Page 390 - Advanced engineering mathematics

370 CHAPTER 12 Vector Integral Calculus

We have a line integral in the plane if C is in the plane and the functions involve only
x and y.

EXAMPLE 12.4
2
Evaluate xy dx − y sin(x)dy if K has coordinate functions x = t , y = t for −1 ≤ t ≤ 2. Here
K
dx = 2tdt and dy = dt
so
2

2
2
xy dx − y sin(x)dy = [t t(2t) − t sin(t )]dt
K −1
2
4 2 66 1
= [2t − t sin(t )]dt = + (cos(4) − cos(1)).
−1 5 2
Line integrals have properties we normally expect of integrals.
1. The line integral of a sum is the sum of the line integrals:

( f + f )dx + (g + g )dy + (h + h )dz
∗
∗
∗
C

∗
∗
∗
= fdx + gdy + hdz + f dx + g dy + h dz.
C C
2. Constants factor through a line integral:

(cf )dx + (cg)dy + (ch)dz = c fdx + gdy + hdz.
C C
b a
For definite integrals, F(x)dx =− F(x)dx. The analogue of this for line integrals is
a b
that reversing the direction on C changes the sign of the line integral. Suppose C is a smooth
curve from P 0 to P 1 .Let C have coordinate functions
x = x(t), y = y(t), z = z(t) for a ≤ t ≤ b.
Define K as the curve with coordinate functions
˜ x(t) = x(a + b − t), ˜y(t) = y(a + b − t), ˜z(t) = z(a + b − t) for a ≤ t ≤ b.
The graphs of C and K are the same, but the initial point of K is the terminal point of C, since
x
(˜(a), ˜y(a), ˜z(a)) = (x(b), y(b), z(b)).
Similarly, the terminal point of K is the initial point of C. We denote a curve K formed from C
in this way as −C. The effect of this reversal of orientation is to change the sign of a line integral.
3.

fdx + gdy + hdz =− fdx + gdy + hdz.
C −C
This can be proved by a simple change of variables in the integrals with respect to t defining
these line integrals.
The next property of line integrals reflects the fact that
b c b
F(x)dx = F(x)dx + F(x)dx
a a c
for definite integrals. A curve C is piecewise smooth if it has a continuous tangent at all but
finitely many points. Such a curve typically has the appearance of the graph in Figure 12.2, with
a finite number of “corners” at which there is no tangent.

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October 14, 2010 14:53 THM/NEIL Page-370 27410_12_ch12_p367-424