1.
Scalar Fields and Vector fields

2.
Definition
• A scalar field is an assignment of a scalar to
each point in region in the space. E.g. the
temperature at a point on the earth is a scalar
field.
• A vector field is an assignment of a vector to
each point in a region in the space. e.g. the
velocity field of a moving fluid is a vector field
as it associates a velocity vector to each point
in the fluid.

3.
Definition
• A scalar field is a map from D to ℜ, where D is
a subset of ℜn.
• A vector field is a map from D to ℜn, where D
is a subset of ℜn.
is a subset of ℜn.
• For n=2: vector field in plane,
• for n=3: vector field in space
• Example: Gradient field

4.
Line integral
• Line integral in a scalar field
• Line integral in a vector field

5.
LINE INTEGRAL IN A SCALAR FIELD
MOTIVATION
A rescue team follows a path in a danger area where for each
position the degree of radiation is defined. Compute the total
amount of radiation gathered by the rescue team along the
path.
path.
RESCUE
BASE

6.
Piecewise Smooth Curves

7.
Piecewise Smooth Curves
A classic property of gravitational fields is that, subject to
certain physical constraints, the work done by gravity on an
object moving between two points in the field is
independent of the path taken by the object.
One of the constraints is that the path must be a piecewise
smooth curve. Recall that a plane curve C given by
smooth curve. Recall that a plane curve C given by
r(t) = x(t)i + y(t)j, a ≤ t ≤ b
is smooth if
are continuous on [a, b] and not simultaneously 0 on (a, b).

8.
Piecewise Smooth Curves
Similarly, a space curve C given by
r(t) = x(t)i + y(t)j + z(t)k, a ≤ t ≤ b
is smooth if
is smooth if
are continuous on [a, b] and not simultaneously 0 on (a, b).
A curve C is piecewise smooth if the interval [a, b] can be
partitioned into a finite number of subintervals, on each of
which C is smooth.

9.
Example 1 – Finding a Piecewise Smooth Parametrization
Find a piecewise smooth parametrization of the
graph of C shown in Figure.

10.
Because C consists of three line segments C1, C2,
and C3, you can construct a smooth parametrization
for each segment and piece them together by
making the last t-value in Ci correspond to the first t-
value in Ci + 1, as follows.
Example 1 – Solution
value in Ci + 1, as follows.

11.
So, C is given by
Example 1 – Solution
Because C1, C2, and C3 are smooth, it follows that C is
piecewise smooth.

12.
Parametrization of a curve induces an
orientation to the curve.
For instance, in Example 1, the curve is
Piecewise Smooth Curves
For instance, in Example 1, the curve is
oriented such that the positive direction is
from (0, 0, 0), following the curve to (1, 2, 1).

13.
Line Integrals
You will study a new type of integral called a line integral
for which you integrate over a piecewise smooth curve C.
To introduce the concept of a line integral, consider the
mass of a wire of finite length, given by a curve C in
space.
The density (mass per unit length) of the wire at the point
(x, y, z) is given by f(x, y, z).

14.
Line Integrals
Partition the curve C by the points
P0, P1, …, Pn
producing n subarcs, as shown in Figure.
producing n subarcs, as shown in Figure.

15.
Line Integrals
The length of the ith subarc is given by ∆si.
Next, choose a point (xi, yi, zi) in each subarc.
If the length of each subarc is small, the total mass
of the wire can be approximated by the sum
of the wire can be approximated by the sum
If you let ||∆|| denote the length of the longest
subarc and let ||∆|| approach 0, it seems
reasonable that the limit of this sum approaches
the mass of the wire.

16.
Line Integrals

17.
Line Integrals
To evaluate a line integral over a plane curve C
given by r(t) = x(t)i + y(t)j, use the fact that
A similar formula holds for a space curve.

18.
Line Integrals
Note that if f(x, y, z) = 1, the line integral gives the arc length of the
curve C. That is,

19.
Evaluate
where C is the line segment shown in Figure.
Example 2 – Evaluating a Line Integral

20.
Begin by writing a parametric form of the
equation of the line segment:
x = t, y = 2t, and z = t, 0 ≤ t ≤ 1.
Example 2 – Solution
Therefore, x'(t) = 1, y'(t) = 2, and z'(t) = 1, which
implies that

21.
So, the line integral takes the following form.
Example 2 – Solution

22.
Line Integrals
For parametrizations given by r(t) = x(t)i +
y(t)j + z(t)k, it is helpful to remember the
form of ds as

23.
• Just as for an ordinary single integral, we can
interpret the line integral of a positive
function as an area.
• In fact, if f(x, y) ≥ 0, represents
the area of one side of the “fence” or “curtain”
shown here,
( )
,
C
f x y ds
∫
shown here,
whose:
– Base is C.
– Height above the point
(x, y) is f(x, y).

24.
• Now, let C be a piecewise-smooth curve.
– That is, C is a union of a finite number of smooth
curves C1, C2, …, Cn, where the initial point of Ci+1
is the terminal point of Ci.

25.
• Then, we define the integral of f along C
as the sum of the integrals of f along each
of the smooth pieces of C:
( )
,
f x y ds
∫ ( )
( ) ( )
( )
1 2
,
, ,
... ,
n
C
C C
C
f x y ds
f x y ds f x y ds
f x y ds
= +
+ +
∫
∫ ∫
∫

26.
LINE INTEGRAL IN A VECTOR FIELD
MOTIVATION
A ship sails from an island to another one along a fixed
route. Knowing all the sea currents, how much fuel will
be needed ?

27.
One of the most important physical applications
of line integrals is that of finding the work done
on an object moving in a force field.
For example, Figure shows an inverse square
force field similar to the gravitational field of the
sun.
sun.

28.
Line Integrals of Vector Fields
To see how a line integral can be used to find
work done in a force field F, consider an object
moving along a path C in the field, as shown in
Figure.
Figure15.13

29.
Line Integrals of Vector Fields
To determine the work done by the force, you need
consider only that part of the force that is acting in the
same direction as that in which the object is moving.
This means that at each point on C, you can consider the
projection F T of the force vector F onto the unit tangent
projection F T of the force vector F onto the unit tangent
vector T.
On a small subarc of length ∆si, the increment of work is
∆Wi = (force)(distance)
≈ [F(xi, yi, zi) T(xi, yi, zi)] ∆si
where (xi, yi, zi) is a point in the ith subarc.

30.
Line Integrals of Vector Fields
Consequently, the total work done is given by the
following integral.
This line integral appears in other contexts and is
the basis of the following definition of the line
integral of a vector field.
Note in the definition that

31.
Line Integrals of Vector Fields

32.
Find the work done by the force field
on a particle as it moves along the helix given by
Example – Work Done by a Force
from the point (1, 0, 0) to (–1, 0, 3π),
as shown in Figure.

33.
Because
r(t) = x(t)i + y(t)j + z(t)k
= cos ti + sin tj + tk
it follows that x(t) = cos t, y(t) = sin t, and z(t) = t.
Example – Solution
So, the force field can be written as

34.
To find the work done by the force field in moving a
particle along the curve C, use the fact that
r'(t) = –sin ti + cos tj + k
and write the following.
Example – Solution

35.
Line Integrals of Vector Fields
For line integrals of vector functions, the
orientation of the curve C is important.
If the orientation of the curve is reversed, the
If the orientation of the curve is reversed, the
unit tangent vector T(t) is changed to –T(t), and
you obtain

36.
Line Integrals in Differential Form
Line Integrals in Differential Form

37.
Line Integrals in Differential Form
A second commonly used form of line integrals is
derived from the vector field notation used in the
preceding section.
If F is a vector field of the form F(x, y) = Mi + Nj, and C is
given by r(t) = x(t)i + y(t)j, then F • dr is often written as
given by r(t) = x(t)i + y(t)j, then F • dr is often written as
M dx + N dy.

38.
Line Integrals in Differential Form
This differential form can be extended to three
variables. The parentheses are often omitted, as
follows.

39.
Example – Evaluating a Line Integral in Differential Form
Let C be the circle of radius 3 given by
r(t) = 3 cos ti + 3 sin tj, 0 ≤ t ≤ 2π
as shown in Figure. Evaluate the line integral

40.
Example – Solution
Because x = 3 cos t and y = 3 sin t, you have dx = –3 sin t
dt and dy = 3 cos t dt. So, the line integral is

41.
Example – Solution

42.
Suppose instead of being a force field, suppose that F represents
the velocity field of a fluid flowing through a region in space.
Under these circumstances, the integral of F .T along a curve in
the region gives the fluid’s flow along the curve.

43.
EXAMPLE: Finding Circulation Around a Circle

44.
Flux Across a Plane Curve
To find the rate at which a fluid is entering or leaving a
region enclosed by a smooth curve C in the xy-plane, we
calculate the line integral over C of F.n, the scalar
component of the fluid’s velocity field in the direction of
the curve’s outward-pointing normal vector.

45.
Notice the difference between flux and
circulation: Flux is the integral of the normal
component of F; circulation is the integral of
the tangential component of F.
the tangential component of F.

46.
How to evaluate Flux of F across C
we choose a smooth
parameterization
that traces the curve C exactly
once as t increases from a to b. We
can find the outward unit normal
vector n by crossing the curve’s
vector n by crossing the curve’s
unit tangent vector T with the
vector k.
• If the motion is clockwise, k×T
points outward;
• if the motion is counterclockwise,
T×k points outward
We choose: n = T × k

47.
Now,
Here the circle on the integral shows that the integration around the closed
curve C is to be in the counterclockwise direction.

48.
EXAMPLE: Finding Flux Across a Circle
Note that the flux of F across the circle is positive, implies the net flow across the curve
is outward. A net inward flow would have given a negative flux.

49.
Path Independence
Under differentiability conditions, a field F is conservative iff it is the
gradient field of a scalar function ƒ; i.e., iff for some ƒ. The
function ƒ then has a special name.

50.
once we have found a potential function ƒ for a field F,
we can evaluate all the work integrals in the domain of
F over any path between A and B by

51.
Connectivity and Simple Connectivity
• All curves are piecewise smooth, that is,
made up of finitely many smooth pieces
connected end to end.
• The components of F have continuous first
• The components of F have continuous first
partial derivatives implies that when
this continuity requirement guarantees that
the mixed second derivatives of the potential
function ƒ are equal.

52.
Simple curve: A curve that doesn’t intersect
itself anywhere between its endpoints.
r(a) = r(b) for a simple closed curve
But r(a) ≠ r(b) when a t1 t2 b

53.
Simply-connected region: A simply-connected region in the
plane is a connected region D such that every simple closed
curve in D encloses only points that are in D.
Intuitively speaking, a simply-connected
region contains no hole and can’t consist
of two separate pieces.
of two separate pieces.
An open connected region means that every point can be connected to every other point
by a smooth curve that lies in the region. Note that connectivity and simple connectivity
are not the same, and neither implies the other. Think of connected regions as being in
“one piece” and simply connected regions as not having any “holes that catch loops.”

54.
INDEPENDENCE OF PATH
Suppose C1 and C2 are two piecewise-smooth curves (which are
called paths) that have the same initial point A and terminal point
B. We have
Note: F is conservative on D is equivalent to saying that
the integral of F around every closed path in D is zero. In
other words, the line integral of a conservative vector field
depends only on the initial point and terminal point of a
curve.

55.
EXAMPLE: Finding Work Done by a Conservative Field

56.
Proof that Part 1 Part 2
Proof that Part 1 Part 2
If we have two paths from A to B, one of
them can be reversed to make a loop.

57.
Finding Potentials for Conservative Fields

58.
EXAMPLE: Finding a Potential Function

59.
Integrating first equation w.r.t. ‘x’
Differentiating it w.r.t. ‘y’ and equating with the
Computing ‘g’ as a function of ‘y’ gives
Computing ‘g’ as a function of ‘y’ gives
Further differentiating ‘f’ w.r.t. to ‘z’ and equating it with
Integrating, we have

60.
EXAMPLE: Showing That a Differential Form Is Exact