Vector Calculus
1
Line Integrals
Mass problem. Find the mass M of a very thin wire whose linear
density function (the mass per unit length) is known.
We model the wire by a smooth curve C between two points P and Q
in 3-space. Given any point (x, y, z) on C, we let f (x, y, z) denote the
corresponding value of the density function.
1. Divide C into n small sections.
Let ∆Mk be the mass of the
kth section, and let ∆sk be the
length of the arc between
Pk−1 and Pk .
2. Choose Pk∗(x∗k , yk∗, zk∗)
on the kth arc
∆Mk ≈ f (x∗k , yk∗, zk∗)∆sk
3. The mass M of the entire wire is
n
n
X
X
M≈
∆Mk ≈
f (x∗k , yk∗, zk∗)∆sk
k=1
k=1
4. Take max ∆sk → 0, and get
ˆ
n
X
M=
lim
f (x∗k , yk∗, zk∗)∆sk =
f (x, y, z) ds
max ∆sk →0
C
k=1
The last term is the notation for the limit of the Riemann sum,
and it is called the line integral of f (x, y, z) with respect to s
along C. The same definition is for f (x, y).
1
• The mass M of the wire is
ˆ
M=
f (x, y, z) ds
C
• The length L of the wire is
ˆ
L=
ds
C
• If C is a curve
´ in the xy-plane and f (x, y) is nonnegative function
on C, then C f (x, y) ds is equal to the area of the “sheet” that
is swept out by a vertical line segment that extends upward from
(x, y) to a height f (x, y) and moves along C
ˆ
∆Ak ≈ f (x∗k , yk∗)∆sk ⇒ A =
f (x, y) ds
C
2
Evaluating line integrals
Let C be smoothly parametrised
~r = x(t)~i + y(t) ~j , a ≤ t ≤ b .
Then
∆sk =
´ tk
tk−1
|~r 0(t)|dt = |~r 0(t∗k )|∆tk
and
ˆ
f (x, y) ds =
C
=
lim
max ∆sk →0
lim
max ∆tk →0
ˆ
n
X
k=1
n
X
f (x∗k , yk∗)∆sk
f (x(t∗k ), y(t∗k ))|~r 0(t∗k )|∆tk
k=1
b
f (x(t), y(t))|~r 0(t)| dt
a
s 2
ˆ b
2
dy
dx
=
f (x(t), y(t))
+
dt
dt
dt
a
=
The line integral does not depend on a parametrisation of C,
in particular on an orientation of C.
Example. Find
´
C (1
+ x2y)ds for
1. C : 21 (t + t2)~i + 21 (t + t2) ~j , 0 ≤ t ≤ 1
2. C : (2 − 2t)~i + (1 − t) ~j , 0 ≤ t ≤ 1
3
Similarly, if C is a curve in 3-space smoothly parametrised
~r = x(t)~i + y(t) ~j + z(t) ~k , then
ˆ
ˆ
b
f (x(t), y(t), z(t))|~r 0(t)| dt
a
s 2 2
ˆ b
2
dx
dy
dz
+
+
dt
=
f (x(t), y(t), z(t))
dt
dt
dt
a
f (x, y, z) ds =
C
Example. Find
´
C (xy
+ z 3)ds for
C : cos t~i + sin t ~j + t ~k , 0 ≤ t ≤ π
√ 4
Answer : 2π /4
Line integrals with respect to x, y and z
Let’s replace ∆sk by ∆xk ( or ∆yk or ∆zk ) in the definition of the line
integral. Then, we get the line integral of f (x, y, z) with respect
to x along C
ˆ
f (x, y, z) dx =
C
lim
max ∆sk →0
ˆ
f (x, y, z) dy =
C
lim
max ∆sk →0
ˆ
f (x, y, z) dz =
C
lim
max ∆sk →0
4
n
X
f (x∗k , yk∗, zk∗)∆xk
k=1
n
X
f (x∗k , yk∗, zk∗)∆yk
k=1
n
X
k=1
f (x∗k , yk∗, zk∗)∆zk
The sign of these line integrals depends on the orientation of C.
Reversing the orientation changes the sign.
Thus, one should find parametric equations for C in which
the orientation of C is in the direction of increasing t, and then
ˆ
ˆ
b
f (x(t), y(t), z(t)) x0(t) dt
f (x, y, z) dx =
C
Example. Find
a
´
C (1
+ x2y)dy for
1. C : 12 (t + t2)~i + 21 (t + t2) ~j , 0 ≤ t ≤ 1
2. C : (2 − 2t)~i + (1 − t) ~j , 0 ≤ t ≤ 1
Let C be a smooth oriented curve, and let −C denote the oriented
curve with opposite orientation but the same points as C. Then
f (x, y) dx = −
f (x, y) dy = −
f (x, y) dx ,
−C
C
f (x, y) dy
−C
C
ˆ
while
ˆ
ˆ
ˆ
ˆ
ˆ
f (x, y) ds = +
f (x, y) ds
−C
C
Convention
ˆ
ˆ
f (x, y) dx + g(x, y) dy =
ˆ
f (x, y) dx +
C
C
g(x, y) dy
C
We have
ˆ
ˆ
(f (x(t), y(t)) x0(t) + g(x(t), y(t)) y 0(t)) dt
f (x, y) dx+g(x, y) dy =
C
b
a
5
Integrating a vector field along a curve
Definition. A vector field in a plane is a function that associates
with each point P in the plane a unique vector F~ (P ) parallel to the
plane
F~ (P ) = F~ (x, y) = f (x, y)~i + g(x, y) ~j
Similarly, a vector field in a 3-space is a function that associates with
each point P in the 3-space a unique vector F~ (P ) in the 3-space
F~ (P ) = F~ (x, y, z) = f (x, y, z)~i + g(x, y, z) ~j + h(x, y, z) ~k
One can say that a vector field is a vector-valued function with the
number of components equal to the number of independent variables
(coordinates).
Introduce
d~r = dx~i + dy ~j + dz ~k
If F~ (x, y, z) = f (x, y, z)~i + g(x, y, z) ~j + h(x, y, z) ~k is a continuous
vector field, and C is a smooth oriented curve, then the line integral
of F~ along C is
ˆ
ˆ
F~ · d~r =
C
ˆC
=
(f ~i + g ~j + h ~k) · (dx~i + dy ~j + dz ~k)
f (x, y, z) dx + g(x, y, z) dy + h(x, y, z) dz
C
If ~r = ~r(t) = x(t)~i + y(t) ~j + z(t) ~k, then
ˆ
ˆ b
F~ · d~r =
F~ (~r(t)) · ~r 0(t) dt
C
a
6
Example.
The curve C is a line segment connecting the points (−π/2 , π) and
(3π/2 , −2π/3). Parameterise C, and evaluate
ˆ
F~ · d~r
C
where
3y
F~ (x, y) = (−6xy + 3π 3 sin 3x)~i − (3x2 + 2π 3 cos ) ~j
2
´
Answer :
F~ · d~r = 47 π 3 ≈ 121.441
C
12
Let t = s, where s is an arc length parameter. Then
ˆ
ˆ
F~ · d~r =
C
ˆ
b
ˆa
b
F~ (~r(s)) · T~ ds
F~ (~r(s)) · ~r 0(s) ds =
a
F~ · T~ ds
=
C
where T~ = ~r 0(s) is a unit tangent vector along C.
F~ · T~ = |F~ | cos θ ⇒ −|F~ | ≤ F~ · T~ ≤ |F~ |
7
Line integrals along piecewise smooth curves
If C is a curve formed from finitely many
smooth curves C1, C2,..., Cn joined end to end,
then
ˆ
ˆ
ˆ
ˆ
=
+
+··· +
C
C1
C2
Cn
Example. Let the curve C between the points
(−π/2 , π) and (3π/2 , −2π/3) be a curve
formed from two line segments C1 and C2,
where C1 is joining (−π/2 , π) and (3π/2 , π),
and C2 is joining (3π/2 , π) and (3π/2 , −2π/3).
Parameterise C1 and C2, and evaluate
ˆ
F~ · d~r
C
where
3y
F~ (x, y) = (−6xy + 3π 3 sin 3x)~i − (3x2 + 2π 3 cos ) ~j
2
8