Section 16.2
Line Integrals
The idea of a line integral is very similar to that of single integrals. If the function f (x) is above
the x-axis on the interval [a, b], then the integral of f (x) over [a, b] is the area under f over the
interval:
Let ⃗r(t) be a curve in the xy plane.
If f (x, y) is a surface that lies above ⃗r(t), we can think about the ”area” between the surface
and the curve.
1
Section 16.2
Specifically, we think of the area enclosed above the curve and below the surface. In the graph
above, the area that we are interested in is that shaded in below the red curve on the surface, and
above the red curve in the xy plane.
To find the value for this area, we evaluate a specific type of integral known as the line integral.
The following theorem allows us to do so:
Definition 3. If f (x, y) is continuous over the curve C = ⃗r(t) = x(t)⃗i + y(t)⃗j in the xy plane, then
the line integral of f over ⃗r(t) from t = a to t = b is
∫
∫
f (x, y) ds =
C
b
f (x(t), y(t))|r′⃗(t)| dt.
a
In order to evaluate a line integral over a curve C, we will need to parameterize the curve C by
writing it as C = x(t)⃗i + y(t)⃗j, then calculate |r′⃗(t)|. Finally, we can set up the integral according
to the theorem and integrate as we would any other single integral.
The theorem above also applies to functions f (x, y, z) over space curves ⃗r(t) = x(t)⃗i + y(t)⃗j +
z(t)⃗k almost exactly, with the obvious changes.
Example:
A piece of tin is cut from a circular cylinder whose base is a circle of radius 3 inches. At any
point (x, y) on the base, the height of the object is given in inches by f (x, y) = 1 + cos(πx)/4. Set
up an integral that will yield the surface area of the piece of tin.
We can graph the base of the circular cylinder in the xy plane by thinking of its defining equation
as x2 + y 2 = 9:
The cylinder from which we wish to cut the tin is graphed below:
2
Section 16.2
The top of the piece of tin should be shaped like f (x, y) = 1 + cos(πx)/4:
We wish to cut along the intersection of the two surfaces:
So the cut we make yields the following shape:
3
Section 16.2
The line integral of f over the region x2 + y 2 = 9 will yield the surface area of this shape.
We must first paramaterize the curve x2 + y 2 = 9; we can think of it as x = 3 cos t and
y = 3 sin t (or ⃗r(t) = 3 cos t⃗i + 3 sin t⃗j) on the interval 0 ≤ t ≤ 2π. Then r′⃗(t) = −3 sin t⃗i + 3 cos t⃗j,
√
√
and |r′⃗(t)| = (−3 sin t)2 + 9 cos2 t = 9 = 3.
Using the parameterization, we rewrite f as f (x(t), y(t)) = 1 + cos(π(3 cos t))/4. So the line
integral is
∫
∫ b
f (x, y) ds =
f (x(t), y(t))|r′⃗(t)| dt
C
a
∫ 2π
=
(1 + cos(π(3 cos t))/4)(3) dt
0
∫ 2π
cos(3π cos t)
=3
1+
dt.
4
0
This is not an elementary integral, but Mathematica can approximate the value:
∫ 2π
cos(3π cos t)
1+
3
dt ≈ 17.99.
4
0
So the surface area of the piece of tin is approximately 17.99 square inches.
Evaluate the line integral of f (x, y) = 2 + x2 y over the upper half of the unit circle x2 + y 2 = 1.
This is actually the example from the graph at the beginning of the section:
4
Section 16.2
The line integral of f over this region will yield the area of the region below f that lies above
the upper half of the circle.
We need to have a parameterization of the curve; we can think of it as x = cos t and y = sin t
′⃗
′⃗
⃗
⃗
⃗
⃗
(or
√ ⃗r(t) = cos ti + sin t√j) on the interval 0 ≤ t ≤ π. Then r (t) = − sin ti + cos tj, and |r (t)| =
2
2
(− sin t) + cos t = 1 = 1.
Using the parameterization, we rewrite f as f (x(t), y(t)) = 2 + cos2 t sin t. So the line integral
is
∫
∫ b
f (x, y) ds =
f (x(t), y(t))|r′⃗(t)| dt
C
a
∫ π
=
(2 + cos2 t sin t)(1) dt
∫0 π
=
2 + cos2 t sin t dt
0
π
1
cos3 t0
3
1 1
= 2π + +
3 3
2
= 2π + .
3
= 2t −
using the substitution u = cos θ
The number 2π + 23 is the area of the region below f that lies above the upper half of the circle.
Suppose that the curve C is made up of several different components, as in the following graph:
We can still evaluate the line integral of a function f over C; if C is made up of different curves
C1 , C2 , . . . , Cn , then the line integral of f over C is just the sum of the line integrals over the
5
Section 16.2
components; we write
∫
∫
f (x, y) ds =
C
∫
∫
f (x, y) ds +
f (x, y) ds + . . . +
C1
C2
f (x, y) ds.
Cn
Example:
∫
Evaluate
8xyz ds, where C consists of the line segments C1 from (0, 0, 3) to (24, 10, 3) followed
C
by the vertical line segment C2 from (24, 10, 3) to (24, 10, 0).
Since we want to evaluate a line integral over two different curves, we will write a separate
integral for each of C1 and C2 .
We need to start by parameterizing the line segments above. To do so, recall that we can find
the equation of a line in space by finding a vector parallel to the line and a point on the line. For
the first line segment C1 , the vector ⟨24, 10, 0⟩ is parallel to the line, so the vector equation for
the line is ⃗r1 (t) = 24t⃗i + 10t⃗j + 3⃗k. Similarly, ⟨0, 0, −3⟩ is parallel to C2 , whose vector equation is
⃗r2 (t) = 24⃗i + 10⃗j + (3 − 3t)⃗k. We will need to know |r′⃗(t)| for each equation: r1′⃗(t) = 24⃗i + 10⃗j, and
r2′⃗(t) = −3⃗k, so that
√
|r1′⃗(t)| = 576 + 100 = 26
and
|r2′⃗(t)| =
√
9 = 3.
The parameterization for C1 is ⃗r1 (t) = 24t⃗i + 10t⃗j + 3⃗k, 0 ≤ t ≤ 1, and the parameterization
for C2 is ⃗r2 (t) = 24⃗i + 10⃗j + (3 − 3t)⃗k, 0 ≤ t ≤ 1. We will need this parameterizations to rewrite f .
In the first case, with x = 24t, y = 10t, and z = 3, f (x, y, z) = xyz becomes 24t · 10t · 3 = 720t2 .
In the second case, with x = 24, y = 10, and z = 3 − 3t, f becomes 24 · 10 · (3 − 3t) = 720 − 720t.
So the first integral is
∫
∫ 1
xyz ds =
720t2 · 26 dt
C1
0
∫
1
t2 dt
= 18720
= 18720
0
3
t 3
1
0
= 6240.
The second integral is
∫
∫
xyz ds =
C2
1
(720 − 720t) · 3 dt
0
∫
= 2160
0
1
1 − t dt
(
t2 )1
= 2160 t −
2 0
= 1080.
6
Section 16.2
So the line integral of f over the two curves is 6240 + 1080 = 7320.
Work
As we have seen, a force acting on points in space can be represented by a vector field. If we
consider in particular the way that the force F⃗ = P (x, y, z)⃗i + Q(x, y, z)⃗j + R(x, y, z)⃗k acts along
a curve ⃗r(t) = g(t)⃗i + h(t)⃗k + j(t)⃗k, it makes sense to think about the work done by the force
in moving a particle along the curve. We can calculate the work quite easily using the following
theorem:
Definition 13. The work done by a force F⃗ = P (x, y, z)⃗i + Q(x, y, z)⃗j + R(x, y, z)⃗k in moving an
object over a smooth curve ⃗r(t) = g(t)⃗i + h(t)⃗k + j(t)⃗k from t = a to t = b is
∫
b
W =
⃗ · r′⃗(t)ds =
F⃗ (r(t))
∫
a
b
F⃗ · T⃗ ds.
a
Recall that
r′⃗(t)
T⃗ =
.
|r′⃗(t)|
The theorem says that work can be thought of as the line integral of the tangential component
of the force. When calculating the integral, we must be careful about the order of integration;
switching the direction of travel along the curve ⃗r will change the sign of the answer.
Calculating the work integral in the given form may be difficult (since we prefer not to work in
terms of the arc length parameter s), so using one of the following equivalent forms may be helpful:
∫
b
W =
F⃗ · T⃗ ds
a
∫
b
=
∫
F⃗ (⃗r(t)) · r′⃗(t)dt
a
F⃗ · d⃗r
=
=
C
∫ b
(
a
∫
=
P
dh
dj )
dg
+Q
+R
dt
dt
dt
dt
b
P dx + Q dy + R dz.
a
Example
Find the work done by the force field F⃗ (x, y) = x2⃗i − xy⃗j in moving a particle along the curve
⃗r(t) = cos t⃗i + sin t⃗j on 0 ≤ t ≤ π2 .
∫ b
To use the formula W =
F⃗ (⃗r(t)) · r′⃗(t)dt, we will need to calculate F⃗ (⃗r(t)) and r′⃗(t), both
a
7
Section 16.2
of which are quite simple to do. The quantity F⃗ (⃗r(t)) is given by
F⃗ (⃗r(t)) = F⃗ (cos t, sin t)
= cos2 t⃗i − cos t sin t⃗j.
Next, we find r′⃗(t):
r′⃗(t) = − sin t⃗i + cos t⃗j.
∫
Finally, to evaluate
b
F⃗ (⃗r(t)) · r′⃗(t)dt, we will need to take the dot product:
a
F⃗ (⃗r(t)) · r′⃗(t) = (cos2 t⃗i − cos t sin t⃗j) · (− sin t⃗i + cos t⃗j)
= − sin t cos2 t − sin t cos2 t
= −2 sin t cos2 t.
So the value for work is
∫
b
W =
F⃗ (⃗r(t)) · r′⃗(t)dt
a
∫
=
π
2
−2 sin t cos2 t dt
0
π
2
cos3 t02
3
2
=− .
3
=
using the substitution u = cos t
8