Math 32B Discussion Session
Week 5 Notes
April 26 and 28, 2016
This week we’re going to define a new type of integral. For the first time, we’ll be
integrating along something other than Euclidean space Rn , and we’ll also integrate objects
other than scalar functions. This week we introduce the line integral of a scalar function
or vector field and explore some of its applications.
Scalar Line Integrals
Suppose C is an oriented curve in R3 , meaning that we’ve determined travel along this curve
in one direction to be positive, and the opposite direction to be negative. Given a continuous
function f (x, y, z), we would like to define the integral of f along C in the usual way that
we define integrals: chop up our domain (in this case C) into small subdomains and for each
of these subdomains approximate f by some fixed scalar. Then multiply this scalar by the
measure of the subdomain (in this case, the length of the interval in C) and add up the
resulting products. We limit this process as usual to obtain an integral.
Rather than work through the details of this definition of the integral, we’ll be guided
by intuition. The primary difference between each of the integrals we’ve defined so far has
been this “measure of the subdomain” quantity. When we defined double integrals, we
concerned ourselves a great deal with dA, and we found that this quantity depends on the
coordinate system chosen for R2 . We had a similar excursion with dV when we considered
triple integrals. Now we want a length element for a curve. But we know how to find the
length of a curve! Given a regular parametrization r(t) of the curve C on an interval [a, b],
the arc length of the portion of C traced out between a and t is given by
Z t
s(t) =
kr0 (u)kdu.
a
The arc length differential is then given by
Z t
ds
d
0
kr (u)kdu dt = kr0 (t)kdt,
ds = dt =
dt
dt
a
where the last equality comes from the fundamental theorem of calculus. In light of this
calculation, the following theorem seems very natural:
Theorem 1. Given a curve C ⊂ R3 and a continuous function f (x, y, z),
Z
Z
f (x, y, z)ds =
C
b
f (r(t))kr0 (t)kdt
a
where r(t) is a parametrization of C on [a, b] so that r0 (t) is continuous.
1
Example. (Section 17.2, Exercise 10) Compute
f (x, y) =
R
C
f (x, y)ds, where
y3
x7
and C is the curve traced out by y = 14 x4 for 1 ≤ x ≤ 2.
(Solution) To use our theorem we need a parametrization of C; thankfully the description
of C gives us a very natural parametrization:
1
r(t) := ( t4 , t) for t ∈ [1, 2].
4
Then r0 (t) = (t3 , 1) and we have
Z
Z 2
Z 2
1
0
f (x, y)ds =
f (r(t))kr (t)kdt =
f ( t4 , t)k(t3 , 1)kdt
4
C
1
1
Z 2 1 4 3√
Z 2 √
(4t )
1
t5 t6 + 1dt.
=
t6 + 1dt =
7
t
64 1
1
We can now make the substitution u = t6 + 1. Then du = 6t5 dt and when t = 1, u = 2,
while when t = 2, u = 65. So we have
Z 2√
Z 65
Z
√
1
1
5
6
f (x, y)ds =
t + 1 · 6t dt =
udu
384 1
384 2
C
65
1 2 3/2
2
=
u
=
653/2 − 23/2 .
384 3
1152
2
Scalar integrals have a variety of applications, including computing the mass of a wire
with varying density or calculating electric potential, but for the most part we’ll focus our
attention on vector line integrals, discussed next.
Vector Line Integrals
Suppose we have a bead on a wire, and this wire lives in some ambient space on which a
vector field is defined. For instance, the wire might be in a stream of flowing water. In
Figure 1, our wire is red and the bead is represented by the green dot on this wire.
Figure 1 can be thought of as an initial state for the bead. The vector field will act on
the bead, pushing it along the wire. But how much will the vector field push the bead along
the wire? This will depend on where the bead is. For example, when the bead is at the point
indicated by the dashed green lines (roughly (1.3,1.1)), the vector field seems to be almost
perpendicular to the wire, so the flow of the vector field won’t push the bead along the wire
at all. At the green dot, on the other hand, the vector field is very nearly tangent to the
2
Figure 1: A bead on a wire.
wire, and the flow of the vector field will push the bead along the wire.
We saw in 32A that an oriented curve admits a unit tangent vector at every point. Our
intuition seems to indicate that when the vector field points in the same direction as this
tangent vector, the bead will be moved in the positive direction, when the vector field points
against the unit tangent vector, the movement is in the negative direction, and when the
vector field is perpendicular to the curve, there is no movement at all. This is reminiscent
of the dot product, which is positive when two vectors point in the same direction, negative
when the point in opposite directions, and zero when they’re orthogonal. It’s unsurprising,
then, that
effect of F on the bead at (x, y) ≈ F(x, y) · T(x, y),
for every point (x, y) on our curve. It stands to reason that integrating this dot product
will give the net effect of F on the bead along the entire curve, as the following definition
indicates.
Definition. Let r(t) be a regular parametrization of the oriented curve C ⊂ R3 which is
positively oriented. The line integral of a vector field F along C is given by
Z
Z
F · dr :=
(F · T)ds,
C
C
where T is the unit tangent vector to C with positive orientation.
The dr in the above expression is purely symbolic; its main purpose is to remind us
that C is oriented, and that we must respect this orientation when determining T. It’s
also important to note that the integral on the right is a scalar line integral, which we
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defined above. Now if we have a parametrization r(t) on the interval [a, b] which is positively
oriented, we saw above that
ds = kr0 (t)kdt.
We also recall from 32A that
T(t) =
r0 (t)
.
kr0 (t)k
Then the quanity on the right in our above definition can be rewritten as
r0 (t)
(F · T)ds = F(r(t)) · 0
kr0 (t)kdt = F(r(t)) · r0 (t)dt.
kr (t)k
This is still a dot product of two vectors (and thus still a scalar quantity), but we have now
made this a single-variable scalar function, making our integral much more tractable. We
summarize this work with the following theorem.
Theorem 2. If r(t) is a regular parametrization of the oriented curve C on [a, b], then
Z
Z
b
F · dr =
C
F(r(t)) · r0 (t)dt.
a
Example. (Section 17.2, Exercise 20) Let F(x, y) = h4, yi, and let C be the quarter circle
given by x2 +Ry 2 = 1, where x ≤ 0 and y ≤ 0, with counterclockwise orientation, as in Figure
2. Compute C F · dr.
Figure 2: The vector field F(x, y) = h4, yi with the curve C.
(Solution) First we need a parametrization of C. One parametrization we can use is
r(t) = (cos(t), sin(t)),
where t ∈ [π, 3π/2]. Then for each t we have
F(r(t)) = F(cos(t), sin(t)) = h4, sin(t)i.
4
We also see that
r0 (t) = h− sin(t), cos(t)i,
so
F(r(t)) · r0 (t) = −4 sin(t) + sin(t) cos(t) = sin(t)(cos(t) − 4).
We then have
Z
Z
3π/2
F · dr =
C
sin(t)(cos(t) − 4)dt.
π
Let’s make the substitution u = cos(t)−4, so that du = − sin(t)dt and our bounds transform
from t = π and t = 3π/2 into u = −5 and −4. Then
2 −5
Z
Z −4
u
9
16 25
udu = −
−
= .
F · dr = −
=−
2 −4
2
2
2
−5
C
Before moving on to some applications of line integrals, we should point out some properties of our new machinery. Line integrals have the nice linearity property that all integrals
enjoy. Additionally, they behave predictably when we reverse the orientation of our curve:
Z
Z
F · dr = − F · dr
C
−C
or when we concatenate curves:
Z
Z
Z
F · dr =
C1 +C2
F · dr +
C1
F · dr.
C2
Applications
In introducing vector line integrals, we started with a vector field moving a bead along a
wire. This setup — where a vector field has some effect on particles along a curve — is very
common in applications, and we call the net effect of the vector field along the curve the
work done by the vector field.
Definition. We define the work done by a vector field F along an oriented curve C ⊂ R3
to be
Z
W := F · dr.
C
For a curve in the plane R2 we can also use line integrals to measure the flux of a vector
field along this curve. We could imagine that the curve in Figure 1 were not a wire but were
instead some permeable membrane through which liquid might pass. In this case, we’d like
to measure the amount of liquid passing through the membrance. Now the vector field will
have largest contribution when it is perpendicular to the curve and smallest contribution
when it is tangent, opposite the relationship from before. By letting n be the unit normal
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Figure 3: The vector field F(x, y) = hx2 , y 2 i with the curve C.
vector to our curve in the positive direction (whatever direction we’ve chosen to be positive),
we see that the contribution of F to our flux at a point (x, y) along the curve is represented
by
F(x, y) · n(x, y).
Notice that this dot product is zero when F is tangent to our curve (and thus not contributing
to the flow of liquid through our curve). The total flux through our membrane is then
obtained by integrating this quantity, as in the definition below.
Definition. We define the flux of a vector field F across an oriented curve C ⊂ R2 to be
Z
Z
C
b
F(r(t)) · N(t)dt.
(F · n)ds =
Flux :=
a
In the latter integral, r(t) is a regular parametrization of C on [a, b] which is positively
oriented, and N(t) is the outward-pointing normal vector1 to r(t):
N(t) := hy 0 (t), −x0 (t)i = n(t)kr0 (t)k.
As usual, n(t) is the unit-length outward-pointing normal vector.
Example. (Section 17.2, Exercise 65) Let F(x, y) = hx2 , y 2 i and let C be the line segment
between (3, 0) and (0, 3), oriented upward, as in Figure 3. Calculate the flux of F across C.
1
Because we’re often interested in the flow out of a region, we typically call the positive direction outwardpointing.
6
(Solution) First we need a parametrization for C. We have one given by
r(t) = (3 − 3t, 3t)
for t ∈ [0, 1]. Notice that r(t) varies from (3, 0) to (0, 3), as desired. Then our outwardpointing normal vector is given by
N(t) = h3, −(−3)i = h3, 3i
and we also have
F(r(t)) = F(3 − 3t, 3t) = h(3 − 3t)2 , (3t)2 i.
So
F(r(t)) · N(t) = 3(3 − 3t)2 + 3(3t)2 = 27(1 − 2t + 2t2 ).
Finally,
Z
Flux = 27
0
1
2
(1 − 2t + 2t )dt = 27 t − t + t2
3
2
2
7
1
0
2
= 27 1 − 1 +
= 18.
3