Chapter 16
Integration in Vector Fields
Note: This module is prepared from Chapter 16 (Thomas’ Calculus, 13th edition) just to help
the students. The study material is expected to be useful but not exhaustive. For detailed study,
the students are advised to attend the lecture/tutorial classes regularly, and consult the text book.
Dr. Suresh Kumar, Associate Professor,
Department of Mathematics, BITS Pilani, Pilani Campus, Rajasthan-333031, INDIA.
E-mail: suresh.kumar@pilani.bits-pilani.ac.in, sukuyd@gmail.com
Appeal: Please do not print this e-module unless it is really necessary.
Dr. Suresh Kumar, Department of Mathematics, BITS Pilani, Pilani Campus
1
Contents
16.0 Scalar and vector fields . . . . . . . . . . . . . . . . . . . . . . . . . . .
16.1 Line integral of a scalar field . . . . . . . . . . . . . . . . . . . . . . .
Line integrals in plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
16.2 Line Integral of a Vector Field: Work, Circulation, and Flux . . .
Work Done . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Flow and Circulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Flux . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
16.3 Path Independence, Conservative Fields, and Potential Functions
16.4 Green’s Theorem in the Plane . . . . . . . . . . . . . . . . . . . . . .
Divergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Spin around an axis: The k-component of curl . . . . . . . . . . . .
Two forms of Green’s Theorem . . . . . . . . . . . . . . . . . . . . . .
16.5 Surface and Area . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Parametrization of a surface . . . . . . . . . . . . . . . . . . . . . . . .
Smooth surface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Area of a smooth surface . . . . . . . . . . . . . . . . . . . . . . . . . .
Area of an explicitly defined smooth surface . . . . . . . . . . . . .
Area of an implicitly defined smooth surface . . . . . . . . . . . . .
16.6 Surface Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Integral over a smooth surface . . . . . . . . . . . . . . . . . . . . . .
Orientable surface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Surface integral of a vector field . . . . . . . . . . . . . . . . . . . . .
Flux . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
16.7 Stokes Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
16.8 Divergence Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . .
SUMMARY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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31
16.0 Scalar and vector fields
Scalar Field: A scalar field is a function f (x, y, z) that yields a scalar to each point of its domain. For example, consider an air medium with temperature at 0 degrees. Suppose a heat
source of temperature 16 degrees is switched on at the origin in the air medium, and it raises the
temperature of surroundings such that the temperature decreases as square of the distance from
the origin. Then temperature in the spherical region x2 + y 2 + z 2 ≤ 16 is given by the function
T (x, y, z) = 16 − x2 − y 2 − z 2 . Obviously, this temperature function is a scalar field.
Vector Field: A vector field is a function that assigns a vector to each point of its domain. A
vector field on a three-dimensional domain in space can have a formula like
→
−
F = M (x, y, z)î + N (x, y, z)ĵ + P (x, y, z)k̂,
where the components M , N and P are scalar functions (scalar fields) of x, y, z. Thus, a vector
→
−
field is a system of vectors, where initial point of each vector F (x, y, z) is (x, y, z).
For example, the velocity vectors in water flow in a channel constitute a vector field. See the
figure below where the streamlines are shown in a contracting channel. Notice that the water
speeds up as the channel narrows and the velocity vectors increase in length.
Figure 1: Water flow in a channel
Another example of vector field is of the gravitational field of earth where each vector points
towards the center of the earth as shown in the figure below.
Figure 2: Gravitational field of earth
3
Ex. Draw the following vector fields which represent the velocity of a gas flowing in the xy-plane.
→
−
(a) Uniform expansion: F (x, y) = xî + y ĵ,.
→
−
(b) Uniform counter-clockwise rotation: F (x, y) = −y î + xĵ.
→
−
(c) Shearing flow: F (x, y) = y î.
→
−
x
î + x2 +y
(d) Whirlpool effect: F (x, y) = x2−y
2 ĵ.
+y 2
Sol.
Ex. Consider the flow of water in xy-plane with uniform velocity, say 2 m/s, in the direction of
positive x-axis. Write the vector field representing this flow, and make its rough sketch.
→
−
Sol. F (x, y) = 2î. Try the rough sketch yourself. At different points, draw vectors of length 2
units each parallel to x-axis in the direction of î.
4
16.1 Line integral of a scalar field
Let C be a smooth curve1 with end points A and B in the domain of a real valued function or a
scalar field f (x, y, z).
Partition the curve C into n subarcs with lengths ∆sk (k = 1, 2, ..., n) such that ∆sk → 0 as
n → ∞. Let (xk , yk , zk ) be a point on the kth subarc. Then the limit
lim
n→∞
n
X
f (xk , yk , zk )∆sk ,
k=1
if exists, defines the line integral of f (x, y, z) along the curve C, and is written as
Z
Z
B
f (x, y, z)ds = lim
f (x, y, z)ds =
n→∞
A
C
n
X
f (xk , yk , zk )∆sk ,
k=1
where ∆sk → 0 as n → ∞.
If the curve C is parametrized by x = x(t), y = y(t), z = z(t), and its end points A and B
correspond to the parameter values t = a and t = b, respectively, then
Z b
Z B
Z
ds
f (x, y, z)ds =
f (x, y, z)ds =
f (x(t), y(t), z(t)) dt,
dt
a
A
C
p
where ds
=
x0 (t)2 + y 0 (t)2 + z 0 (t)2 .
dt
Note that if the function f is continuous on the smooth curve C, then the line integral of f
can be shown to exist along the curve C.
Further, if C is piecewise smooth curve, that is, C consists of a finite number of smooth arcs C1 ,
C2 , ....., Cn joined end to end, then the line integral of f along C in the sum of the line integrals
of f along the smooth arcs making C, that is,
Z
Z
Z
Z
f ds =
f ds +
f ds + ..... +
f ds.
C
C1
C2
Cn
−
Smooth Curve: A curve C given by →
r (t) = x(t)î + y(t)ĵ + z(t)k̂, a ≤ t ≤ b is smooth if
non-zero in [a, b].
1
5
−
d→
r
dt
is continuous and
Ex. Integrate f (x, y, z) = x − 3y 2 + z over the line segment C joining the origin to the point
(1, 1, 1).
Sol. A simple parametrization of the given path C reads as
→
−
r (t) = tî + tĵ + tk̂,
∴
0 ≤ t ≤ 1.
−
√
√
d→
r
ds
=
= 1 + 1 + 1 = 3.
dt
dt
It follows that
Z
Z
f (x, y, z)ds =
0
C
1
ds
f (t, t, t) dt =
dt
Z
1
√
(t − 3t2 + t) 3dt = 0.
0
Ex. Integrate f (x, y, z) = x2 + y + z along the curve C consisting of the sides of the triangle
(traversed counterclockwise) with vertices A(1, 0, 0), B(0, 1, 0) and C(0, 0, 1).
Sol. Here C is not smooth but it consists of three smooth parts which are the line segments AB,
BC and CA.
Z
Z
Z
Z
∴
f ds =
f ds +
f ds +
f ds.
C
Let us first solve
AB
R
AB
BC
CA
f ds. The parametric equation of the line AB is
y−0
z−0
x−1
=
=
= t,
1
−1
0
p
√
that is, x = t + 1, y = −t, z = 0. So ds
= x0 (t)2 + y 0 (t)2 + z 0 (t)2 = 2. Further, along AB, we
dt
find f = (t + 1)2 − t = t2 + t + 1. Also, from the relations x = t + 1, y = −t, z = 0, it is easy to
find that the point A(1, 0, 0) corresponds to t = 0 and the point B(0, 1, 0) corresponds to t = −1.
Thus,
Z
Z −1
√
5√
f ds =
(t2 + t + 1) 2 dt = −
2.
6
AB
0
Likewise, the line integrals along the sides BC and CA can be determined.
Line integrals in plane
Here we discuss an interesting geometrical meaning of line integral. Suppose z = f (x, y) is a nonnegative and continuous function representing a curve in space with projection or domain curve C
in the XY-plane as shown in figure below.
Then the line integral of f (x, y) along C,
Z
f (x, y)ds = lim
C
n→∞
n
X
f (xk , yk )∆sk ,
k=1
where ∆sk → 0 as n → ∞, gives the area of the portion of the cylindrical surface or “wall”R beneath
b
or under the space curve z = f (x, y) ≥ 0. It is clearly the generalization of the integral a f (x)dx
(area under the plane curve y = f (x) ≥ 0 from x = a to x = b above the X-axis) where the interval
[a, b] happens to be the projection of the plane curve y = f (x) on the X-axis.
6
Ex. Find area of the curved surface of the cylinder x2 + y 2 = a2 , 0 ≤ z ≤ h.
Sol. Here z = f (x, y) = h, x2 + y 2 = a2 is a non-negative and continuous function representing a
circular curve in space (the plane z = h) with projection or domain curve C : x2 + y 2 = a2 in the
XY-plane. So the required area is the area beneath the circular curve x2 + y 2 = a2 lying in the
plane z = h. The domain curve C : x2 + y 2 = a2 in the XY-plane is parametrized by x = a cot t,
y = a sin t, 0 ≤ t ≤ 2π.
Z 2π p
Z
h a2 cos2 t + a2 sin2 t dt = 2πah.
f (x, y)ds =
∴ Area =
0
C
16.2 Line Integral of a Vector Field: Work, Circulation, and
Flux
Let C be a smooth curve with end points A and B in the domain of a vector valued function or
→
−
→
−
vector field F (x, y, z). Then line integral of F (x, y, z) along the curve C is defined as the line
→
−
integral of its tangential component F .T̂ along the curve C, that is,
Z
Z
Z
−
→
− d→
→
− →
→
−
r
F.
F .d−
r,
F .T̂ ds =
ds =
ds
C
C
C
−
where →
r = xî + y ĵ + z k̂ is position vector of any point (x, y, z) on the curve C.
−
If the curve C is parametrized by →
r (t) = x(t)î + y(t)ĵ + z(t)k̂, and its end points A and B
correspond to the parameter values t1 and t2 , respectively, then
Z
Z
Z t2
−
→
−
→
− →
→
− d→
r
−
F .T̂ ds =
F .d r =
F.
dt.
dt
C
C
t1
In the following, we mention some physical applications of line integral.
Work Done
→
−
If a particle moves along a smooth curve C from A to B under the action of force filed F (x, y, z),
then the work done in moving the particle from A to B along C is given by
Z
Z
Z B
→
−
→
− →
→
− →
−
W =
F .T̂ ds =
F .d r =
F .d−
r.
C
C
A
7
→
−
Ex. Find the work done by the force field F = xî + y ĵ + z k̂ in moving an object along the curve
−
C parametrized by →
r (t) = cos(πt)î + t2 ĵ + sin(πt)k̂, 0 ≤ t ≤ 1.
Sol. The work done is the line integral
Z 1
Z 1
−
→
− d→
1
r
dt =
2t3 dt = .
F.
dt
2
0
0
Flow and Circulation
→
−
If F (x, y, z) represents the velocity field in a fluid flow, then the flow along a smooth curve C
→
−
extending from A to B in domain of F (x, y, z) is given by
Z
Flow =
→
−
F .T̂ ds =
C
Z
→
− →
F .d−
r =
Z
B
→
− →
F .d−
r.
A
C
→
−
In case, C is a closed curve, then the above integral is called circulation of F (x, y, z) around C,
and is denoted by
I
I
→
−
→
− →
F .T̂ ds =
F .d−
r.
C
C
→
−
−
Ex. Find the circulation of the field F = (x − y)î + xĵ around the circle C given by →
r (t) =
cos tî + sin tĵ, 0 ≤ t ≤ 2π.
Sol. The circulation of the given field is
Z 2π
Z 2π
−
→
− d→
r
F.
(1 − sin t cos t)dt = 2π.
dt =
dt
0
0
Flux
→
−
Consider the planar flow of a fluid with continuous velocity field F (x, y) in the xy-plane. If a
→
−
smooth and simple closed curve C lies in the domain of F (x, y), and n̂ is a outward pointing
→
−
unit normal vector on C, then the outward flux of F across C is given by line integral of normal
→
−
→
−
component F .n̂ of F along C, that is,
I
→
−
Outward Flux =
F .n̂ds.
C
If the curve C is traversed counterclockwise in the xy-plane as viewed from the tip of positive
z-axis (see Figure 3), then we have
−
d→
r
dx
dy
dx
dy
n̂ = T̂ × k̂ =
× k̂ =
î + ĵ × k̂ = − ĵ + î.
ds
ds
ds
ds
ds
→
−
→
−
dy
dx
− N . So
. If F (x, y) = M (x, y)î + N (x, y)ĵ, then F .n̂ = M
ds
ds
I
I
→
−
Outward Flux =
F .n̂ds =
M dy − N dx.
C
C
8
Figure 3:
→
−
−
Ex. Find the flux of the field F = (x − y)î + xĵ across the circle C given by →
r (t) = cos tî +
sin tĵ, 0 ≤ t ≤ 2π.
Sol. Here M = x − y and N = x. On C, x = cos t and y = sin t. So the flux of the given field is
Z 2π
Z 2π
cos2 tdt = π.
(M dy − N dx) =
0
0
9
16.3 Path Independence, Conservative Fields, and Potential
Functions
→
−
→
−
Consider a vector field F defined in an open region D2 . If the line integral of F along any smooth
R →
RB→
− −
− −
r does not
path C extending from a point A to a point B inside D, that is, C F .d→
r = A F .d→
depend on the path C joining A and B rather it depends solely on the end points A and B of
→
−
→
−
the path, then the line integral of F is said to be path independent in D, and F is said to be
conservative field in D.
→
−
Suppose there exits a differentiable function φ such that F = ∇φ in D. Then φ is called
→
−
−
potential of the vector field F in D. Suppose a smooth curve C parameterized by →
r (t) =
x(t)î + y(t)ĵ + z(t)k̂ lies in D, and its end points A and B correspond to the parameter values t1
and t2 , respectively. Along the curve C, φ(x(t), y(t), z(t)) is a differentiable function of t. So we
have
−
−
→
− d→
dφ dx dφ dy dφ dz
d→
r
r
dφ
=
+
+
= ∇φ.
= F.
.
dt
dx dt
dy dt
dz dt
dt
dt
It follows that
Z
Z
→
− →
−
F .d r =
C
B
→
− →
F .d−
r =
A
Z
t2
t1
Z t2
−
→
− d→
r
dφ
F.
dt =
dt = φ(x(t), y(t), z(t))]tt21 = φ(B) − φ(A).
dt
t1 dt
→
−
This shows that line integral of F is path independent in D. The above result is known as the
fundamental theorem of line integrals. We notice that the path independence of the line integral
→
−
→
−
of F is ensured provided F is gradient field of some differentiable function φ. In this regard, we
shall state some useful results. But before that let us see a practical example of conservative field
from the world of Physics.
The most prominent examples of conservative forces are the gravitational force and the electric
force associated to an electrostatic field. According to Newton’s law of gravitation, the gravitational
→
−
force F acting on a mass m due to a mass M located at distance r, obeys the equation
→
−
GM m
F = − 2 r̂,
r
where G is the gravitational constant and r̂ is a unit vector pointing from M towards m. The
force of gravity is conservative because
→
−
F = −∇φ,
where
GM m
r
is the gravitational potential energy.
φ=−
Simply connected regions
A region D is said to be connected if any two points of D can be joined by a polygonal path inside
D. Further, a connected region D is said to be simply connected if any closed curve inside D
encloses points of D only.
2
Open region is a region which contains a neighbourhood of its every point. In other words, all points of an
open region are its interior points.
10
Conservative fields are gradient fields.
→
−
→
−
Let F be a continuous vector field in an open connected region D. Then F is conservative if and
→
−
only if F = ∇φ for some differentiable function φ.
Loop property of conservative fields
→
−
→
−
Let F I
be a continuous vector field in an open connected region D. Then F is conservative if and
→
− →
F .d−
r = 0 for every loop or closed curve C in D.
only if
C
Component test for conservative fields
→
−
Let F = M (x, y, z)î + N (x, y, z)ĵ + P (x, y, z)k̂ be a vector field having continuous first order
→
−
partial derivatives in a simply connected open region D. Then F is conservative if and only if
∂M
∂N ∂N
∂P ∂P
∂M
=
,
=
,
=
.
∂y
∂x ∂z
∂y ∂x
∂z
→
−
→
−
→
−
By verifying the above conditions on components of F , we find whether F is conserved. If F
→
−
is conserved, then there exists some differentiable function φ such that F = ∇φ or
M (x, y, z)î + N (x, y, z)ĵ + P (x, y, z)k̂ =
∂φ
∂φ
∂φ
î +
ĵ +
k̂,
∂x
∂y
∂z
or
M=
∂φ
∂φ
∂φ
, N=
, P =
.
∂x
∂y
∂z
To determine φ, we solve the above equations as illustrated in the following example.
→
−
Ex. Show that the field F = y î + xĵ + 4k̂ is conservative. Find its potential and hence evaluate
the line integral
Z
(2,3,−1)
ydx + xdy + 4dz
(1,1,1)
Sol. Here M = y, N = x and P = 4. So we have
∂M
∂N ∂N
∂P ∂P
∂M
=1=
,
=0=
,
=0=
.
∂y
∂x ∂z
∂y ∂z
∂x
11
This shows that the given field is conservative.
→
−
→
−
Now, let φ be potential of F so that F = ∇φ. So we have
∂φ
= y,
∂x
∂φ
= x,
∂y
∂φ
= 4.
∂z
We shall use these three equations to determine φ. From first equation, we get
φ(x, y, z) = xy + g(y, z),
which gives
∂φ
∂g
=x+
.
∂y
∂y
In view of second equation, we have
∂g
= 0.
∂y
This shows that g is a function of z alone, and therefore
φ(x, y, z) = xy + h(z).
∂φ
dh
=0+
= 4, or h(z) = 4z + C.
∂z
dz
= 4. Thus, finally we get
in view of the third equation ∂φ
∂z
∴
φ(x, y, z) = xy + 4z + C.
Since the given vector field is conservative, so the required line integral is independent of the path
from (1, 1, 1, ) to (2, 3, −1), and equals
φ(2, 3, −1) − φ(1, 1, 1) = 2 + C − (5 + C) = −3.
Note. We can solve the above integral without using/finding the potential φ because the line integral of the conservative field is path independent. So we can use any path (for example, straight
line joining the end points) for solving the integral. Solve the above line integral by using the
parametrization of straight line passing through the points (2, 3, −1), (1, 1, 1), and verify the answer −3.
→
−
Ex. Show that the field F = (ex cos y + yz)î + (xz − ex sin y)ĵ + (xy + z)k̂ is conservative over its
natural domain. Find its potential.
Sol. φ(x, y, z) = ex cos y + xyz + 12 z 2 + C.
16.4 Green’s Theorem in the Plane
In this section we derive a method for computing a work or flux integral over a closed curve C in
→
−
the plane when the field F is not conservative. This method, known as Green’s Theorem, allows
us to convert the line integral into a double integral over the region enclosed by C. The discussion
is given in terms of velocity fields of fluid flows (a fluid is a liquid or a gas) because they are easy
to visualize. However, Green’s Theorem applies to any vector field, independent of any particular
interpretation of the field, provided the assumptions of the theorem are satisfied. We introduce
two new ideas for Green’s Theorem: divergence and curl.
12
Divergence
→
−
Consider planar flow of a fluid in the xy-plane. Suppose F (x, y) = M (x, y)î + N (x, y)ĵ is the
velocity field of the fluid having continuous first order partial derivatives in a region R of the
xy-plane.
Let A(x, y) be a point in R, and ABCD be a small rectangle traversed counterclockwise
completely lying inside R, with edges parallel to the coordinate axes having lengths ∆x and
∆y. To be more precise, let AB = ∆x is bottom edge; CD = ∆x is the top edge; AD = ∆y is
the left edge and BC = ∆y is the right edge. We also assume that the velocity field components
M and N do not change their signs throughout the small rectangle ABCD. Then the amount of
fluid coming out of the bottom edge AB = ∆x per unit time is given by
→
−
F (x, y).(−ĵ)∆x = −N (x, y)∆x.
Likewise, the flow rate of the fluid across the top edge CD = ∆x of the rectangle is
→
−
F (x, y + ∆y).ĵ∆x = N (x, y + ∆y)∆x.
The flow rate of the fluid across the left edge AD = ∆y of the rectangle is
→
−
F (x, y).(−î)∆y = −M (x, y)∆y,
while the flow rate of the fluid across the right edge BC = ∆y of the rectangle reads as
→
−
F (x + ∆x, y).î∆y = M (x + ∆x, y)∆y.
Summing the flow rates across the opposite pairs gives
∂N
∆y∆x
∂y
∂M
∆x∆y
Right and Left: M (x + ∆x, y)∆y − M (x, y)∆y ≈
∂x
Top and Bottom: N (x, y + ∆y)∆x − N (x, y)∆x ≈
So the net flow rate across the entire rectangle or the flux across the rectangle ≈
∂M
∂x
∂N
.
∂y
∂M
∂x
+
∂N
∂y
∆x∆y.
So the flux per unit area or flux density for the rectangle ≈
+
In the limit (∆x, ∆y) → (0, 0), we get the flux density at the point A(x, y), and we call it the
→
−
divergence of F at the point A(x, y). Formally, we have the following definition of divergence.
13
→
−
The divergence (flux density) of a vector field F (x, y) = M (x, y)î + N (x, y)ĵ at a point (x, y)
is
→
−
∂N
∂M
+
.
div F =
∂x
∂y
If a fluid is compressible, then the divergence of its velocity field measures to what extent it is
expanding or contracting at each point. Intuitively, if a fluid is expanding at a point (x, y), the
lines of flow would diverge there, and since the fluid would be flowing out of a small rectangle about
(x, y), the divergence of the velocity field at (x, y) would be positive. In case of compression, the
divergence would be negative. The points of expansion and compression in a fluid are sometimes
called as sources (something coming out) and sinks (something loosing into), respectively. Of
course, there would be no sources and sinks in an incompressible fluid, and divergence of the
velocity field of such a fluid would be zero at every point.
→
−
The divergence of a more general vector field F = M (x, y, z)î + N (x, y, z)ĵ + P (x, y, z)k̂ at a
point (x, y, z) is defined as
→
−
∂M
∂N
∂P
div F =
+
+
.
∂x
∂y
∂z
→
−
An easy to use notation and formula for computing divergence of F = M (x, y, z)î + N (x, y, z)ĵ +
P (x, y, z)k̂ read as
→
−
→
−
∂
∂
∂
∂M
∂N
∂P
div F = ∇. F = î
+ ĵ
+ k̂
.(M î + N ĵ + P k̂) =
+
+
.
∂x
∂y
∂z
∂x
∂y
∂z
Spin around an axis: The k-component of curl
→
−
Consider planar flow of a fluid in the xy-plane. Suppose F (x, y) = M (x, y)î + N (x, y)ĵ is the
velocity field of the fluid having continuous first order partial derivatives in a region R of the
xy-plane.
Let A(x, y) be a point in R, and ABCD be a small rectangle traversed counterclockwise
completely lying inside R, with edges parallel to the coordinate axes having lengths ∆x and
∆y. To be more precise, let AB = ∆x is bottom edge; CD = ∆x is the top edge; AD = ∆y is
the left edge and BC = ∆y is the right edge. We also assume that the velocity field components
M and N are positive throughout the small rectangle ABCD. Then the amount of fluid flowing
along the bottom edge AB = ∆x per unit time is given by
→
−
F (x, y).î∆x = M (x, y)∆x.
14
Likewise, the flow rate of the fluid along the top edge CD = ∆x of the rectangle is
→
−
F (x, y + ∆y).(−î)∆x = −M (x, y + ∆y)∆x.
The flow rate of the fluid along the left edge DA = ∆y of the rectangle is
→
−
F (x, y).(−ĵ)∆y = −N (x, y)∆y,
while the flow rate of the fluid along the right edge BC = ∆y of the rectangle reads as
→
−
F (x + ∆x, y).ĵ∆y = N (x + ∆x, y)∆y.
Summing the flow rates across the opposite pairs gives
Top and Bottom: −M (x, y + ∆y)∆x + M (x, y)∆x ≈ −
Right and Left: N (x + ∆x, y)∆y − N (x, y)∆y ≈
∂M
∆y∆x
∂y
∂N
∆x∆y
∂x
So the net flow rate or the counterclockwise circulation rate around the rectangular boundary
∂N
∂M
≈
−
∆x∆y.
∂x
∂y
∂N
∂M
−
.
∂x
∂y
→
−
In the limit (∆x, ∆y) → (0, 0), we get the circulation density of F at the point A(x, y).
Physically, it is the measure of how a floating paddle wheel with axis perpendicular to the xyplane spins at a point (x, y) in the fluid flowing in the xy-plane.
− ∂M
, in fact, is the k-component of a more general vector known as curl of
The expression ∂N
∂x
∂y
→
−
F . Formally, we have the following definition of curl of a vector field.
→
−
The curl of a vector field F = M (x, y, z)î + N (x, y, z)ĵ + P (x, y, z)k̂ at a point (x, y, z) is
→
−
∂P
∂N
∂M
∂P
∂N
∂M
curl F =
−
î +
−
ĵ +
−
k̂.
∂y
∂z
∂z
∂x
∂x
∂y
→
−
An easy to use notation and formula for computing curl of F = M (x, y, z)î + N (x, y, z)ĵ +
P (x, y, z)k̂ read as
So the circulation density for the rectangle ≈
→
−
→
−
curl F = ∇ × F =
î
ĵ
k̂
∂
∂x
∂
∂y
∂
∂z
M N
P
=
∂P
∂N
−
∂y
∂z
15
î +
∂M
∂P
−
∂z
∂x
ĵ +
∂N
∂M
−
∂x
∂y
k̂.
Note that the determinant here is not the determinant of numbers. The determinant notation is
used just to facilitate the calculation of curl.
→
−
→
−
A vector field F is said to be irrotational if ∇ × F = 0.
→
−
For example, the vector field F = xî + y ĵ + z k̂ is irrotational. Verify!
Next verify that curl of a gradient vector is zero, that is, ∇ × ∇f = 0.
16
Ex. The following vector fields represent the velocity of a gas flowing in the xy-plane (see the
corresponding figures also).
→
−
(a) Uniform expansion or compression: F (x, y) = cxî + cy ĵ, c > 0.
→
−
(b) Uniform rotation: F (x, y) = −cy î + cxĵ, c > 0.
→
−
(c) Shearing flow: F (x, y) = y î.
→
−
x
î + x2 +y
(d) Whirlpool effect: F (x, y) = x2−y
2 ĵ.
+y 2
Find the divergence and curl of each vector field and interpret its physical meaning.
→
−
Sol. (a) ∇. F = 2c: The gas is undergoing uniform expansion. In a small circular nbd of any
point in the vector field domain, the size/magnitude of vectors entering the nbd is smaller than
the size of vectors leaving the nbd, thus an expansion.
→
−
∇ × F = 0: The gas is not circulating at very small scales.
→
−
(b) ∇. F = 0: The gas is neither expanding nor compressing.
→
−
∇ × F = 2ck̂: It indicates uniform counterclockwise rotation at every point. In a small circular
nbd of any point in the vector field domain, the size/magnitude of vectors entering the nbd is same
as the size of vectors leaving the nbd, thus no expansion.
→
−
(c) ∇. F = 0: The gas is neither expanding nor compressing.
→
−
∇ × F = −k̂: It indicates uniform clockwise rotation at every point.
→
−
(d) ∇. F = 0: The gas is neither expanding nor compressing.
→
−
∇ × F = 0: The circulation density is 0 at every point away from the origin (where the vector
field is undefined and the whirlpool effect is taking place), and the gas is not circulating at any
point for which the vector field is defined.
17
Two forms of Green’s Theorem
We have the following two forms of the Green’s theorem.
Flux-Divergence or Normal Form
→
−
Let F (x, y) = M (x, y)î+N (x, y)ĵ be a vector field having continuous first order partial derivatives
in an open region in xy-plane containing a piecewise smooth simple closed curve C enclosing a
→
−
region R. Then the outward flux of F across the curve C equals the double integral of the
→
−
divergence of F over the region R, that is,
I
I
ZZ →
−
∂M
∂N
F .n̂ds =
M dy − N dx =
+
dxdy.
∂x
∂y
C
C
R
Circulation-Curl or Tangential Form
→
−
Let F (x, y) = M (x, y)î+N (x, y)ĵ be a vector field having continuous first order partial derivatives
in an open region in xy-plane containing a piecewise smooth simple closed curve C enclosing a
→
−
region R. Then the counterclockwise circulation of F around the curve C equals the double
→
−
integral of curl F .k̂ over R, that is,
ZZ I
I
→
−
∂N
∂M
M dx + N dy =
F .T̂ ds =
−
dxdy.
∂x
∂y
R
C
C
→
−
Ex. Verify both forms of the Green’s theorem for the vector field F = (x − y)î + xĵ and the region
−
R enclosed by the circle C given by →
r (t) = cos tî + sin tĵ, 0 ≤ t ≤ 2π.
Sol. Here M = x − y and N = x. On C, x = cos t and y = sin t. So
I
Z 2π
(M dy − N dx) =
cos2 tdt = π.
C
0
Next,
ZZ R
∂N
∂M
+
∂x
∂y
→
−
F .n̂ds =
I
I
∴
C
ZZ
dxdy =
(1 + 0)dxdy = π.
R
ZZ M dy − N dx = π =
C
R
∂M
∂N
+
∂x
∂y
dxdy
This verifies normal form of Green’s theorem.
Similarly, it can be verified that
I
I
ZZ →
−
∂N
∂M
F .T̂ ds =
M dx + N dy = 2π =
−
dxdy.
∂x
∂y
C
R
C
Ex. Use Green’s theorem to find area of the circle x2 + y 2 = a2 .
Sol. Let C be the boundary of the given circle traversed counter-clock wise enclosing the region
→
−
R of the circle. Let us choose, F = −y î + xĵ so that M = −y, N = x and ∂N
− ∂M
= 2. It follows
∂x
∂y
that
ZZ ZZ
∂N
∂M
−
dxdy = 2
dxdy,
∂x
∂y
R
R
18
which is two times the area of the circle x2 + y 2 = a2 . Next, by the tangential form of the Green’s
theorem, we have
I
ZZ ∂M
∂N
−
dxdy.
M dx + N dy =
∂x
∂y
C
R
So area of the given circle is
ZZ
I
I
Z
1
1
1 2π
dxdy =
[(−a sin t)(−a sin t)+(a cos t)(a cos t)]dt = πa2 .
M dx+N dy =
(−ydx+xdy) =
2
2
2
R
0
C
C
Note. The formula
ZZ
I
1
dxdy =
(−ydx + xdy)
2 C
R
gives the area of any region R in xy-plane enclosed by a counter-clock wise traversed curve C. It
may also be noted that we get the same formula for area when we choose M = x and N = y in
the normal form of Green’s theorem.
19
16.5 Surface and Area
Parametrization of a surface
−
We know →
r (t) = xî + y ĵ + z k̂ is the position vector of any point (x, y, z) in xyz-space. In case, x,
−
y, z are continuous functions of a single variable t (a ≤ t ≤ b), then →
r (t) = x(t)î + y(t)ĵ + z(t)k̂
represents a curve in space with end points (x(a), y(a), z(a)) and (x(b), y(b), z(b)). So the curve is
parametrized by the single parameter t.
−
Likewise, →
r (u, v) = x(u, v)î + y(u, v)ĵ + z(u, v)k̂ represents a surface in space parametrized by
the two parameters u and v.
−
For example, →
r (φ, θ) = a sin φ cos θî+a sin φ sin θĵ +a cos φk̂, 0 ≤ φ ≤ π, 0 ≤ θ ≤ 2π represents
2
2
the sphere x + y + z 2 = a2 .
−
If we set φ = π/2, then →
r (π/2, θ) = a cos θî + a sin θĵ, 0 ≤ θ ≤ 2π represents the circle
x2 + y 2 = a2 , as expected.
−
Obviously, the partial derivative →
rθ (π/2, θ) = −a sin θî + a cos θĵ, 0 ≤ θ ≤ 2π is tangent vector
to the circle x2 + y 2 = a2 .
−
−
In general, the vectors →
ru and →
rv evaluated at some point (u0 , v0 ) of the plane of parameters u
−
and v lie in the tangent plane to the surface →
r (u, v) = x(u, v)î + y(u, v)ĵ + z(u, v)k̂ at the point
−
−
(x(u0 , v0 ), y(u0 , v0 ), z(u0 , v0 )) of the surface provided →
ru and →
rv are non-zero at (u0 , v0 ). It implies
→
−
→
−
→
−
that ru × rv is normal vector to the surface r (u, v).
Smooth surface
−
−
−
The parametrized surface →
r (u, v) = x(u, v)î + y(u, v)ĵ + z(u, v)k̂ is said to be smooth if →
ru and →
rv
→
−
→
−
→
−
are continuous with ru × rv 6= 0 at every point of the parameters domain under consideration.
→
−
−
−
−
The condition that →
ru × →
rv 6= 0 in the definition of smoothness implies that the vectors →
ru and
→
−
rv are non-zero and are not parallel, so these always determine a plane tangent to the surface. We
relax this condition on the boundary of the domain, but this does not affect the area computations.
Area of a smooth surface
−
Consider a smooth surface S parametrized by →
r (u, v) = x(u, v)î + y(u, v)ĵ + z(u, v)k̂, where u
and v belong to some region R in the uv-plane. Divide the region R into meshes by choosing
lines parallel to u− and v-axes. Let there be n complete rectangles with dimensions ∆ui and
∆vi (i = 1, 2, ...., n). Let ∆σi be the area of the patch on the surface S corresponding to the ith
rectangle. We approximate the area ∆σi by the area of the parallelogram described by the vectors
−
−
∆ui →
ru and ∆vi →
rv , that lies in the tangent plane to the surface.
So we have
−
−
−
−
∆σ ≈ |∆u →
r × ∆v →
r | = |→
r ×→
r |∆u ∆v .
i
i u
i v
u
v
i
So the approximate area of the surface S =
(∆ui , ∆vi ) → (0, 0), the sum
n
X
i
n
X
−
−
|→
ru × →
rv |∆ui ∆vi . In the limit n → ∞ such that
i=1
−
−
|→
ru × →
rv |∆ui ∆vi defines the double integral
−
−
|→
ru × →
rv |dudv,
R
i=1
and gives the area of the surface S, that is,
ZZ
ZZ
−
−
Area of S =
dσ =
|→
ru × →
rv |dudv.
S
ZZ
R
20
Ex. Find the surface area of the sphere x2 + y 2 + z 2 = a2 .
Sol. We use the following well known parametrization:
→
−
r (φ, θ) = a sin φ cos θî + a sin φ sin θĵ + a cos φk̂, 0 ≤ φ ≤ π, 0 ≤ θ ≤ 2π.
−
−
Then we find, →
rφ × →
rθ = a2 sin2 φ cos θî + a2 sin2 φ sin θĵ + a2 sin φ cos φk̂, so
−
−
|→
rφ × →
rθ | = a2 sin φ.
Z
∴
2π
Z
Area of the sphere =
0
π
a2 sin φ dφdθ = 4πa2 .
0
Ex. Find the area of the curved surface of the cylinder x2 + y 2 = 16, z = 0, z = 3.
−
Sol. The given cylindrical surface is parametrized by →
r (θ, z) = 4 cos θî + 4 sin θĵ + z k̂, 0 ≤ θ ≤ 2π,
→
−
→
−
0 ≤ z ≤ 3. So x = 4 cos θ, y = 4 sin θ, | rθ × rz | = |4 cos θî + 4 sin θĵ| = 4, and hence
Z 2π Z 3
4 dθdz = 24π.
∴ Area =
0
0
Area of an explicitly defined smooth surface
Suppose a smooth surface S is explicitly defined, say, z = f (x, y). Then its vector equation reads
as
→
−
r (x, y) = xî + y ĵ + f (x, y)k̂,
which is clearly a parametrization of the surface in terms of two parameters x and y. Obviously,
the ranges of these parameters are given by the domain or the projected region R of the surface
in the XY-plane.
ZZ
ZZ
ZZ q
→
−
→
−
Area of S =
dσ =
| rx × ry |dxdy =
fx2 + fy2 + 1 dxdy.
S
R
R
21
−
Remark: In case the surface S lies in XY-plane, we have z = 0 and →
r (x, y) = xî + y ĵ. Therefore,
ZZ
ZZ
ZZ
ZZ
ZZ
→
−
→
−
Area of S =
dσ =
| rx × ry |dxdy =
|î × ĵ| dxdy =
|k̂| dxdy =
dxdy,
S
R
R
R
R
as expected.
Ex. Find the surface area of paraboloid z = x2 + y 2 , 0 ≤ z ≤ 4.
Sol. Here f (x, y) = x2 + y 2 , and the projected region R in the xy-plane of the given surface is the
circle x2 + y 2 = 4. So the required surface area equals to
ZZ p
Z 2π Z 2 √
ZZ q
π √
2
2
2
2
fx + fy + 1 dxdy =
4x + 4y + 1 dxdy =
4r2 + 1 rdrdθ = (17 17−1).
6
R
0
0
R
Ex. Find the surface area of the sphere x2 + y 2 + z 2 = a2p
using its parametrization in x and y.
Sol.p
From the given equation of the sphere, we have z = ± a2 − x2 − y 2 . So for upper hemisphere
z = a2 − x2 − y 2 , the parameterization is
p
→
−
r (x, y) = xî + y ĵ + a2 − x2 − y 2 k̂,
p
and for the lower hemisphere z = − a2 − x2 − y 2 , the parameterization reads
p
→
−
r (x, y) = xî + y ĵ − a2 − x2 − y 2 k̂.
In both cases, (x, y) belong to the circular region x2 + y 2 ≤ a2 . Next, for the upper hemisphere,
x
y
a
−
−
|→
rx × →
ry | = − p
î − p
î + k̂ = p
.
a2 − x 2 − y 2
a2 − x 2 − y 2
a2 − x 2 − y 2
So surface area of upper hemisphere is
ZZ
ZZ
→
−
→
−
| rx × ry | dxdy =
x2 +y 2 ≤1
x2 +y 2 ≤1
a
p
dxdy =
a2 − x 2 − y 2
Z
0
2π
Z
1
√
0
Likewise, for the lower hemisphere, we find
x
y
a
−
−
|→
rx × →
ry | = p
î + p
î + k̂ = p
.
a2 − x 2 − y 2
a2 − x 2 − y 2
a2 − x 2 − y 2
22
ar
drdθ = 2πa2 .
a2 − r 2
So surface area of lower hemisphere is
ZZ
ZZ
→
−
→
−
| rx × ry | dxdy =
x2 +y 2 ≤1
x2 +y 2 ≤1
a
p
dxdy =
a2 − x 2 − y 2
Z
0
2π
Z
0
1
√
ar
drdθ = 2πa2 .
− r2
a2
2
Finally, total surface area of the sphere is 4πa , as expected.
Area of an implicitly defined smooth surface
Consider an implicitly defined smooth surface S by
F (x, y, z) = c,
where c is a constant. Suppose R is the shadow or vertical projection on one of the suitable
coordinate planes, that is, xy, yz or zx-plane such that any line perpendicular to the plane region
R does not intersect the surface S in more than one point. This would be possible if the surface
does not fold back over itself over the plane of projection. Suppose the projection region R lies
in the xy-plane. Then an theorem of advanced calculus, called implicit function theorem, tells us
that the surface S is the graph of a differentiable function z = f (x, y), although this function may
not explicitly be known. Then the given surface in parametrized form is
→
−
r (x, y) = xî + y ĵ + f (x, y)k̂.
∂z
Fx
∂f
∂z
Fy
∂f
−
−
k̂ = î +
k̂ = î − k̂, →
ry = î +
k̂ = ĵ +
k̂ = ĵ − k̂.
∴ →
rx = î +
∂x
∂x
Fz
∂y
∂y
Fz
It follows that
1
∇F
→
−
−
rx × →
ry =
(Fx î + Fy ĵ + Fz k̂) =
.
Fz
∇F.k̂
So the surface area differential is
|∇F |
−
−
dσ = |→
rx × →
ry |dxdy =
dxdy.
|∇F.k̂|
Thus, the surface area of the implicit surface F (x, y, z) = c, when its projected region R lies in
xy-plane, reads as
ZZ
|∇F |
dxdy.
R |∇F.k̂|
Suppose, F (x, y, z) = c is expressible as z = f (x, y), and the function f (x, y) is known. Then,
F (x, y, z) − c = z − f (x, y) = 0, and therefore Fx = −fx , Fy = −fy and Fz = 1. So the above
surface area formula can also be written as
ZZ q
fx2 + fy2 + 1 dxdy.
R
Likewise, if the projected region R is in the yz-plane, we have
ZZ
ZZ q
|∇F |
Surface Area =
dydz =
fy2 + fz2 + 1 dydz.
R |∇F.î|
R
If the projected region R is in the zx-plane, then
ZZ
ZZ p
|∇F |
Surface Area =
dzdx =
fz2 + fx2 + 1 dzdx.
R |∇F.ĵ|
R
23
16.6 Surface Integrals
Integral over a smooth surface
Let G(x, y, z) be a function defined in an open region containing a smooth surface S. Divide the
surface S into n patches with areas ∆σi (i = 1, 2, ..., n) such that ∆σi → 0 in the limit n → ∞.
n
X
Choose a point (xi , yi , zi ) on the ith patch of S and construct the sum
G(xi , yi , zi )∆σi , which
i=1
defines the integral of G(x, y, z) over the surface S in the limit n → ∞, that is,
ZZ
G(x, y, z)dσ = lim
n→∞
S
n
X
G(xi , yi , zi )∆σi ,
i=1
provided the limit exists finitely.
−
If the surface S is parametrized by →
r (u, v) = x(u, v)î + y(u, v)ĵ + z(u, v)k̂, where u and v
belong to some region R in the uv-plane, then
ZZ
ZZ
−
−
G(x, y, z)dσ =
G(x(u, v), y(u, v), z(u, v))|→
ru × →
rv |dudv.
S
R
If the surface S is explicitlyp
defined by z = f (x, y) and R is its projected region in the xy-plane,
−
−
then dσ = |→
rx × →
ry |dxdy = fx2 + fy2 + 1dxdy, and therefore
ZZ
ZZ
q
G(x, y, f (x, y)) fx2 + fy2 + 1 dxdy.
G(x, y, z)dσ =
R
S
p
Ex. Integrate 4x2 + 4y 2 + 1 over the surface of the paraboloid z = x2 + y 2 , 0 ≤ z ≤ 2.
Sol. The projected
region R in the xy-plane of the given surface is the circle x2 + y 2 = 2. So the
p
integral of 4x2 + 4y 2 + 1 equals to
ZZ p
ZZ p
q
p
2
2
2
2
4x + 4y + 1 fx + fy + 1 dxdy =
4x2 + 4y 2 + 1 4x2 + 4y 2 + 1 dxdy
R
Z
R
2π
√
Z
=
0
2
(4r2 + 1) rdrdθ = 2π(4 + 1) = 10π.
0
Remark: In case a surface S consists of a number of smooth surfaces say S1 , S2 , ........, Sn , then
ZZ
ZZ
ZZ
ZZ
G(x, y, z)dσ =
G(x, y, z)dσ +
G(x, y, z)dσ + ..... +
G(x, y, z)dσ.
S
S1
S2
Sn
For example, a cuboid consists of six smooth surfaces which are the faces of the cuboid.
24
Orientable surface
A smooth surface S is said to be orientable, if it possesses continuous unique normal vector at each
point. For example, the surface of a sphere is orientable. A Mobius band is not orientable surface.
Take a strip of paper, give half twist and join the two ends of the strip to get a Mobius band. By
moving continuously along the surface, we can reach to the opposite of starting point. It means
there exist two different normal vectors at each point of the surface. So it is not orientable.
If the surface is orientable, conventionally we consider unit normal vector n̂ drawn outwards to
the surface, and say that the surface is oriented in the direction of n̂.
Surface integral of a vector field
→
−
Let F be a continuous vector field in an open region containing a smooth surface S oriented in
→
−
the direction of n̂. Then the integral of F over the surface S is defined by
ZZ
→
−
F .n̂dσ.
S
−
If the surface S is parametrized by →
r (u, v) = x(u, v)î + y(u, v)ĵ + z(u, v)k̂, where u and v belong
to some region R in the uv-plane, then we have
ZZ
ZZ
→
− →
→
−
−
F .(−
ru × →
rv )dudv
F .n̂dσ =
R
S
−
−
−
−
−
−
since n̂ = (→
ru × →
rv )/|→
ru × →
rv | and dσ = |→
ru × →
rv |dudv. Note that it is very useful formula for
finding surface integral of the vector field over the surfaces, which can be parametrized easily, for
example the spherical and cylindrical surfaces.
Flux
→
−
Let F be a continuous vector field in an open region containing a smooth surface S oriented in
→
−
the direction of n̂. Then the the flux of F across S is defined as
ZZ
ZZ
→
−
→
− →
−
Flux =
F .n̂dσ =
F .(−
ru × →
rv )dudv.
S
R
Ex. Find the flux of water flowing with velocity 5 cm/s out of the circular mouth of a tap with
radius 2 cm.
Sol. Suppose the direction of flow of water is along negative z-axis. Then the velocity field is
given by
→
−
F = 0î + 0ĵ − 5k̂.
25
The circular mouth cross-section is parametrized by
→
−
r (r, θ) = r cos θî + r sin θĵ + ck̂, 0 ≤ r ≤ 2, 0 ≤ θ ≤ 2π,
assuming the mouth cross-section lying in some plane z = c.
−
−
We find →
rθ × →
rr = −rk̂, the vector normal to the circular mouth cross-section of the tap
pointing along negative z-axis.
ZZ
Z 2 Z 2π
→
− →
−
→
−
5rdθdr = 20π cubic cm/s.
∴ Flux =
F .( rθ × rr )dθdr =
0
R
0
Note that water coming out of the circular mouth of the tap in 1 second is in the shape of a
water cylinder of height 5 cm, and radius of base 2 cm. So its volume 20π cubic cm is exactly the
amount of water coming out in one second, which is the flux.
One more point should be noted that conventionally the vector normal to the surface is taken
along the direction of flow in case of a fluid.
→
−
Ex. Find the outward flux of F = 4xî − 2y 2 ĵ + z 2 k̂ across the curved surface of the cylinder
x2 + y 2 = 16, z = 0, z = 3.
−
Sol. The given cylindrical surface is parametrized by →
r (θ, z) = 4 cos θî + 4 sin θĵ + z k̂, 0 ≤ θ ≤ 2π,
→
−
→
−
0 ≤ z ≤ 3. So x = 4 cos θ, y = 4 sin θ, rθ × rz = 4 cos θî + 4 sin θĵ, and hence
Z 3 Z 2π
ZZ
→
− →
−
→
−
(16 cos θî − 32 sin2 θĵ).(4 cos θî + 4 sin θĵ)dθdz = 192π.
F .( rθ × rz )dθdz =
Flux =
0
R
0
−
−
Important Remark: In the above example, the normal vector →
rθ × →
rz = 4 cos θî + 4 sin θĵ at
any point of the cylindrical surface in the first octant makes an acute angle with positive x-axis
or y-axis since its x and y components are positive when θ varies from 0 to π/2 in the first
octant. So this normal vector points outwards to the cylindrical surface. The normal vector
→
−
−
rz × →
rθ = −4 cos θî − 4 sin θĵ points inward to the cylindrical surface. Likewise, the normal vector
→
−
→
−
rφ × rθ = a2 sin2 φ cos θî + a2 sin2 φ sin θĵ + a2 sin φ cos φk̂ points outwards to the spherical surface
x2 + y 2 + z 2 = a2 . Notice that the z component a2 sin φ cos φ of the vector is positive at all points
of the sphere above the xy-plane where φ varies from 0 to π/2.
By convention, the orientation of a curved surface is given by the outward pointing normal
vector. For outward flux across a curved surface, we need to calculate outward unit normal vector.
−
−
In general, →
ru × →
rv gives a normal vector to the surface parametrized by u and v. We need to fix
its sign so that it points outwards to the given surface as discussed above.
16.7 Stokes Theorem
→
−
Let F = M î + N ĵ + P k̂ be a vector field having continuous first order partial derivatives in an
open region containing a piecewise smooth oriented surface S with piecewise smooth boundary C.
→
−
Then the circulation of F around C in the direction counterclockwise with respect to the unit
→
−
normal vector n̂ on S equals the integral of ∇ × F .n̂ over S, that is,
I
ZZ
→
− →
→
−
−
F .d r =
∇ × F .n̂dσ.
C
S
26
→
−
Notice that if F = M î + N ĵ and S is the region enclosed by C in the xy-plane, then the above
theorem reduces to the tangential form of Green’s theorem.
→
−
→
−
If F = ∇φ, then it is easy to verify that ∇ × F = 0. Therefore, curl of a conservative vector
→
−
field always vanishes. On the other hand, Stokes theorem can be used to show that if ∇ × F = 0
H →
− −
in a simply connected open region D, then C F .d→
r = 0 for any piecewise smooth closed path C
→
−
inside the region D, which in turn implies that F is conservative in D.
→
−
Ex. Verify Stoke’s theorem for the vector field F = y î − xĵ over the surface of S : x2 + y 2 + z 2 =
9, z ≥ 0 with bounding circle x2 + y 2 = 9 in the XY-plane.
Sol. First we calculate the counterclockwise circulation around C (as viewed from above) using
−
the parametrization →
r (t) = 3 cos t î + 3 sin t ĵ, 0 ≤ t ≤ 2π:
I
Z 2π
→
− →
F .d−
r =
(3 sin tî − 3 cos tĵ).(−3 sin t î + 3 cos t ĵ)dt = −18π.
0
C
→
−
To determine S ∇ × F .n̂dσ, we proceed as follows.
→
−
First, we find that ∇ × F = −2k̂.
Next, the surface S of the hemisphere x2 + y 2 + z 2 = 9, z ≥ 0 is parametrized by
→
−
r (φ, θ) = 3 sin φ cos θî + 3 sin φ sin θĵ + 3 cos φk̂, 0 ≤ φ ≤ π/2, 0 ≤ θ ≤ 2π.
−
−
Also, →
rφ × →
rθ = 9 sin2 φ cos θî + 9 sin2 φ sin θĵ + 9 sin φ cos φk̂.
It follows that
ZZ
Z 2π Z π/2
Z 2π Z π/2
→
−
→
− →
−
→
−
∇ × F .n̂dσ =
F .( rφ × rθ )dφdθ =
(−2)(9 sin φ cos φ)dφdθ = −18π.
RR
S
0
0
0
Hence the stoke’s theorem is verified.
27
0
RR
→
−
Alternative approach: To determine S ∇ × F .n̂dσ, we consider the projection of S over
p
XY-plane. Clearly the projected region is R : x2 + y 2 ≤ 9. Here z = f (x, y) = 1 − x2 − y 2 . So
we have
s
q
x2
3
y2
3
fx2 + fy2 + 1 =
= .
+
+1= p
2
2
2
2
9−x −y
9−x −y
z
9 − x2 − y 2
→
−
ĵ+z k̂
Next we find ∇ × F = −2k̂ and n̂ = √xî+y
= xî+y3ĵ+zk̂ . So we have
2
2
2
x +y +z
→
−
∇ × F .n̂dσ =
ZZ
S
ZZ
→
− q 2
∇ × F .n̂ fx + fy2 + 1dxdy =
ZZ
R
x2 +y 2 ≤9
−2z 3
dxdy = −2(9π) = −18π.
3 z
Attempt the following problems.
1. Verify Stoke’s theorem for the vector field
→
−
F = (2x − y)î − yz 2 ĵ − y 2 z k̂ over the upper half surface of x2 + y 2 + z 2 = 1, bounded by its
projection on the XY-plane.
Ans. π.
H →
− −
2. Using Stoke’s theorem, evaluate C F .d→
r , where
→
−
2
2
F = y î + x ĵ − (x + z)k̂ and C is boundary of the triangle with vertices (0, 0, 0), (1, 0, 0)
and (1, 1, 0).
Ans. 1/3.
16.8 Divergence Theorem
→
−
Let F = M î + N ĵ + P k̂ be a vector field having continuous first order partial derivatives in an
→
−
open region containing a piecewise smooth oriented closed surface S. Then the flux of F across S
→
−
in the direction of the outward drawn unit normal vector n̂ on S equals the integral of ∇. F over
the region D enclosed by S, that is,
ZZ
ZZZ
→
−
→
−
F .n̂dσ =
∇. F dV.
S
D
→
−
Notice that if F = M î + N ĵ and D is the region enclosed by S in the xy-plane, then the above
theorem reduces to the normal form of Green’s theorem.
→
−
Ex. Verify Divergence theorem for the expanding vector field F = xî + y ĵ + z k̂ over the surface
S of the sphere x2 + y 2 + z 2 = a2 .
Sol. The surface of the sphere is x2 + y 2 + z 2 = a2 is parametrized by
→
−
r (φ, θ) = a sin φ cos θî + a sin φ sin θĵ + a cos φk̂, 0 ≤ φ ≤ π, 0 ≤ θ ≤ 2π.
On the surface of this sphere,
→
−
F = xî + y ĵ + z k̂ = a sin φ cos θî + a sin φ sin θĵ + a cos φk̂.
−
−
Also, →
rφ × →
rθ = a2 sin2 φ cos θî + a2 sin2 φ sin θĵ + a2 sin φ cos φk̂. So we have
→
− →
−
F .(−
rφ × →
rθ ) = a3 sin φ
28
→
−
F .n̂dσ =
ZZ
Z
2π
0
S
Z
π
→
− →
−
F .(−
rφ × →
rθ )dφdθ =
Z
2π
Z
0
0
x2 +y 2 +z 2 ≤a2
0
a3 sin φdφdθ = 4πa3 .
0
→
−
Next we find ∇. F = ∇.(xî + y ĵ + z k̂) = 3. So
Z 2π Z
ZZZ
ZZZ
→
−
3dV = 3
∇. F dV =
D
π
0
π
Z
0
a
4
ρ2 sin φdρdφdθ = 3. πa3 = 4πa3 .
3
Hence the divergence theorem is verified.
Ex. Use divergence theorem to find volume of the sphere x2 + y 2 + z 2 = a2 .
→
−
Sol. Suppose S is the surface of the given sphere enclosing the region D. Let us choose, F = z k̂
→
−
so that ∇. F = 1. So we get
ZZZ
ZZZ
→
−
∇. F dV =
dV,
D
D
which is the volume of the given sphere. Now, by divergence theorem, we have
ZZZ
ZZ
→
−
→
−
∇. F dV =
F .n̂dσ.
D
S
ZZZ
ZZ
=⇒
dV =
D
Z
Z
2π
Z
2π
Z
(z k̂).n̂dσ =
S
0
π
−
−
(a cos φk̂).(→
rφ × →
rθ )dφdθ
0
π
4
(a cos φ)(a2 sin φ cos φ)dφdθ = πa3 .
3
0
0
Attempt the following problems.
→
−
1. Verify divergence theorem for F = (x2 − yz)î + (y 2 − zx)ĵ + (z 2 − xy)k̂ over the cuboidal
region given by 0 ≤ x ≤ 1, 0 ≤ y ≤ 2, 0 ≤ z ≤ 3.
Ans. 36.
→
−
2. Verify divergence theorem for F = 4xî − 2y 2 ĵ + z 2 k̂ over the region bounded by the cylinder
x2 + y 2 = 16, z = 0, z = 3.
Ans. 0 + 192π + 144π = 336π. Here 0, 192π and 144π are integrals over the lower circular
face, curved face and the upper circular face of the given cylinder, respectively.
=
29
→
−
3. Using divergence theorem, integrate F = (2x + 3z)î − (xz + y)ĵ + (y 2 + 2z)k̂ over the surface
of the sphere having center at origin and radius 3.
Ans. 108π.
30
SUMMARY
• Line integral of a continuous scalar field f (x, y, z) along a smooth curve
−
C:→
r (t) =< x(t), y(t), z(t) >, a ≤ t ≤ b is given by
Z
Z b
−
d→
r
dt,
f (x, y, z)ds =
f (x(t), y(t), z(t))
dt
C
a
where
−
d→
r
=
dt
s
dx
dt
2
+
dy
dt
2
+
dz
dt
2
.
H
In case, C is a closed curve, we denote the line integral by C f (x, y, z)ds, that is we
put a circle on the integral sign as an indication of the closed curve.
• Area under a space curve C 0 : z = f (x, y) over its shadow or projection or domain curve
−
C:→
r (t) =< x(t), y(t) >, a ≤ t ≤ b in xy-plane is given by
s 2
Z b
Z b
Z
2
→
−
dx
dy
dr
f (x(t), y(t))
dt =
+
dt.
f (x(t), y(t))
f (x, y)ds =
dt
dt
dt
a
a
C
→
−
• Line integral of the tangential component of a continuous vector field F (x, y, z) along
−
a smooth curve C : →
r (t) =< x(t), y(t), z(t) >, a ≤ t ≤ b is given by
Z b
Z
Z
−
→
−
→
− →
→
−
d→
r
−
F (x(t), y(t), z(t)).
F .d r =
F .T̂ ds =
dt.
dt
a
C
C
→
−
If F is a force field, and C is the trajectory traversed by a particle due to the action of
→
−
→
−
→
−
F , then the above integral gives the work done on the particle by the force F . If F is
velocity field of a fluid, then this integral gives the measure of flow rate along the curve
H →
−
→
−
C. In case, C is a closed curve, the integral C F .T̂ ds is called circulation of F .
→
−
−
In the two dimensional case, F (x, y) = M (x, y)î + N (x, y)ĵ, C : →
r (t) =< x(t), y(t) >,
a ≤ t ≤ b, we have
Z
Z b
→
−
dx
dy
F .T̂ ds =
M (x(t), y(t)) + N (x(t), y(t))
dt.
dt
dt
C
a
→
−
• Line integral of the normal component of a continuous vector field F (x, y, z) along a
−
smooth curve C : →
r (t) =< x(t), y(t), z(t) >, a ≤ t ≤ b is given by
Z
Z b
−
→
−
→
−
d→
r
F .n̂ds =
F (x(t), y(t), z(t)).n̂
dt.
dt
C
a
→
−
−
In the two dimensional case, F (x, y) = M (x, y)î + N (x, y)ĵ, C : →
r (t) =< x(t), y(t) >,
a ≤ t ≤ b (counter-clockwise traversed simple closed curve), we have
Z
Z b
→
−
dy
dx
F .n̂ds =
M (x(t), y(t)) − N (x(t), y(t))
dt.
dt
dt
C
a
31
•
•
•
•
→
−
If F is velocity field of a fluid, and n̂ is the outward drawn unit normal vector to a
H →
−
closed curve C in xy-plane, then the integral C F .n̂ds gives the outward flow rate or
→
−
flux of F across the curve C.
→
−
A vector field F is said to be conservative in an open region D if its line integral
R →
−
F .T̂ ds is path independent in D, that is, if we consider any two points say A and B
C
inside D, then whatever smooth curve C we choose inside D to join A and B, the value
R →
R →
−
−
of line integral C F .T̂ ds remains unaltered. In other words, C F .T̂ ds depends only
on the end points A and B, and not on the smooth curve C joining A and B inside D.
→
−
A continuous vector field F in an open connected region D is conservative if and only
→
−
if F = ∇φ for some differentiable scalar field φ. The scalar field φ is called potential of
→
−
F.
→
−
A continuous vector field F in an open connected region D is conservative if and only
H →
−
if C F .T̂ ds = 0 along every closed curve inside D.
→
−
A vector field F = M (x, y, z)î + N (x, y, z)ĵ + P (x, y, z)k̂ having continuous first order
partial derivatives in a simply connected open region D is conservative if and only if
∂M
∂N ∂N
∂P ∂P
∂M
=
,
=
,
=
.
∂y
∂x ∂z
∂y ∂x
∂z
→
−
• The divergence of a vector field F = M (x, y, z)î + N (x, y, z)ĵ + P (x, y, z)k̂ is given by
→
−
→
−
∂M
∂M
∂M
div F = ∇. F =
+
+
.
∂x
∂y
∂z
→
−
→
−
If F is the velocity field of a gas or a fluid, then a positive value of div F at a point
implies the expansion of gas at that point while a negative value refers to compression.
→
−
If div F = 0, then the fluid is incompressible.
→
−
• The curl of a vector field F = M (x, y, z)î + N (x, y, z)ĵ + P (x, y, z)k̂ is given by
→
−
→
−
curl F = ∇× F =
î
ĵ
k̂
∂
∂x
∂
∂y
∂
∂z
M N
P
=
∂N
∂P
−
∂y
∂z
∂M
∂P
î+
−
∂z
∂x
∂N
∂M
ĵ+
−
∂x
∂y
k̂.
→
−
→
−
If F is the velocity field of a gas or a fluid, then curl F is a measure of rotation of the
→
−
→
−
gas. In case, if curl F = 0, then the field F is irrotational.
→
−
• Green’s theorem (Normal Form): Let F (x, y) = M (x, y)î + N (x, y)ĵ be a vector
field having continuous first order partial derivatives in an open region in xy-plane
containing a piecewise smooth simple closed curve C enclosing a region R. Then the
→
−
→
−
outward flux of F across the curve C equals the double integral of the divergence of F
over the region R, that is,
I
I
ZZ →
−
∂M
∂N
F .n̂ds =
M dy − N dx =
+
dxdy.
∂x
∂y
C
C
R
32
→
−
• Green’s theorem (Tangential Form): Let F (x, y) = M (x, y)î + N (x, y)ĵ be a
vector field having continuous first order partial derivatives in an open region in xyplane containing a piecewise smooth simple closed curve C enclosing a region R. Then
→
−
the counterclockwise circulation of F around the curve C equals the double integral of
→
−
curl F .k̂ over R, that is,
I
ZZ I
→
−
∂N
∂M
F .T̂ ds =
M dx + N dy =
−
dxdy.
∂x
∂y
C
C
R
−
• The parametrized surface S : →
r (u, v) = x(u, v)î+y(u, v)ĵ +z(u, v)k̂ is said to be smooth
→
−
→
−
→
−
−
−
if ru and rv are continuous with →
ru × →
rv 6= 0 at every point of the parameters domain
−
−
−
under consideration. Further, →
ru × →
rv is a normal vector to the surface →
r (u, v).
→
−
• The integral of a scalar field f (x, y, z) over a smooth surface S : r (u, v) = x(u, v)î +
y(u, v)ĵ + z(u, v)k̂ is given by
ZZ
ZZ
−
−
f (x, y, z)dσ =
f (x(u, v), y(u, v), z(u, v))|→
ru × →
rv |dudv,
S
R
where R is the region of the parameters u and v.
• In case, the smooth surface S is given in explicit form, say, z = g(x, y), then use
ZZ
ZZ
q
f (x, y, z)dσ =
f (x, y, g(x, y)) gx2 + gy2 + 1dxdy,
S
R
where R is the projected region of S in the xy-plane (the domain of g(x, y)).
• In case, the smooth surface S is given in implicit form, say, F (x, y, z) = 0, and we
are able to project this surface on one of the coordinate planes, say, xy-plane to some
region R such that any line perpendicular to the xy-plane meets the surface S in at
most one point, then by implicit function theorem z should be explicit function of x, y,
say z = g(x, y), even if we are not able to find the algebraic expression of g(x, y) from
F (x, y, z) = 0. Also, we can use
ZZ
ZZ
|∇F |
dxdy.
f (x, y, g(x, y))
f (x, y, z)dσ =
|∇F.k̂|
S
R
p 2
|∇F |
Note that |∇F.
=
gx + gy2 + 1.
k̂|
• In case f (x, y, z) = 1,RRthe surface integral gives the area of the surface S. So surface
area of S is given by S dσ.
→
−
• The integral of a continuous vector field F over a smooth surface S oriented in the
direction of n̂ is given by
ZZ
→
−
F .n̂dσ.
S
→
−
It gives the flux of F across the surface S along n̂.
−
If the surface S is parametrized by →
r (u, v) = x(u, v)î + y(u, v)ĵ + z(u, v)k̂, where u and
v belong to some region R in the uv-plane, then we have
ZZ
ZZ
→
−
→
− →
−
F .n̂dσ =
F .(−
ru × →
rv )dudv.
S
R
33
• A surface S in the explicit form z = f (x, y) is parametrized by
→
−
r (x, y) = xî + y ĵ + f (x, y)k̂, where x and y belong to the projection region R of the
surface S in the xy-plane. Note that the
p region R is the domain of the function f (x, y).
−
−
Also, |→
rx × →
ry | = | − fx î − fy ĵ + k̂| = fx2 + fy2 + 1.
→
−
The integral of a vector field F (x, y, z) over the surface S : z = f (x, y) is given by
ZZ
ZZ
ZZ
→
−
→
− →
→
−
−
→
−
F .n̂dσ =
F .( rx × ry )dxdy =
F (x, y, f (x, y)).(−fx î − fy ĵ + k̂)dxdy.
S
R
R
• The surface of the sphere is x2 + y 2 + z 2 = a2 is parametrized by
→
−
r (φ, θ) = a sin φ cos θî + a sin φ sin θĵ + a cos φk̂, 0 ≤ φ ≤ π, 0 ≤ θ ≤ 2π.
−
−
Also, |→
rφ × →
rθ | = |a2 sin2 φ cos θî + a2 sin2 φ sin θĵ + a2 sin φ cos φk̂| = a2 sin φ.
→
−
The integral of a vector field F (x, y, z) over the surface S of the sphere x2 + y 2 + z 2 = a2
is given by
Z 2π Z π
ZZ
→
− →
→
−
−
F .(−
rφ × →
rθ )dφdθ
F .n̂dσ =
0
S
2π
Z
π
Z
=
0
0
→
−
F (a sin φ cos θ, a sin φ sin θ, a cos φ).(a2 sin2 φ cos θî+a2 sin2 φ sin θĵ+a2 sin φ cos φk̂)dφdθ.
0
• The curved surface S1 of the cylinder x2 + y 2 = a2 , 0 ≤ z ≤ h is parametrized by
→
−
r (θ, z) = a cos θî + a sin θĵ + z k̂, 0 ≤ θ ≤ 2π, 0 ≤ z ≤ h.
−
−
Also, |→
rθ × →
rz | = |a cos θî + a sin θĵ| = a.
→
−
The integral of a vector field F (x, y, z) over the curved surface S1 of the cylinder
x2 + y 2 = a2 , 0 ≤ z ≤ h is given by
ZZ
Z h Z 2π
Z h Z 2π
→
−
→
− →
→
−
−
→
−
F .n̂dσ =
F .( rθ × rz )dθdz =
F (a cos θ, a sin θ, z).(a cos θî+a sin θĵ)dθdz.
S1
0
0
0
0
• The upper circular surface S2 of the cylinder x2 + y 2 = a2 , 0 ≤ z ≤ h is the circle
x2 + y 2 = a2 that lies in the plane z = h. So this circular region is parametrized by
→
−
r (r, θ) = r cos θî + r sin θĵ + hk̂, 0 ≤ r ≤ a, 0 ≤ θ ≤ 2π.
−
−
Also, |→
rr × →
rθ | = |r cos θî + r sin θĵ| = r.
→
−
The integral of a vector field F (x, y, z) over the upper circular surface S2 of the cylinder
x2 + y 2 = a2 , 0 ≤ z ≤ h is given by
ZZ
Z 2π Z a
Z 2π Z a
→
−
→
− →
→
−
−
→
−
F .n̂dσ =
F .( rr × rθ )drdθ =
F (r cos θ, r sin θ, h).(r cos θî+r sin θĵ)drdθ.
S2
0
0
0
0
→
−
Likewise, the integral of F (x, y, z) over the lower circular surface S3 , that is, x2 +y 2 = a2 ,
z = 0, of the cylinder x2 + y 2 = a2 , 0 ≤ z ≤ h is given by
→
−
F .n̂dσ =
ZZ
S3
Z
0
2π
Z
0
a
→
− →
−
F .(−
rr × →
rθ )drdθ =
Z
2π
0
34
Z
0
a
→
−
F (r cos θ, r sin θ, 0).(r cos θî+r sin θĵ)drdθ.
→
−
• Stoke’s Theorem: Let F = M î + N ĵ + P k̂ be a vector field having continuous
first order partial derivatives in an open region containing a piecewise smooth oriented
→
−
surface S with piecewise smooth boundary C. Then the circulation of F around C in
the direction counterclockwise with respect to the unit normal vector n̂ on S equals the
→
−
integral of ∇ × F .n̂ over S, that is,
ZZ
I
→
− →
→
−
−
F .d r =
∇ × F .n̂dσ.
C
S
→
−
Notice that if F = M î + N ĵ and S is the region enclosed by C in the xy-plane, then
the above theorem reduces to the tangential form of Green’s theorem.
In the above theorem, it is important to note that the unit vector n̂ points normal to
the surface as per the chosen orientation of the curve C. A simple rule is the right hand
thumb rule to avoid any confusion. If the two fingers describe the orientation of C, then
the thumb points in the direction of n̂.
→
−
• Divergence Theorem: Let F = M î + N ĵ + P k̂ be a vector field having continuous
first order partial derivatives in an open region containing a piecewise smooth oriented
→
−
closed surface S. Then the flux of F across S in the direction of the outward drawn
→
−
unit normal vector n̂ on S equals the integral of ∇. F over the region D enclosed by S,
that is,
ZZZ
ZZ
→
−
→
−
∇. F dV.
F .n̂dσ =
D
S
→
−
Notice that if F = M î + N ĵ and D is the region enclosed by S in the xy-plane, then
the above theorem reduces to the normal form of Green’s theorem.
Note: To check whether the normal vector points outwards or inwards to the closed
surface, just calculate the normal vector at some point of the surface and see its direction
with respect to the surface. For example, the normal vector to the sphere x2 +y 2 +z 2 = 4
is
→
−
−
r ×→
r = 4 sin2 φ cos θî + 4 sin2 φ sin θĵ + 4 sin φ cos φk̂.
φ
θ
Now, let us choose the point (2, 0, 0) on the sphere. In spherical coordinates, it corresponds to r = 2, φ = π/2, and θ = 0. So we find
→
−
−
r ×→
r = 4î,
φ
θ
which clearly points along positive x-axis, and outwards to the spherical surface.
35