LECTURE 6.

LINE INTEGRALS

Multivariable calculus II
MATH 324E -K72K SP TA
Hanoi National University of Education - 2024
Outlines

Summarizing of the previous lecture

Line integrals of first kind

Line integrals in plane
Line integrals in space

Line integrals of second kind (vector line integrals)

Path independence

Exercises
Line integrals in plane

We consider a plane curve C given by the parametric equations

x = x(t), y = y (t), a≤t≤b (1)

or, equivalently, by the vector equation

r (t) = x(t)i + y (t)j,

and we assume that C is a smooth curve. [This means that r ′ is

continuous and r ′ (t) ̸= 0 ].
If we divide the parameter interval [a, b] into n subintervals [ti−1 , ti ]
of equal width and we let xi = x(ti ) and yi = y (ti ), then the
corresponding points Pi (xi , yi ) divide C into subarcs with lengths
∆s1 , ∆s2 , . . . , ∆sn (See Figure 1.)
 Figure:

We choose any point Pi∗ (xi∗ , yi∗ ) in the ith subarc. (This corresponds
to a point ti∗ in [ti−1 , ti ].)
Now if f is any function of two variables whose domain includes the
curve C , we evaluate f at the point (xi∗ , yi∗ ), multiply by the length
∆si of the subarc, and form the sum
n
X
f (xi∗ , yi∗ )∆si ,
i=1

which is similar to a Riemann sum. Then we take the limit of these

sums and make the following definition by analogy with a single
integral.
Definition
If f is defined on a smooth curve C given by Equations (1), then
the line integral of f along C is
Z n
X
f (x, y )ds = lim f (xi∗ , yi∗ )∆si ,
n→∞
C i=1

if this limit exists.

We remember that the length of C is
Z b r 
dx 2  dy 2
L= + dt.
a dt dt

A similar type of argument can be used to show that if is a continuous

function, then the limit in Definition 1 always exists and the following
formula can be used to evaluate the line integral
Z Z b r

 dx 2  dy 2
f (x, y )ds = f x(t), y (t) + dt. (2)
a dt dt
C

The value of the line integral does not depend on the parametrization
of the curve, provided that the curve is traversed exactly once as t
increases from a to b.
If s(t) is the length of C between r (a) and r (t) , then
r 
ds dx 2  dy 2
= + .
dt dt dt
We can use the parametric equations to express x and y in terms of
t and write ds as
r 
dx 2  dy 2
ds = + dt.
dt dt
Just as for an ordinary single integral, we can interpret the line inte-
gral of a positive function as an area.
 R
In fact, if f (x, y ) ≥ 0, f (x, y )ds represents the area of one side
C
of the “fence” or “curtain” in Figure 2, whose base is C and whose
height above the point (x, y ) is f .

Figure:
Example
Evaluate (2 + x 2 y )ds, where C is the upper half of the unit circle
R
C
x 2 + y 2 = 1.
Solution. We first need parametric equations to represent C . Recall
that the unit circle can be parametrized by means of the equations
x = cos t, y = sin t,and the upper half of the circle is described by
the parameter interval 0 ≤ t ≤ π. Therefore
Z Z π q
(2 + x 2 y )ds = (2 + cos2 t sin t) (x ′ (t))2 + (y ′ (t))2 dt
0
C
Z π p
= (2 + cos2 t sin t) sin2 t + cos2 tdt
0
π
cos3 t iπ
Z h
= (2 + cos2 t sin t)dt = 2t −
0 3 0
2
=2π + .
3
Suppose now that C is a piecewise-smooth curve; that is, C is a
union of a finite number of smooth curves C1 , C2 , . . . , Cn , where,
as illustrated in Figure 3, the initial point of Cn+1 is the terminal
point of C1 .Then we define the integral of f along as the sum of the
integrals of f along each of the smooth pieces of C :
Z Z Z Z
f (x, y )ds = f (x, y )ds + f (x, y )ds + · · · + f (x, y )ds.
C C1 C2 Cn

Figure:
 R
Any physical interpretation of a line integral f (x, y )ds depends on
C
the physical interpretation, of the function f . Suppose that ρ(x, y )
represents the linear density at a point (x, y ) of a thin wire shaped
like a curve C . Then the mass of the part of the wire from Pi−1 to
Pi in Figure 1 is approximately ∗ ∗
P ρ(x∗i , y∗i )∆si and so the total mass
of the wire is approximately ρ(xi , yi )∆si .
By taking more and more points on the curve, we obtain the mass
m of the wire as the limiting value of these approximations:
n
X Z
m = lim ρ(xi∗ , yi∗ )∆si = ρ(x, y )ds.
n→∞ C
i=1

The center of mass of the wire with density function is located at

the point (x, y ), where

1 1
Z Z
x= xρ(x, y )ds, y= y ρ(x, y )ds (3)
m C m C
Example
A wire takes the shape of the semicircle x 2 + y 2 = 1, y ≥ 0, and is
thicker near its base than near the top. Find the center of mass of
the wire if the linear density at any point is proportional to its
distance from the line y = 1.
Solution

We use the parametrization x = cos t, y = sin t, 0 ≤ t ≤ π, and find

that ds = dt. The linear density is ρ(x, y ) = k(1 − y ), where k is a
constant, and so the mass of the wire is
Z Z π
 π
m= k(1 − y )ds = k(1 − sin t)dt = k t + cos t 0 = k(π − 2).
C 0
We have
1 1
Z Z
y= y ρ(x, y )ds = yk(1 − y )ds
m C k(π − 2) C
Z π
1 1  1 1 π
= (sin t − sin2 t)dt = − cos t − t + sin 2t 0
π−2 0 π−2 2 4
4−π
=
2(π − 2)

By symmetry we see that x = 0, so the center of mass is

 4−π 
0, .
2(π − 2)
Line integrals in space

We now suppose that is a smooth space curve given by the parametric

equations

x = x(t), y = y (t), z = z(t), a ≤ t ≤ b,

or by a vector equation r (t) = x(t)i + y (t)j + z(t)k.

If f is a function of three variables that is continuous on some region
containing C , then we define the line integral of f along C (with
respect to arc length) in a manner similar to that for plane curves:
Z n
X
f (x, y , z)ds = lim f (xi∗ , yi∗ , zi∗ )∆si .
C n→∞
i=1
We evaluate it using a formula
r 
Z Z b
dx 2  dy 2  dz 2
f (x, y , z)ds = f x(t), y (t), z(t) + + dt.
a dt dt dt
C
(4)
Observe that the integrals in both Formulas (2) and (4) can be
written in the more compact vector notation
Z b
f (r (t))|r ′ (t)|dt.
a
Z
The notation f (r )d|r | thus can be used for the scalar line inte-
C
grals (w.r.t. arc length).
For f (x, y , z) = 1 we get
Z Z b
ds = |r ′ (t)|dt = L
C a

where L is the length of the curve C .

Example
R
Evaluate C y sin zds , where C is the circular helix given by the
equations x = cos t, y = sin t, z = t, 0 ≤ t ≤ 2π.

Figure: The circular helix

Solution

We have
Z Z 2π r 
dx 2  dy 2  dz 2
y sin zds = (sin t) sin t + + dt
C 0 dt dt dt
Z 2π p
= sin2 t sin2 t + cos2 t + 1dt
0
√ Z 2π 1
= 2 (1 − cos 2t)dt
2
√ 0
2 1 2π √
= t − sin 2t 0 = 2π.
2 2
Line integrals of second kind (vector line integrals)

Now we are introduce something perhaps a little different from what

we have seen to now—integrals with vector valued integrands. Specif-
ically, suppose C is a plane (space) curve and f : C → R2 (
f : C → R3 ) is a function from C into the Euclidean space R2
(R3 ). The location function for a point M = (x, y , z) on C is
r (t) = (x(t), y (t)) (r (t) = (x(t), y (t), z(t)),R respectively), with
a ≤ t ≤ b. We are going to define an integral C f (r ) · dr .
Why should we care about such a thing? Again, let’s think about a
physical model. You learned in fifth grade physics that the work done
by a force F acting through a distance d is simply the product Fd.
The force F and the displacement d are, of course, really vectors,
and we saw earlier in life that the "product" of the two is actually the
scalar, or dot, product of the two vectors. Now, in general, neither
of these quantities will be constant, and we will have a variable force
F (r ) acting along a curve C in plane (space). How do we compute
the work done in this situation? Let’s see. Once more, we partition
the curve by choosing a sequence of points {r0 , r1 , . . . , rn } on the
curve, with r0 being the initial point and rn being the final point.
Now, of course, there is an orientation, or direction, specified on the
curve. One may think of specifying an orientation by simply putting
an arrow on the curve—it thus makes sense to speak of the initial
point and the terminal point of the curve. Exactly as in the scalar
integrand case, we choose a point ri∗ on the subarc joining ri−1 to
ri , and evaluate F (ri∗ ). Now then, the work done in going from ri−1
to ri is approximately the scalar product F (ri∗ ) · (ri − ri−1 ). Add all
these up for an approximation to the total work done:
n
X
S= F (ri∗ ) · (ri − ri−1 ).
i=1
The course should be obvious now; we take finer and finer partitions,
and the limiting value of the sums is the integral
Z
F (r ) · dr .

This integral too is called a line integral. To prevent confusion, we

sometimes speak of scalar line integrals and vector line integrals. How
to find such a vector integral should be clear from the discussion of
scalar line integrals. We let r (t), a ≤ t ≤ b, be a vector description of
C . (Here r (a) is the initial point and r (b) is the terminal point.) The
discussion proceeds almost exactly as it did in the previous section
and we get
Z Zb
dr
F (r ) · dr = F (r (t)) · dt.
C dt
a
In R2 , the components of F is P and Q, i.e. F = (P, Q). The vector
r = (x, y ) = (x(t), y (t)); dr = (dx, dy ), and we have the following
form of vector line integrals in plane:
Z Z
F (r ) · dr = Pdx + Qdy
C C
Z b
P(x(t), y (t))x ′ (t) + Q(x(t), y (t))y ′ (t) dt.
 
=
a
In R3 , the components of F is P, Q and R, i.e. F = (P, Q, R). The
vector r = (x, y , z) = (x(t), y (t), z(t)); dr = (dx, dy , dz), and we
have the following form of vector line integrals in space:
Z Z
F (r ) · dr = Pdx + Qdy + Rdz
C C
Z b
P(x(t), y (t), z(t))x ′ (t) + Q(x(t), y (t), z(t))y ′ (t)+

=
a
+ R(x(t), y (t), z(t))z ′ (t) dt.

In general, a given parametrization r (t), a ≤ t ≤ b, determines an
orientation of a curve C , with the positive direction corresponding
to increasing values of the parameter.
If −C denotes the curve consisting of the same points as C but with
the opposite orientation, then we have
Z Z
F · dr = − F · dr .
−C C

But if we integrate with respect to arc length, the value of the line
integral does not change when we reverse the orientation of the curve:
Z Z
fds = fds.
−C C

This is because ∆si is always positive, whereas ∆xi , ∆yi and ∆zi
change sign when we reverse the orientation of C .
Example
y 2 dx + xdy , where
R
Evaluate C
(a) C = C1 is the line segment from (−5, −3) to (0, 2).
(b) C = C2 is the arc of the parabola x = 4 − y 2 from (−5, −3)
to (0, 2).

Figure:
Solution

a) A parametric representation for the line segment is

x = 5t − 5, y = 5t − 3, 0 ≤ t ≤ 1.

Then dx = 5dt, dy = 5dt and we obtain

Z Z 1
2
y dx + xdy = (5t − 3)2 (5dt) + (5t − 5)(5dt)
C1 0
Z 1
=5 (25t 2 − 25t + 4)dt
0
h 25t 3 25t 2 i1 5
=5 − + 4t = − .
3 2 0 6
b) Since the parabola is given as a function of y , let’s take y as the
parameter and write C2 as

x = 4 − y 2, y = y, −3 ≤ y ≤ 2.

Then dx = −2ydy . Thus

Z Z 2
2
y dx + xdy = y 2 (−2y )dy + (4 − y 2 )dy
C2 −3
Z 2
= (−2y 3 − y 2 + 4)dy
−3
h y4 y3 i2 5
= − − + 4y = 40 .
2 3 −3 6
Notice that we got different answers in parts (a) and (b) of Example
5 even though the two curves had the same endpoints. Thus, in
general, the value of a line integral depends not just on the endpoints
of the curve but also on the path.
Example
Find I = C (xy + z 2 )dx + (x + z)dy + 2yzdz, where C is the
R

straight line from the origin to the point (1, 2, 3).

Solution
R The given integral can be rewritten in the following vector
form: C [(xy + z 2 )i + (x + z)j + 2yzk] · dr .
The line C has a vector description r (t) = ti + 2tj + 3tk. Thus,
dr
= i + 2j + 3k, and so
dt
Z 1
I = [(2t 2 + 9t 2 )i + (t + 3t)j + 12t 2 k] · (i + 2j + 3k)dt
0
Z 1 Z 1
2 2 59
= (11t + 8t + 36t )dt = (47t 2 + 8t)dt = .
0 0 3
Example
R
Evaluate C ydx + zdy + xdz , where C consists of the line segment
C1 from (2, 0, 0) to (3, 4, 5), followed by the vertical line segment
C2 from (3, 4, 5) to (3, 4, 0).

Figure:
Path independence
Z
Suppose we evaluate the vector line integral F (r ) · dr , where C
C
is a curve from the point p to the point q. Let r (t), a ≤ t ≤ b, be
a vector description of C . Then, of course, we have r (a) = p and
r (b) = q . As we have already seen,
Z Z b
dr
F (r ) · dr = F (r (t)) · dt.
C a dt

Now let us make the very special assumption that there exists a real-
valued (or scalar) function g : R3 → R such that the derivative, or
gradient, of g is the integrand F :

∇g = F .
Next let’s use the Chain Rule to compute the derivative of the com-
position h(t) = g (r (t)):

dr dr
h′ (t) = ∇g · = F (r (t)) · .
dt dt
This is precisely the integrand in our line integral:
Z Z b
dr
F (r ) · dr = F (r (t)) ·
C a dt
Z b
= h′ (t)dt = h(b) − h(a) = g (p) − g (q).
a

This is a very exciting result and calls for some meditation. Note
that the curve C has completely disappeared from the answer. The
value of the integral depends only on the values of the function g at
the endpoints; the path from p to q does not affect the answer. The
line integral is path independent.
The result is aesthetically pleasing and is clearly the lineal descendant
of the fundamental theorem of calculus we learned before.
A moment’s reflection on the examples we have seen should con-
vince us that a lot of integrals are not path independent, thus many
very nice functions F (or vector fields ) are not the gradient of any
function. A function F that is the gradient of a function g is said to
be conservative and the function g is said to be a potential function
for F .
Let’s suppose the domain D of the function F : D → R3 is open
and connected (Thus any two points in D may be joined by a nice
path.) We have just seen that if there exists a function g : D → R
such that F = ∇g , then the integral of F between any two points
of D does not depend on the path between the two points. It turns
out, as we shall see, that the converse of this is true. Specifically, if
every integral of F in D is path independent, then there is a function
g such that F = ∇g . Let’s see why this is so.
Choose a point p = (x0 , y0 , z0 ) ∈ D. Now define g (s) = g (x, y , z)
to be the integral from p to s along any curve joining these points.
We are assuming path independence of the integral, so it matters
not what curve we choose. So, let’s compute the partial derivative
∂g
∂x . The domain D is open and hence includes an open ball centered
at s = (x, y , z) ∈ D.
Choose a point q = (x1 , y , z) in such an open ball, and let L be the
straight line segment from s to q . Then, of course, L lies in D. Now
let’s integrate F from p to s by going along any curve C from p to
q and then along L from q to s :
Z Z
g (s) = g (x, y , z) = F (r ) · dr + F (r ) · dr .
C L

∂
R
The first integral on the right does not depend on x, and so ∂x C F (r )·
dr = 0.
Thus Z
∂g ∂
= F (r ) · dr .
∂x ∂x L
R
We clearly need to find L F (r ) · dr . This is easy. Suppose

F (r ) = f1 (r )i + f2 (r )j + f3 (r )k.

A vector description of L is simply r (t) = ti + yj + zk, x1 ≤ t ≤ x.

dr R
Thus = i, and our line integral becomes simply L F (r ) · dr =
Rx dt
x1 f1 (t, y , z)dt.
We note that now
Z Z x
∂ ∂
F (r ) · dr = f1 (t, y , z)dt = f1 (x, y , z).
∂x L ∂x x1

Hence
∂g
= f1 .
∂x
∂g ∂g
Similarly, we get ∂y = f2 and ∂z = f3 . It gives us thus the desired
result: F = ∇g .
Exercises

1. Find the exact value of C x 3 y 2 zds , where C is the curve with

R

parametric equations x = e −t cos 4t, y = e −t sin 4t, z = e −t , 0 ≤

t ≤ 2π.
2. A thin wire is bent into the shape of a semicircle x 2 + y 2 =
4, x ≥ 0. If the linear density is a constant k, find the mass and
center of mass of the wire.
3. Find the mass and center of mass of a wire in the shape of the
helix x = t, y = cos t, z = sin t, 0 ≤ t ≤ 2π, if the density at any
point is equal to the square of the distance from the origin.
4. Prove that
R if F : D → R3 , where D is openHand connected,
and every C F (r ) · dr is path independent, then F (r ) · dr = 0
P
for every closed path in D. (A closed path, or curve, is one with no
endpoints.) [Physicists andH others like to use a snake sign with a little
circle superimposed on it to indicate that the path of integration
is closed.]
 3
H5. Prove that if F : D → R , where D is open and connected,
R and
F (r ) · dr = 0 for every closed path in D, then every C F (r ) · dr is
P
path independent.
6. a) Find a potential function g for the function F (r ) = yzi + xzj +
xyk. R
b) Evaluate the line integral F C F (r ) · dr , where C is the curve

r (t) = (e t sin t)i + t 2 e 3t j + cos3 tk, 0 ≤ t ≤ 1.

7. Evaluate [(e x sin y + 3y )i + (e x cos y + 2x − 2y )j] · dr , where

H
E
E is the ellipse 4x 2 + y 2 = 4 oriented clockwise.
[Hint: Find the gradient of g (x, y , z) = e x sin y + 2xy − y 2 .]