Now we compute the left side of Equation 2 by breaking up
as the union of the four
curves
,
,
, and
shown in Figure 3. On
we take
as the parameter and write
the parametric equations as
,
,
. Thus
Observe that
goes from right to left but
goes from left to right, so we can write
the parametric equations of
as
,
,
. Therefore
On
or
(either of which might reduce to just a single point),
is constant, so
and
Hence
D
2
y
C
P dx
yy
D
P
y
dA
3
y
C
Q dy
yy
D
Q
x
dA
D
D
x
,
y
a
x
b
,
t
1
x
y
t
2
x
t
1
t
2
4
yy
D
P
y
dA
y
b
a
y
t
2
x
t
1
x
P
y
x
,
y dy dx
y
b
a
P x
,
t
2
x
P x
,
t
1
x
dx
C
C
1
C
2
C
3
C
4
C
1
x
x
x y
t
1
x
a
x
b
y
C
1
P x
,
y dx
y
b
a
P x
,
t
1
x
dx
C
3
C
3
C
3
x
x y
t
2
x
a
x
b
y
C
3
P x
,
y dx
y
C
3
P x
,
y dx
y
b
a
P x
,
t
2
x
dx
C
2
C
4
x
dx
0
y
C
2
P x
,
y dx
0
y
C
4
P x
,
y dx
y
C
P x
,
y dx
y
C
1
P x
,
y dx
y
C
2
P x
,
y dx
y
C
3
P x
,
y dx
y
C
4
P x
,
y dx
y
b
a
P x
,
t
1
x
dx
y
b
a
P x
,
t
2
x
dx
George Green
Green’s Theorem is named after the self-
taught English scientist George Green
(1793–1841). He worked full-time in his father’s
bakery from the age of nine and taught himself
mathematics from library books. In 1828 he
published privately
An Essay on the Application
of Mathematical Analysis to the Theories of
Electricity and Magnetism
, but only 100 copies
were printed and most of those went to his
friends. This pamphlet contained a theorem
that is equivalent to what we know as Green’s
Theorem, but it didn’t become widely known
at that time. Finally, at age 40, Green entered
Cambridge University as an undergraduate
but died four years after graduation. In 1846
William Thomson (Lord Kelvin) located a copy
of Green’s essay, realized its significance, and
had it reprinted. Green was the first person to
try to formulate a mathematical theory of elec-
tricity and magnetism. His work was the basis
for the subsequent electromagnetic theories of
Thomson, Stokes, Rayleigh, and Maxwell.
FIGURE 3
y
x
0
a
b
D
C¡
y=g™(x)
y=g¡(x)
C™
C£
C¢

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Chapter 7 / Exercise 35

**Calculus of a Single Variable: Early Transcendental Functions**

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CHAPTER 16
Comparing this expression with the one in Equation 4, we see that
Equation 3 can be proved in much the same way by expressing
as a type II region (see
Exercise 30). Then, by adding Equations 2 and 3, we obtain Green’s Theorem.
Evaluate
, where
is the triangular curve consisting of the
line segments from
to
, from
to
, and from
to
.
SOLUTION
Although the given line integral could be evaluated as usual by the methods of
Section 16.2, that would involve setting up three separate integrals along the three sides
of the triangle, so let’s use Green’s Theorem instead. Notice that the region
enclosed by
is simple and
has positive orientation (see Figure 4). If we let
and
, then we have
Evaluate
, where
is the circle
.
SOLUTION
The region
bounded by
is the disk
, so let’s change to polar
coordinates after applying Green’s Theorem:
In Examples 1 and 2 we found that the double integral was easier to evaluate than the
line integral. (Try setting up the line integral in Example 2 and you’ll soon be convinced!)
But sometimes it’s easier to evaluate the line integral, and Green’s Theorem is used in the
reverse direction. For instance, if it is known that