is parametrized by a variable t that describes
motion along the curve at a variable speed kx0 (t)k.
Rather than reparameterizing the path in terms of
s, another way to directly give the line integral is
Line integrals
Math 131 Multivariate Calculus
Z
D Joyce, Spring 2014
b
Z
f (x(t)) kx0 (t)k dt.
f ds =
x
a
Preview. We’ll start our study of line integrals.
For the next few meetings we’ll study line integrals. Example 1. Evaluate the scalar line integral
First, we’ll see just what line integrals are, both
Z
scalar line integrals and vector line integrals. Sec(3x + xy + z 3 ) ds
x
ond, we’ll discuss Green’s theorem. Green’s theorem relates a double integral over a plane region to
a line integral around the boundary of that plane where x is the path x(t) = (cos 4t, sin 4t, 3t) for
region. Finally, we’ll look at conservative vector t ∈ [0, 2π].
fields and curls.
Vector line integrals. The vector line integrals
Scalar line integrals. So, what is a line integral?
we’re going to look at are integrals of vectors of
It’s
Z b an integral along a curve. An ordinary integral, a vector field F dotted with unit tangent vectors
f (s) ds, is the integral along the line segment T for the curves. Since the integrand is actually
a
a scalar, the dot product of vectors, the value of
[a, b]. We’ll generalize this integral by changing the
these integrals is also a scalar.
line segment [a, b] of the x-axis to a segment of any
A vector line integral of a vector-valued function
curve in Rn . When we do that, we’ll take the inten
m
n
grand f (x) to be a scalar-valued function Rn → R. F : R → R along a path x : [a, b] → R is the
(Soon, we’ll look at vector-valued functions, too.) integral
Z b
The natural parameterization of a curve is by its
F(x(t)) · x0 (t) dt
arclength s that we discussed a while ago. For that
a
parameterization, a path x : R → Rn in Rn has the
Z
property that kx0 (s)k = 1 for all s. It means that a which we’ll also denote
F · ds. Note that the ds
point is travelling the path at unit speed. At each
x
point x(s) on the path, the function f has a value, in this notation is a vector, not the scalar ds we
Z b
just used for the scalar line integrals.
f (x(s)). Just as the ordinary integral,
f (s) ds,
These vector line integrals can be given in terms
a
of
the unit tangent vector T of a path. Recall that
sums the values of f along the line segment [a, b]
Z b
T is defined by
on the x-axis, the line integral,
f (x(s)) ds, sums
a
dx
x0 (t)
the values of f along the path x(s) as s varies from
=
.
T(t) = 0
kx (t)k
ds
a to b. This line integral is variously denoted
Z
Z
Z b
The vector line integral equals
f = f ds =
f (x(s)) ds.
x
x
a
Typically, a path is not given already
parametrized by its arclength s.
Instead, it
Z
(F · T) ds,
x
1
R
Then we can rewrite the vector integral x F · ds as
Z
Z b
F · ds
F(x(t)) · x0 (t) dt
x
a
Z b
Z b
x0 (t)
=
F(x(t)) · x0 (t) dt
=
F(x(t)) · 0
kx0 (t)k dt
a
kx (t)k
Z b
Za
(M x0 (t) + N y 0 (t) + P z 0 (t)) dt
=
=
(F · T) ds
a
x
Z b
dx
dy
dz
=
M
dt + N
dt + P
dt
Thus, if we interpret the vector differential ds as
dt
dt
dt
a
Z b
being the product T ds, we can literally define the
vector line integral as
=
M dx + N dy + P dz
and here’s the proof.
a
Z
Z
dx
Here,
the
differential
dx
is
a
notation
for
dt.
x
x
dt
That last integral is usually abbreviated
This formulation doesn’t depend on the path, that
Z
is it doesn’t depend on the speed a moving point
M dx + N dy + P dz.
goes along the curve. But it does depend on which
x
end of the curve it starts as. If you traverse the
The part of the expression after the integral sign,
curve in the other direction, the value of the integral
namely,
is negated. Thus, the vector line integral depends
M dx + N dy + P dz,
on the orientation of the curve.
is called a differential form. In this course, we’re
taking differential forms as being just notational deExample 2. Evaluate the vector line integral
vices, but they can be abstracted in a way to make
Z
them elements of some abstract algebraic object.
(3z, y 2 , 6z) · ds
x
Example 3. Evaluate the line integral
Z
over the path x(t) = (cos t, sin t, t/3) for t ∈ [0, π].
(x2 − y) dx + (x − y 2 ) dy
F · ds =
F · T ds.
C
Differential forms. When the integrand F and
the path x are given by their coordinate functions,
vector integrals are written using what are called
differential forms. Let’s stick to R3 for a while.
Let the vector field F be given coordinatewise as
F = (M, N, P ), that is,
where C is the line segment from (1, 1) to (3, 5).
This will require choosing a parametrization of that
line segment.
Math 131 Home Page at
http://math.clarku.edu/~djoyce/ma131/
F(x, y, z) = (M (x, y, z), N (x, y, z), P (x, y, z))
and the path x be given, as usual, as
x(t) = (x(t), y(t), z(t)).
2