At the beginning of the supplement to lecture 17, we asked a handful of natural
questions. Let’s get around to answering them. In class, we took a slightly more
physical interpretation when discovering so called line and path integrals. In particular, we thought of path integrals as representing masses of wires, and line integrals
as representing work done by a vector field on a particle traversing a curve. Here,
we’ll develop these integrals with the idea of average value in mind.
Let f(x, y) be a continuous function, and let C be a curve in the plane (of finite
length). We could approximate the average value of f on C by partitioning C into
n segments, say Ci , choosing sample points within these segments, say (xi , yi ), and
taking the following sum
n
X
`(Ci )
1 X
=
f(xi , yi )`(Ci )
`(C)
`(C)
n
f(xi , yi )
i=1
i=1
where `(Ci ) is the length of Ci . This approximation gets better as n → ∞. So we
make the following natural definition.
Definition 1. Let f be continuous and let C be a curve of finite length in the plane.
Then
Z
f ds := lim
n→∞
C
n
X
f(xi , yi )`(Ci ).
i=1
We call the above integral the path integral of f over C. Moreover, the average
value of f along C is given by
1
`(C)
Z
f ds.
C
How to we actually compute these path integrals? Suppose r(t) is a differentiable
function that parameterizes C, say on the interval [a, b]. Perhaps we can carry out
our integration on the much simpler space [a, b] rather than on C, much like a
change of variables. Indeed, partition [a, b]
a = t0 < t1 < · · · tn = b.
This induces a partition of C into segments Ci with endpoints r(ti−1 ) and r(ti ).
Let t∗i ∈ [ti−1 , ti ] so that we may take
xi = x(t∗i )
yi = y(t∗i )
where r(t) = hx(t), y(t)i. As for `(Ci ), notice that1
p
`(Ci ) ≈
(x(ti ) − x(ti−1 ))2 + (y(ti ) − y(ti−1 ))2
s
2 2
x(ti ) − x(ti−1 )
y(ti ) − y(ti−1 )
=
+
(ti − ti−1 )
ti − ti−1
ti − ti−1
q
≈
x 0 (t∗i )2 + y 0 (t∗i )2 ∆t
1Assuming continuity of the derivative of r(t).
1
2
Putting everything together, we get
n
X
f(xi , yi )`(Ci ) ≈
i=1
n
X
f(r(t∗i ))kr 0 (t∗i )k∆t.
i=1
In fact, we have
Z
f ds = lim
n
X
n→∞
C
Zb
f(r(t∗i ))kr 0 (t∗i )k∆t =
f(r(t))kr 0 (t)k dt.
a
i=1
Now, suppose F is a continuous vector field, and let C be a smooth, oriented2
curve of finite length in the plane. We wish to know the average tangential component of F along C. Let T(x, y) be a tangent vector to C at (x, y) pointing in the
direction of C. Recall that the component of F on T is given computed by
F·T
.
kTk
So we could approximate this average by partitioning C into n segments, say Ci ,
choosing sample points within these segments, say (xi , yi ), and taking the following
sum
n
X
F(xi , yi ) · T(xi , yi ) `(Ci )
·
.
kT(xi , yi )k
`(C)
i=1
This approximation gets better as n → ∞. So we make the following natural
definition.
Definition 2. Let F be continuous and let C be a smooth curve of finite length in
the plane. Then
Z
F · dr := lim
C
n→∞
n
X
F(xi , yi ) · T(xi , yi )
i=1
kT(xi , yi )k
· `(Ci ).
We call the above integral the line integral of F along C. Moreover, the average
tangential component of F along C is given by
Z
1
F · dr.
`(C) C
2This simply means that C has a direction.
3
How do we actually compute these line integrals? Suppose r(t) is a differentiable
function that parameterizes C, say on the interval [a, b], that agrees with the orientation of C. This means that if the direction of C is from the start point (x0 , y0 )
to the end point (x1 , y1 ), then r(a) = (x0 , y0 ) and r(b) = (x1 , y1 ). Partition [a, b]
a = t0 < t1 < · · · < tn = b.
This induces a partition of C into segments Ci with endpoints r(ti−1 ) and r(ti ).
Let t∗i ∈ [ti−1 , ti ] so that we may take
xi = x(t∗i )
yi = y(t∗i )
where r(t) = hx(t), y(t)i. As for T, we may take
T(xi , yi ) = r 0 (t∗i ),
what’s more, we already know that
`(Ci ) ≈ kr 0 (t∗i )k ∆t.
Putting everything together we get
n
X
F(xi , yi ) · T(xi , yi )
kT(xi , yi )k
i=1
· `(Ci )
≈
n
X
F(r(t∗ )) · r 0 (t∗ )
i
i=1
=
n
X
i
kr 0 (t∗i )k
kr 0 (t∗i )k ∆t
F(r(t∗i )) · r 0 (t∗i ) ∆t.
i=1
In fact, we have
Z
F · dr = lim
C
n→∞
n
X
Zb
F(r(t)) · r 0 (t) dt.
F(r(t∗i )) · r 0 (t∗i ) ∆t =
i=1
a
Example 1. Compute the average tangential component of F along C where
−x
−y
F(x, y) =
,
(x2 + y2 )3/2 (x2 + y2 )3/2
and C is the unit circle oriented in the clockwise direction.
Solution. Note that C is properly oriented and parameterized by
x(t) = cos(−t)
y(t) = sin(−t)
with 0 6 t 6 2π. Recall that
cos2 (−t) + sin2 (−t) = 1.
4
So the average tangential component of F along C is given by
Z
Z
1
1 2π
h− cos(−t), − sin(−t)i · h− sin(−t), cos(−t)i dt
F · dr =
`(C) C
2π 0
Z
1 2π
cos(−t) sin(−t) − sin(−t) cos(−t) dt
=
2π 0
= 0.
Of course, since F is always perpendicular to C, we could have guessed the result.
But it’s sometimes reassuring to see the math match up with your intuition.
R
1
Problem 1. Compute 2π
C F · dr where
y
−x
F(x, y) =
,
x2 + y2 x2 + y2
and C is the circle of radius r oriented counterclockwise.
R
1
Problem 2. Compute 2π
C F · dr where F is the same as the previous problem,
but C is the boundary of [−1, 1]2 oriented counterclockwise. Hint: You will have to
break up the integral over the four straightlines comprising C.
R
1
Problem 3. Compute 2π
C F·dr where F is the same as the previous two problems,
but C is the boundary of [1, 2] × [0, 1] oriented counterclockwise.
R
1
(Challenge) Problem 4. Compute 2π
C F·dr where F is the same as the previous
three problems, but C is the ellipse
x2
y2
+
=1
a2 b2
oriented counterclockwise.
(Challenge) Problem 5. Compute
1
2π
R
C
F · dr where, again, F is the same as
the previous four problems, but C is the circle of radius 1 centered at (2, 0) oriented
counterclockwise.