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Line Integrals of Vector
Fields
In the previous two sections we looked at line integrals of functions
...
We’ll start with the vector field,
F ( x, y, z ) = P ( x, y, z ) i + Q ( x, y, z ) j + R ( x, y, z ) k
and the three-dimensional, smooth curve given by
r (t ) = x (t ) i + y (t ) j + z (t ) k
a≤t ≤b
The line integral of F along C is
∫ F id r = ∫ a F ( r ( t ) )ir′ ( t ) dt
b
C
Note the notation in the left side
...
Also, F r ( t ) is a shorthand for,
(
)
F ( r (t )) = F ( x (t ) , y (t ) , z (t ))
We can also write line integrals of vector fields as a line integral with respect to arc length as
follows,
∫ F id r = ∫ F iT ds
C
C
where T ( t ) is the unit tangent vector and is given by,
r′ (t )
T (t ) =
r′ (t )
If we use our knowledge on how to compute line integrals with respect to arc length we can see
that this second form is equivalent to the first form given above
...
Let’s take a look at a couple of examples
...
Solution
Okay, we first need the vector field evaluated along the curve
...
r ′ ( t ) = i + 2t j + 3t k
2
Finally, let’s get the dot product taken care of
...
Solution
We’ll first need the parameterization of the line segment
...
We’ve been using the two
dimensional version of this over the last couple of sections
...
r ( t ) = (1 − t ) −1, 2, 0 + t 3, 0,1
= 4t − 1, 2 − 2t , t ,
0 ≤ t ≤1
So, let’s get the vector field evaluated along the curve
...
r′ ( t ) = 4, −2,1
The dot product is then,
F ( r ( t ) )ir ′ ( t ) = 4 ( 4t 2 − t ) − ( 2t − 2t 2 ) = 18t 2 − 6t
The line integral becomes,
1
2
F
i
d
r
=
18
t
− 6t dt
∫
∫
0
C
= ( 6t − 3t
3
2
)
1
0
=3
Let’s close this section out by doing one of these in general to get a nice relationship between line
integrals of vector fields and line integrals with respect to x, y, and z
...
This also allows us to say the following about reversing the direction of the path with line
integrals of vector fields
...