Page 389 - Advanced engineering mathematics

12.1 Line Integrals 369

fdx + gdy + hdz
C
b

dx dy dz
= f (x(t), y(t), z(t)) + g(x(t), y(t), z(t)) + h(x(t), y(t), z(t)) dt.
a dt dt dt

fdx + gdy + hdz is a number obtained by replacing x, y and z in f (x, y, z), g(x, y, z)
C
and h(x, y, z) with the coordinate functions x(t), y(t) and z(t) of C, replacing

dx = x (t)dt,dy = y (t)dt, and dz = z (t)dt,
and integrating the resulting function of t from a to b.
EXAMPLE 12.2
z
We will evaluate xdx − yz dy + e dz if C is the curve with coordinate functions
C
3
2
x = t , y =−t, z = t for 1 ≤ t ≤ 2.
First,
2
dx = 3t dt,dy =−dt, and dz = 2tdt.
z
Put the coordinate functions of C into x, −yz and e to obtain

z
xdx − yz dy + e dz
C

2

2
3
2
t
2
= t (3t ) − (−t)(t )(−1) + e (2t) dt
1
2
2
5
t
3
= [3t − t + 2te ]dt
1
111
4
= + e − e.
4
EXAMPLE 12.3

Evaluate xyz dx − cos(yz)dy + xz dz along the straight line segment L from (1,1,1) to
C
(−2,1,3).
Parametric equations of L are
x = 1 − 3t, y = 1, z = 1 + 2t for 0 ≤ t ≤ 1.
Then
dx =−3dt,dy = 0 and dz = 2dt.
The line integral is

xyz dx − cos(yz)dy + xz dz
C
1
= [(1 − 3t)(1 + 2t)(−3) − cos(1 + 2t)(0) + (1 − 3t)(1 + 2t)(2)]dt
0
1 3

2
= (−1 + t + 6t )dt = .
2
0
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October 14, 2010 14:53 THM/NEIL Page-369 27410_12_ch12_p367-424