let S be a surface parameterized by ~r = ~r (s, t). Let ~r0 =
~r (s0, t0)be a fixed point on S. We define a vector field in st-space as on page 72:
G1= ~F ·∂~r
∂s G2= ~F ·∂~r
∂t.
(a) Let ~a = ∂~r
∂s(~r0), ~b =∂~r
∂t(~r0).Show that
∂G1
∂t (~r0)−∂G2
∂s (~r0) =
grad( ~F · ~a ) ·~b − grad( ~F ·~b ) · ~a . (b) Use Problem 30 on page 961 of the textbook to show
curl ~F ·∂~r
∂s ×∂~r
∂t =∂G2
∂s −∂G1
∂t .
