If is a vector field and , , is a path, the line integral of along is
Here is what this looks like in two dimensions.
The vector field gives a vector at each point of the plane. As you move along the path, at each point you compute the dot product of the velocity vector with the vector from the field at that point. Integrate to "add up" the results; the total is the line integral.
Notice that I'm writing instead of , the differential for a path integral. Here's the difference:
One is a vector, the other is a scalar: uses the velocity vector, while uses the length of the velocity vector.
Example. Compute for :
(a) Along the path for .
(b) Along the path for .
(a) The path is the part of lying in the first quadrant --- a quarter circle. while , so the quarter circle is traversed counterclockwise.
The velocity is
Write the vector field in terms of t:
Then
(I used a double angle formula from trigonometry.) Therefore,
(b) is the same quarter circle as in (a), but
The extra factor of 2 means the path is traversed twice as fast as (a). The field is
So
Therefore,
Traversing the path twice as rapidly made no difference. This is true in general: If you traverse the same path in the same direction at different speeds, the line integral does not change.
Example. Compute for :
(a) Along the path for .
(b) Along the path for .
(a) The path is the part of the parabola from to . I have
Hence,
The integral is
(b) If I change the path to , I am now traversing from to --- the same path as in (a), but in the opposite direction.
I have
Hence,
Therefore,
This is true in general: If you traverse the same path in the opposite direction, the line integral is multiplied by -1.
Differential notation.
If is a vector field and for , is a path, then
So
If I formally multiply the in, I get
You will sometimes see line integrals written in this form.
You can do the computation converting , , and to functions of t using . Replace , , by , , , and integrate the whole thing with respect to t.
Alternatively, you can use one of the coordinate variables x, y, or z as the parameter.
Example. Let be the segment joining to . Compute
The segment is
That is,
The parameter range is .
Now
Plugging everything into the integral, I get
Example. Compute where C is the part of the curve going from to .
I'll do everything in terms of x. I have , so becomes . The curve extends from to . Substituting and , I get
If represents a force field and is the tranjectory of an object moving in the field, the work done by the object in moving along the path through the field is
Example. Suppose a force field is given by
Find the work done by a particle moving along a path consisting of the segment from to , followed by the arc of the circle from counterclockwise to .
The segment from to may be parametrized by
Thus,
Hence,
On this segment,
So
The work done is
The arc of the circle from to may be parametrized by
Thus,
Hence,
On this arc,
So
The work done is
Copyright 2018 by Bruce Ikenaga