Lecture 20: Line Integrals of Scalar-valued
Functions - §13.2
April 24, 2012 (Tue)
Lecture 20: Line Integrals of Scalar-valued Functions - §13.2
Line Integrals of Scalar-valued Functions
Let f (x, y , z) be a continuous function defined on a smooth 1 curve
C parametrized by a vector function ~r (t) =< x(t), y (t), z(t) > for
a ≤ t ≤ b. Then the line integral of f (x, y , z) along the curve C is
Z
Z b
f (x, y , z) ds =
f (x(t), y (t), z(t))|~r 0 (t)| dt
C
a
p
where |~r 0 (t)| = (x 0 (t))2 + (y 0 (t))2 + (z 0 (t))2 .
I
Example:
~r (t) is continuously differentiable and ~r 0 (t) 6= 0 for all t.
1
Lecture 20: Line Integrals of Scalar-valued Functions - §13.2
Line Integrals of Scalar-valued Functions
Let f (x, y , z) be a continuous function defined on a smooth 1 curve
C parametrized by a vector function ~r (t) =< x(t), y (t), z(t) > for
a ≤ t ≤ b. Then the line integral of f (x, y , z) along the curve C is
Z
Z b
f (x, y , z) ds =
f (x(t), y (t), z(t))|~r 0 (t)| dt
C
a
p
where |~r 0 (t)| = (x 0 (t))2 + (y 0 (t))2 + (z 0 (t))2 .
I
Example:
R
1. Evaluate C (x 2 + y 2 ) ds, where C is the upper half of the
counterclockwise-oriented unit circle.
~r (t) is continuously differentiable and ~r 0 (t) 6= 0 for all t.
1
Lecture 20: Line Integrals of Scalar-valued Functions - §13.2
Line Integrals of Scalar-valued Functions
Let f (x, y , z) be a continuous function defined on a smooth 1 curve
C parametrized by a vector function ~r (t) =< x(t), y (t), z(t) > for
a ≤ t ≤ b. Then the line integral of f (x, y , z) along the curve C is
Z
Z b
f (x, y , z) ds =
f (x(t), y (t), z(t))|~r 0 (t)| dt
C
a
p
where |~r 0 (t)| = (x 0 (t))2 + (y 0 (t))2 + (z 0 (t))2 .
I
Example:
R
1. Evaluate C (x 2 + y 2 ) ds, where C is the upper half of the
counterclockwise-oriented unit circle.
2. In Ex 1, what will happen if we take a different
parametrization?
~r (t) is continuously differentiable and ~r 0 (t) 6= 0 for all t.
1
Lecture 20: Line Integrals of Scalar-valued Functions - §13.2
Line Integrals of Scalar-valued Functions
Let f (x, y , z) be a continuous function defined on a smooth 1 curve
C parametrized by a vector function ~r (t) =< x(t), y (t), z(t) > for
a ≤ t ≤ b. Then the line integral of f (x, y , z) along the curve C is
Z
Z b
f (x, y , z) ds =
f (x(t), y (t), z(t))|~r 0 (t)| dt
C
a
p
where |~r 0 (t)| = (x 0 (t))2 + (y 0 (t))2 + (z 0 (t))2 .
I
Example:
R
1. Evaluate C (x 2 + y 2 ) ds, where C is the upper half of the
counterclockwise-oriented unit circle.
2. In Ex 1, what will happen if we take a different
parametrization?
R
3. Evaluate C (y − x) ds, where C consists of the line C1 from
(0, 0) to (2, 1) followed by the line C2 from (2, 1) to (4, -7).
~r (t) is continuously differentiable and ~r 0 (t) 6= 0 for all t.
1
Lecture 20: Line Integrals of Scalar-valued Functions - §13.2
Line Integrals of Scalar-valued Functions
Let f (x, y , z) be a continuous function defined on a smooth 1 curve
C parametrized by a vector function ~r (t) =< x(t), y (t), z(t) > for
a ≤ t ≤ b. Then the line integral of f (x, y , z) along the curve C is
Z
Z b
f (x, y , z) ds =
f (x(t), y (t), z(t))|~r 0 (t)| dt
C
a
p
where |~r 0 (t)| = (x 0 (t))2 + (y 0 (t))2 + (z 0 (t))2 .
I
Example:
R
1. Evaluate C (x 2 + y 2 ) ds, where C is the upper half of the
counterclockwise-oriented unit circle.
2. In Ex 1, what will happen if we take a different
parametrization?
R
3. Evaluate C (y − x) ds, where C consists of the line C1 from
(0, 0) to R(2, 1) followed by the line C2 from (2, 1) to (4, -7).
4. Evaluate C y sin z ds, where C is the circular helix given by
the equations x = cos t, y = sin t, z = t, 0 ≤ t ≤ 2π.
~r (t) is continuously differentiable and ~r 0 (t) 6= 0 for all t.
1
Lecture 20: Line Integrals of Scalar-valued Functions - §13.2
Remarks
1. Line integrals of scalar-valued functions depend on the image
curve and not on the particular parametrization.
2. If C is a given curve, we denote −C is tracing out the curve
in the opposite direction. Then,
Z
Z
f (x, y , z) ds =
f (x, y , z) ds
C
−C
3. Practice problems: # 3, 7 on p 740, §13.2
Lecture 20: Line Integrals of Scalar-valued Functions - §13.2