Complex Analysis - Maths KU

y
π
t = gives
2
(0,4)
C
t = 0 gives (4, 0)
Figure 5.4 Path C of integration
x
5.1 Real Integrals 239
(iii) The line integral of G along C with respect to arc length s
is
�
n�
G(x, y) ds = lim G(x ∗ k,y ∗ k)∆sk. (5)
C
�P �→0
k=1
It can be proved that if G is continuous on C, then the three types of line
integrals defined in (3), (4), and (5) exist. We shall assume continuity of G
as matter of course. The curve C is referred to as the path of integration.
Method of Evaluation–C Defined Parametrically The
line integrals in Definition 5.2 can be evaluated in two ways, depending on
whether the curve C is defined by a pair of parametric equations or by an
explicit function. Either way, the basic idea is to convert a line integral to a
definite integral in a single variable. If C is smooth curve parametrized by
x = x(t), y = y(t), a ≤ t ≤ b, then replace x and y in the integral by the
functions x(t) and y(t), and the appropriate differential dx, dy, ords by
x ′ (t) dt, y ′ �
(t) dt, or [x ′ (t)] 2 +[y ′ (t)] 2 �
dt.
[x ′ (t)] 2 +[y ′ (t)] 2 dt is called the differential of the arc length.
The term ds =
In this manner each of the line integrals in Definition 5.2 becomes a definite
integral in which the variable of integration is the parameter t. That is,
�
� b
G(x, y) dx = G (x(t),y(t)) x
C
a
′ (t) dt, (6)
�
� b
G(x, y) dy = G (x(t),y(t)) y
C
a
′ (t) dt, (7)
�
� b
�
G(x, y) ds = G (x(t),y(t)) [x ′ (t)] 2 +[y ′ (t)] 2 dt. (8)
C
a
EXAMPLE 1 C Defined Parametrically
Evaluate (a) �
C xy2dx, (b) �
C xy2dy, and (c) �
C xy2ds, where the path of integration
C is the quarter circle defined by x = 4 cos t, y = 4 sin t,
0 ≤ t ≤ π/2.
Solution The path C of integration is shown in color in Figure 5.4. In each
of the three given line integrals, x is replaced by 4 cos t and y is replaced by
4 sin t.
(a) Since dx = −4 sin tdt, we have from(6):
�
xy
C
2 � π/2
dx = (4 cos t) (4 sin t)
0
2 (−4 sin tdt)
= −256
� π/2
0
sin 3 �
1
t cos tdt= −256
4 sin4 �π/2 t = −64.
0