Theory Supplement Section L
L
63
PROOFS OF LINE INTEGRAL THEOREMS
In this section we give proofs of the Fundamental Theorem of Calculus for Line Integrals and
Green’s Theorem.
Proof of the Fundamental Theorem of Calculus for Line Integrals
Suppose that C is a curve from point P to point Q. Then the Fundamental Theorem of Calculus for
Line Integrals says
Z
grad f · dr = f (Q) − f (P ).
C
The Fundamental Theorem for Line Integrals can be derived from the Fundamental Theorem for
ordinary definite integrals.
Suppose that (x(t), y(t)), for a ≤ t ≤ b, is a parameterization of C, with endpoints P =
(x(a), y(a)) and Q = (x(b), y(b)). Thus, the values of f along C are given by the single variable
function h(t) = f (x(t), y(t)).
Using this parameterization of C we have
Z
C
grad f · d~r =
=
Z
Z
b
a
b
(fx (x(t), y(t))~i + fy (x(t), y(t))~j ) · (x0 (t)~i + y 0 (t)~j )dt
(fx (x(t), y(t))x0 (t) + fy (x(t), y(t))y 0 (t))dt.
a
By the chain rule
∂f dx ∂f dy
dh
=
+
= fx x0 (t) + fy y 0 (t).
dt
∂x dt
∂y dt
By the Fundamental Theorem of Calculus, this gives us
Z
C
grad f · d~r =
Z
b
h0 (t)dt
a
= h(b) − h(a) = f (Q) − f (P ).
Proof of Green’s Theorem
In this section we will give a proof of Green’s Theorem based on the change of variables formula
for double integrals. Suppose that C is a simple closed curve surrounding a region R in the plane
and oriented so that the region is on the left as we move around the curve. Assume the vector field
F~ , defined at every point of R, is given in components by
F~ (x, y) = F1 (x, y)~i + F2 (x, y)~j ,
where F1 and F2 are continuously differentiable. We wish to show that
Z
C
F~ · d~r =
Z R
∂F1
∂F2
−
∂x
∂y
dx dy.
64
Theory Supplement Section L
y
C3
d
C4
C2
R
c
C1
x
b
Figure L.46: A rectangular region R with boundary C broken
into C1 , C2 , C3 , and C4
a
Proof for Rectangles
We prove Green’s Theorem first when R is a rectangular region, as shown in Figure L.46. The
line integral in Green’s theorem can be written as
Z
Z
Z
Z
Z
~
~
~
~
F~ · d~r
F · d~r +
F · d~r +
F · d~r +
F · d~r =
C
=
=
Z
Z
C2
C1
b
F1 (x, c) dx +
a
d
c
Z
C3
d
c
F2 (b, y) dy −
(F2 (b, y) − F2 (a, y)) dy +
Z
b
Z
C4
b
a
F1 (x, d) dx −
Z
d
F2 (a, y) dy
c
(−F1 (x, d) + F1 (x, c)) dx.
a
On the other hand, the double integral in Green’s theorem can be written as an iterated integral.
We evaluate the inside integral using the Fundamental Theorem of Calculus.
Z Z
Z
∂F2
∂F1
∂F1
∂F2
dx dy =
−
−
dx dy +
dx dy
∂x
∂y
∂y
R
R ∂x
R
Z bZ d
Z dZ b
∂F1
∂F2
−
dx dy +
dy dx
=
∂y
c
a
a ∂x
c
Z d
Z b
=
(F2 (b, y) − F2 (a, y)) dy +
(−F1 (x, d) + F1 (x, c)) dx.
c
a
Since the line integral and the double integral are equal, we have proved Green’s theorem for rectangles.
Proof for Regions Parameterized by Rectangles
t
y
D3
d
D4
C3
D2
T
C4
c
D1
a
R
C2
C1
s
b
x
Figure L.47: A curved region R in the xy-plane corresponding to a rectangular region T in the st-plane
Now we prove Green’s Theorem for a region R which can be transformed into a rectangular
region. Suppose we have a smooth change of coordinates
x = x(s, t),
y = y(s, t).
Theory Supplement Section L
65
Consider a curved region R in the xy-plane corresponding to a rectangular region T in the stplane, as in Figure L.47. We suppose that the change of coordinates is one-to-one on the interior of
T.
We prove Green’s theorem for R using Green’s theorem for T and the change of variables
formula for double integrals given on page 823. First we express the line integral around C
Z
F~ · d~r ,
C
as a line integral in the st-plane around the rectangle D = D1 + D2 + D3 + D4 . In vector notation,
the change of coordinates is
~r = ~r (s, t) = x(s, t)~i + y(s, t)~j
and so
∂~r
∂~r
F~ · d~r = F~ (~r (s, t)) ·
ds + F~ (~r (s, t)) ·
dt.
∂s
∂t
~ on the st-plane with components
We define a vector field G
∂~r
G1 = F~ ·
∂s
∂~r
G2 = F~ ·
.
∂t
and
~ ·d~u .
Then, if ~u is the position vector of a point in the st-plane, we have F~ ·d~r = G1 ds+G2 dt = G
Problem 5 at the end of this section asks you to show that the formula for line integrals along
parameterized paths leads to the following result:
Z
Z
~ · d~u .
F~ · d~r =
G
C
D
In addition, using the product rule and chain rule we can show that
∂y ∂G2
∂G1
∂F2
∂F1 ∂x
∂s ∂s −
=
−
∂y .
∂s
∂t
∂x
∂y ∂x
∂t ∂t
(See Problem 6 at the end of this section.) Hence, by the change of variables formula for double
integrals on page 823,
Z Z Z ∂y ∂F1
∂F1 ∂x
∂G2
∂G1
∂F2
∂F2
∂s ∂s −
−
ds
dt
=
−
dx dy =
ds dt.
∂y ∂x
∂y
∂x
∂y ∂x
∂s
∂t
T
R
T
∂t ∂t
Thus we have shown that
and that
Z R
Z
C
∂F1
∂F2
−
∂x
∂y
F~ · d~r =
dx dy =
Z
D
~ · d~u
G
Z T
∂G1
∂G2
−
∂s
∂t
ds dt.
The integrals on the right are equal, by Green’s Theorem for rectangles; hence the integrals on the
left are equal as well, which is Green’s Theorem for the region R.
Pasting Regions Together
Lastly we show that Green’s Theorem holds for a region formed by pasting together regions
which can be transformed into rectangles. Figure L.48 shows two regions R 1 and R2 that fit together
66
Theory Supplement Section L
y
R
C1
−C
R1
C
R2
C2
x
Figure L.48: Two regions R1 and R2
pasted together to form a region R
to form a region R. We break the boundary of R into C1 , the part shared with R1 , and C2 , the part
shared with R2 . We let C be the part of the boundary of R1 which it shares with R2 . So
Boundary of R = C1 + C2 ,
Boundary of R1 = C1 + C,
Boundary of R2 = C2 + (−C).
Note that when the curve C is considered as part of the boundary of R2 , it receives the opposite
orientation from the one it receives as the boundary of R1 . Thus
Z
Boundary of R1
F~ · d~r +
Z
Boundary of R2
Z
Z
F~ · d~r +
F~ · d~r
C1 +C
C2 +(−C)
Z
Z
Z
Z
F~ · d~r
F~ · d~r −
F~ · d~r +
F~ · d~r +
=
C
C2
C
C1
Z
Z
F~ · d~r
F~ · d~r +
=
C2
C1
Z
=
F~ · d~r .
F~ · d~r =
Boundary of R
So, applying Green’s Theorem for R1 and R2 , we get
Z R
∂F2
∂F1
−
∂x
∂y
dx dy =
=
=
Z
Z
Z
R1
∂F1
∂F2
−
∂x
∂y
Boundary of R1
Boundary of R
dx dy +
Z
~
F · d~r +
Z
R2
Boundary of R2
∂F1
∂F2
−
∂x
∂y
dx dy
F~ · d~r
F~ · d~r ,
which is Green’s Theorem for R. Thus, we have proved Green’s Theorem for any region formed by
pasting together regions that are smoothly parameterized by rectangles.
Example 1
Let R be the annulus (ring) centered at the origin with inner radius 1 and outer radius 2. Using polar
coordinates, show that the proof of Green’s Theorem applies to R. See Figure L.49.
Solution
In polar coordinates, x = r cos t and y = r sin t, the annulus corresponds to the rectangle in the
rt-plane 1 ≤ r ≤ 2, 0 ≤ t ≤ 2π. The sides t = 0 and t = 2π are pasted together in the xy-plane
along the x-axis; the other two sides become the inner and outer circles of the annulus. Thus R is
formed by pasting the ends of a rectangle together.
Theory Supplement Section L
t
2π
67
y
t=0
t = 2π
1
2
x
r=1
r=2
r
1
2
Figure L.49: The annulus R in the xy-plane and the corresponding rectangle 1 ≤ r ≤ 2, 0 ≤ t ≤ 2π
in the rt-plane
Problems for Section L
1. Let R be the annulus centered at (−1, 2) with inner radius 2 and outer radius 3. Show that R can be parameterized by a rectangle.
2. Let R be the region under the first arc of the graph of
the sine function. Show that R can be parameterized by
a rectangle.
3. Let f (x) and g(x) be two smooth functions, and suppose
that f (x) ≤ g(x) for a ≤ x ≤ b. Let R be the region
f (x) ≤ y ≤ g(x), a ≤ x ≤ b.
(a) Sketch an example of such a region.
(b) For a constant x0 , parameterize the vertical line segment in R where x = x0 . Choose your parameterization so that the parameter starts at 0 and ends at
1.
(c) By putting together the parameterizations in part (b)
for different values of x0 , show that R can be parameterized by a rectangle.
4. Let f (y) and g(y) be two smooth functions, and suppose
that f (y) ≤ g(y) for c ≤ y ≤ d. Let R be the region
f (y) ≤ x ≤ g(y), c ≤ y ≤ d.
(a) Sketch an example of such a region.
(b) For a constant y0 , parameterize the horizontal line
segment in R where y = y0 . Choose your parameterization so that the parameter starts at 0 and ends
at 1.
(c) By putting together the parameterizations in part (b)
for different values of y0 , show that R can be parameterized by a rectangle.
5. Use the formula for calculating line integrals by parameterization to prove the statement on page 65:
Z
C
~ · d~r =
F
Z
D
~ · d~
G
u.
6. Use the product rule and the chain rule to prove the formula on page 65:
∂G1
∂G2
−
=
∂s
∂t
∂F2
∂F1
−
∂x
∂y
∂x
∂s
∂x
∂y
∂s
∂y
∂t ∂t
.