Math 251 Section 14.2 Line Integrals
Let
C : {( x (t ), y (t )) : a  t  b} be a smooth curve and f ( x , y ) a continuous function defined
t
along the curve. Recall the arclength function is
s (t )   ( x ' (u )) 2  ( y ' (u )) 2 du
a
The integral with respect to arclength of
f ( x , y ) along C is
b
2
2
 f ( x, y ) ds   f (( x (t ), y (t )) ( x ' (t ))  ( y ' (t )) dt .
C
a
Example 1: The lower edge of a curved piece of metal of uniform thickness is in the shape of the curve,
x ( t )  t 3  2, y ( t )  t 2
C:
is given by
0  t  1.
The height of the metal at any point along this curve
f ( x , y )  y . The area of one side is given by
 f ( x, y ) ds . Evaluate this integral.
C
Example 2:
Evaluate
C
is the boundary of the triangle with vertices, (0,0), (3,0) and (3,4).
 xyds .
C
Line Integrals of Vector Fields

Let


as

F ( x, y )  P ( x, y ) i  Q ( x, y ) j


r (t )  x (t ) i  y (t ) j

. The work done in applying this force along the curve is



W   F  T ds
be a force field along a smooth curve given in vector form
where
r '(t )
T 
C


is the unit tangent vector. Since
ds | r '(t ) | dt ,
| r '(t ) |

Examples: Evaluate

 F  dr
C





1.
F ( x, y )  y i  x j ,
2.
F ( x, y , z )  ( y  z ) i  x 2 j  4 y 2 k ,

2


r (t )  t i  t j

2

0  t 1




r (t )  t i  t 2 j  t 4 k
0  t 1

Note:

 F  dr   P ( x (t ), y (t )) x ' (t ) dt  Q (( x (t ), y (t )) y ' (t ) dt   Pdx  Qdy
C
C
C
b
We define
 Pdx   P ( x (t ), y (t )) x ' (t ) dt
C
a
b
and
 Qdy   Q ( x (t ), y (t )) y ' (t ) dt
C
a