Savitribai Phule Pune University A Text Book for S.Y.B.Sc./S.Y.B.A. Mathematics (2013 Pattern) PAPER II(A)-MT 222(A) MULTIVARIABLE CALCULUS Panel of Authors S.S. Munot (Convenor) Dr.S.B.Gaikwad S.A.Ghule Editors Dr.P.M. Avhad Dr.S.A.Katre Conceptualized by Board of Studies (BOS) in Mathematics, Savitribai Phule Pune University, Pune.
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Savitribai Phule Pune University
A Text Book for S.Y.B.Sc./S.Y.B.A. Mathematics (2013 Pattern)
PAPER II(A)-MT 222(A)
MULTIVARIABLE CALCULUS
Panel of Authors
S.S. Munot (Convenor)Dr.S.B.Gaikwad
S.A.Ghule
Editors
Dr.P.M. Avhad Dr.S.A.Katre
Conceptualized by Board of Studies (BOS) in Mathematics, Savitribai Phule Pune University, Pune.
i
Page ii
Preface
This text book is an initiative by the BOS Mathematics, Savitribai Phule Pune University. The
syllabus of Savitribai Phule Pune University is always considered commendable, with a list of re-
puted books in the subject recommended for reference. Many times teachers face difficulty in doing
justice to every aspect covered collectively by all these books. So, while preparing new syllabus for
S.Y.B.Sc./S.Y.B.A. it was thought the University should prepare textbooks for Mathematics with
the following objectives:
1. Uniformising notations, definitions and to focus on important aspects of revised syllabus which
need to be stressed and understood.
2. Collecting all relevant topics, problems, questions prescribed in syllabus.
3. Providing a ready reference to teachers and students.
The book is written in accordance with the new prescribed syllabus of S.Y.B.Sc., S.Y.B.A. Mathe-
matics (2013 Pattern), for the Paper II(A)- MT 222(A): Multivariable Calculus II (Second Term).
The syllabus deals with important topics in Mathematics, Vector Calculus.
This book consists of a detail introduction at the beginning of each chapter, several illustrative
examples and problems for practice with hints and solutions given at the end of each section. A
proper understanding of all the illustrative examples would go a long way in making the subject fully
comprehensible.
In case of queries/suggestions, send an email to: sumati.munot @ gmail.com
We hope our endeavor will benefit both students and teachers.
-Authors
ii
Page iii
Acknowledgment
We sincerely thank the following University authorities for their constant motivation, guidance
and valuable help in the preparation of this book.
• Dr. W. N. Gade, Hon. Vice Chancellor, Savitribai Phule Pune University, Pune.
• Dr. V. B. Gaikwad, Director BCUD, Savitribai Phule Pune University, Pune.
• Dr. K. C. Mohite, Dean, Faculty of Science, Savitribai Phule Pune University, Pune.
• Dr. B. N. Waphare, Professor, Department of Mathematics, Savitribai Phule Pune Univer-
sity, Pune.
• Dr. M. M. Shikare, Professor, Department of Mathematics, Savitribai Phule Pune University,
Pune.
• Dr. V. S. Kharat, Professor, Department of Mathematics, Savitribai Phule Pune University,
Pune.
• Dr. V. V. Joshi, Professor, Department of Mathematics, Savitribai Phule Pune University,
Pune.
• Mr. Dattatraya Kute, Senate Member, Savitribai Phule Pune University; Manager, Savit-
ribai Phule Pune University Press.
• All the staff of Savitribai Phule Pune University press.
iii
Page iv
Syllabus: Paper II(A) MT 222(A):Multivariable Calculus
(1) Vector Valued Functions: . . . [14]
1.1 Vector valued function .
1.2 Limit and continuity of vector function .
1.3 Derivative of vector function and motion .
1.4 Differentiations rules.
1.5 Constant vector function and its necessary and sufficient condition .
1.6 Integration of vector function of a scalar variable .
1.6 Arc length and unit tangent vector T .Curvature and unit normal vector N.
(2) Line Integrals: . . . [16]
2.1 Definition and evaluation of line integral.
2.2 Properties of line integrals .
2.3 Vector fields,work,circulation and flux across smooth curves.
If r(t), a ≤ t ≤ b is a smooth curve in the domain of a continuous velocity field F,then the flow along
the curve is given by
Flow=
∫ b
a
F ·Tds =
∫ b
a
F · drdtdt
The integral
∫ b
a
F ·Tds is called a flow integral.If the curve is a closed loop,then the flow is called
the circulation around the curve.
Example 2.8. A velocity field is F=xi + zj + yk.Find the flow along the curve r(t) = cos ti +
sin tj + tk, 0 ≤ t ≤ π/2.
32
Line Integrals Page 33 Work,Circulation and Flux
Solution. By the definition of flow
Flow =
∫ b
a
F · Tds
=
∫ π/2
0
F · drdtdt
=
∫ π/2
0
(cos ti + tj + sin tk)((− sin t)i + (cos t)j + k)dt
=
∫ π/2
0
(− sin t cos t+ t cos t+ sin t)dt
=π − 1
2
Flux Across a Plane Curve
If C is a smooth curve in the domain of a continuous vector field F(x, y) = M(x, y)i + N(x, y)j in
the plane,and if n is the outward pointing unit normal vector on C,the flux of F across C is given
by∫C
F · ndsIt is easy to see that
F · n = M(x, y)dy
ds−N(x, y)
dx
ds
Flux of F across C =
∫C
M dy-N dx
Example 2.9. Find the flux of F=(x-y)i+xj across the circle x2 + y2 = 1
Solution. Suppose is the circle x2 + y2 = 1 .The parametric form is C : x = cos t, y = sin t, 0 ≤t ≤ 2π.Here M = x− y = cos t− sin t, N = x = cos t
Flux of F across C is given by
Flux =
∫C
Mdy −Ndx
=
∫ 2π
0
{(cos t− sin t) cos t+ cos t sin t}dt
=
∫ 2π
0
cos2 tdt
=
∫ 2π
0
1 + cos 2t
2dt
=π
The flux of F across the circle is π.
33
Path Independence,Potential Functions Page 34 Line Integrals
Problem Set II
(1) Find the work done by the force field F(x,y)= xyi + x2j on a particle that moves along the
curve C : x = y2 from (0, 0) to (1, 1).
(2) Find the circulation and flux of the vector fields F1(x,y)= xi + yj and F2(x,y)= −yi + xj
around and across each of the following curves.
(a) The circle r(t)= cos ti + sin tj, (0 ≤ t ≤ 2π).
(b) The ellipse r(t)= cos ti + 4 sin tj, (0 ≤ t ≤ 2π).
(3) Find the work done by F over the curve in the direction of increasing t
(a) F = xyi + yj− yzk and r(t)= ti + t2j + tk, 0 ≤ t ≤ 1
(b) F = zi + xj + yk and r(t)= sin ti + cos tj + tk, 0 ≤ t ≤ 2π
4. Path Independence,Potential Functions
We will study properties of vector fields that relate to the work they perform on particles moving
along various curves.For certain kinds of vector fields work done on a moving particle along a curve
depends only on the endpoints of the curve and not on the curve itself.Such vector fields are of
special importance in physics and engineering. Let F be a field defined on an open region D in
space.Suppose that for any two points A and B in D the work
∫ B
A
F · dr done in moving from A to
B is the same over all paths from A to B. Then the integral
∫F · dr is path independent in D
and the field F is conservative on D.
Gradient Fields
The gradient field of a differentiable function f(x, y, z) is the field of vectors∂f
∂xi +
∂f
∂yj +
∂f
∂zk.
We denote it as ∇f .
∇f =∂f
∂xi +
∂f
∂yj +
∂f
∂zk
Example 2.10. Find the gradient field of f(x, y, z) = xyz.
Solution. The gradient field of f is
∇f =∂f
∂xi +
∂f
∂yj +
∂f
∂zk = yzi + xzj + xyk
34
Line Integrals Page 35 Path Independence,Potential Functions
Example 2.11. Sketch the gradient field of φ(x, y) = x+ y.
Solution. The gradient field of φ is
∇φ =∂φ
∂xi +
∂φ
∂yj = i + j
Potential Function:If F is a field defined on D and F= ∇f for some scalar function f on D, then
f is called a potential function for F.
Theorem 2.1. Let F be a vector field whose components are continuous throughout an open
connected region D in space.If there exists a differentiable function f such that F = ∇f ,then for all
points A and B in D the value of
∫ B
A
F · dr is independent of the path from A to B in D.
Proof. Suppose that A and B are two points in D.
C : r(t) = g(t)i + h(t)j + k(t)k, a ≤ t ≤ b is a smooth curve in D joining A and B. Along the curve
C,f is a differentiable function of t.By chain rule,we get
df
dt=∂f
∂x
dx
dti +
∂f
∂y
dy
dtj +
∂f
∂z
dz
dtk
=(∂f
∂xi +
∂f
∂yj +
∂f
∂zk) · (dx
dti +
dx
dtj +
dx
dtk)
=∇f · drdt
=F · drdt, becauseF = ∇f
Hence,
∫C
F · dr =
∫ b
a
F · drdtdt
=
∫ b
a
df
dtdt
= [f(g(t), h(t), k(t))]ba
=f(B)− f(A)
35
Path Independence,Potential Functions Page 36 Line Integrals
Therefore the value integral depends only on the values of f at A and B and not on the path in
between.
Remark 2.7. The converse of this theorem is true;i·e· if for all points A and B in D the value
of
∫ B
A
F · dr is independent of the path from A to B in D,then there exists a differentiable function
f such that F = ∇f .
Remark 2.8. If the integral is independent of the path from A to B,then its value is
∫ B
A
F ·dr =
f(B)− f(A).
Theorem 2.2. The following statements are equivalent:
1.∫
F · dr = 0 around every closed loop in D.
2. The field F is conservative on D.
Proof. (1)⇒ (2)
To show that the field F is conservative,for this it is sufficient to show that for any two points A and
B in D the integral of F · dr has the same value over any two paths C1 and C2 from A to B.
We reverse the direction on C2 to make a path −C2 from B to A.Observe that together C1 and −C2
make a closed loop C(see the fig). We have,
∫C1
F·dr−∫C2
F·dr =
∫C1
F·dr+
∫−C2
F·dr =
∫C
F·dr = 0
Therefore
∫C1
F · dr =
∫C2
F · dr
Hence the integrals over C1 and C2 give the same value.
(2)⇒(1)
Now assume that the field F is conservative.We want to show that the integral of F · dris zero over
any closed loop C.Choose any two different points A and B on C.We use these points to break C in
two pieces,C1 from A in to B followed by C2 from B back to A.
36
Line Integrals Page 37 Path Independence,Potential Functions
∫C
F · dr =
∫C1
F · dr +
∫C2
F · dr =
∫ B
A
F · dr−∫ B
A
F · dr = 0
Remark 2.9. Let F= M(x, y, z)i+N(x, y, z)j+P (x, y, z)k be a field whose component functions
have continuous first order partial derivatives. Then F is conservative if and only if∂P
∂y=∂N
∂z,∂M
∂z=∂P
∂xand
∂N
∂x=∂M
∂yWhen we know that F is conservative,we usually find a potential function for F.It requires solving
the equation
∇f = F or∂f
∂xi +
∂f
∂yj +
∂f
∂zk = Mi +Nj + Pk
Therefore f can be obtained by integrating the equations∂f
∂x= M,
∂f
∂y= N,
∂f
∂z= P .
Example 2.12. Show that F=(ex cos y + yz)i + (xz − ex sin y)j + (xy + z)k is conservative and
find a potential function for it.
Solution. Comparing F=(ex cos y + yz)i + (xz − ex sin y)j + (xy + z)k with F= M i +N j + Pk
we have
M = ex cos y + yz,N = xz − ex sin y, P = xy + z∂P
∂y= x =
∂N
∂z,∂M
∂z= y =
∂P
∂x,∂N
∂x= −ex sin y + z =
∂M
∂yBy Remark 2.9 F is conservative and we can find a potential function f by integrating the equations∂f
∂x= ex cos y + yz,
∂f
∂y= xz − ex sin y,
∂f
∂z= xy + z
We integrating the first equation with respect to x,keeping y and z constant.
f(x, y, z) = ex cos y + xyz + g(y, z)
We write the constant of integration as a function of y and z.Now we can find∂f
∂yfrom this equation
and equate it with the expression∂f
∂y.
−ex sin y + xz +∂g
∂y= xz − ex sin y
37
Path Independence,Potential Functions Page 38 Line Integrals
From the last equation we get,∂g
∂y= 0.Therefore g is a function of z alone,say h(z).
f(x, y, z) = ex cos y + xyz + h(z)
We now calculate∂f
∂zfrom this equation and
equate it to the expression∂f
∂z.
xy +dh
dz= xy + z.It gives
dh
dz= z.So h(z) = z2/2 + C
Hence f(x, y, z) = ex cos y + xyz + z2/2 + k.
Remark 2.10. Let F= M(x, y)i+N(x, y)j be a field whose component functions have continuous
first order partial derivatives. Then F is conservative if and only if∂M
∂y=∂N
∂xWhen we know that F is conservative,we usually find a potential function for F.It requires solving
the equation
∇f = F or∂f
∂xi +
∂f
∂yj = Mi +Nj
Therefore f can be obtained by integrating the equations∂f
∂x= M,
∂f
∂y= N .
Example 2.13. Show that F=2xy3i + (1 + 3x2y2)j is conservative and find a potential function
for it.
Solution. Comparing F=2xy3i + (1 + 3x2y2)j with F= M i +N j we have
M = 2xy3, N = 1 + 3x2y2.We get,∂M
∂y= 6xy2 =
∂N
∂xBy Remark 2.10 F is conservative and we can find a potential function f by integrating the equations∂f
∂x= 2xy3,
∂f
∂y= 1 + 3x2y2
We integrating the first equation with respect to x,keeping y constant.f(x, y) = x2y3 + g(y). We
write the constant of integration as a function of y.Now we can find∂f
∂yfrom this equation and
equate it with the expression∂f
∂y.Therefore 3x2y2 +
dg
dy= 1 + 3x2y2
From the last equation we get,dg
dy= 1.Therefore g(y) = y + k.Hence f(x, y) = x2y3 + y + k. We will
define two important operations on a vector field: the divergence and the curl
Definition 2.2. If F(x, y, z) = f(x, y, z)i + g(x, y, z)j + h(x, y, z)k,then the divergence of F
is denoted by divF and is defined by
divF =∂f
∂x+∂g
∂y+∂h
∂z
38
Line Integrals Page 39 Path Independence,Potential Functions
The del operator allows us to express the divergence of a vector field
F(x, y, z) = f(x, y, z)i + g(x, y, z)j + h(x, y, z)k
divF = ∆ · F =∂f
∂x+∂g
∂y+∂h
∂z
Definition 2.3. If F(x, y, z) = f(x, y, z)i+g(x, y, z)j+h(x, y, z)k,then the curl of F is denoted
by curlF and is defined by
curlF = (∂h
∂y− ∂g
∂z)i + (
∂f
∂z− ∂h
∂x)j + (
∂g
∂x− ∂f
∂y)k
We express the curlF as
curlF=∆× F=
∣∣∣∣∣∣∣i j k∂
∂x
∂
∂y
∂
∂zf g h
∣∣∣∣∣∣∣Example 2.14. Find the divergence and the curl of the vector field
F(x, y, z) = x2yi + 2y3zj + 3zk
Solution. By the definition of divergence
divF =∂
∂x(x2y) +
∂
∂y(2y3z) +
∂
∂z(3z)
divF = 2xy + 6y2z + 3
By the definition of curl
curlF=∆× F=
∣∣∣∣∣∣∣i j k∂
∂x
∂
∂y
∂
∂zx2y 2y3z 3z
∣∣∣∣∣∣∣curlF = −2y3i− x2k
Problem Set III
(1) Find the gradient fields of the following functions
(a) f(x, y, z) = ez − ln(x2 + y2)
(b) f(x, y, z) = ln√x2 + y2 + z2
(c) f(x, y, z) = xy + yz + xz
(2) In each of the following determine whether F is a conservative vector field.If so,find a po-
tential function for it.
(a) F(x, y) = xi + yj.
(b) F(x, y) = x2yi + 5xy2j.
(c) F(x, y) = (cos y + y cosx)i + (sinx− x sin y)j.
(3) Show that each of the following integral is independent of the path,and use Remark 2.8 to
find its value.
(a)
∫ (2,3,−6)
(0,0,0)
2xdx+ 2ydy + 2zdz.
39
Green’s Theorem in Plane Page 40 Line Integrals
(b)
∫ (1,2,3)
(0,0,0)
2xydx+ (x2 − z2)dy − 2yzdz.
(c)
∫ (0,1,1)
(1,0,0)
sin y cosxdx+ cos y sinxdy + dz.
5. Green’s Theorem in Plane
Theorem 2.3. (Green’s Theorem) Let R be a simply connected plane region whose boundary
is a simple,closed,piecewise smooth curve C
oriented counter clockwise.If M and N are continuous functions
of x and y having continuous partial derivatives in R,then∮C
(Mdx+Ndy) =
∫∫R
(∂N
∂x− ∂M
∂y
)dxdy
Proof. Let C be a smooth simple closed curve in the xy−plane with the property that lines
parallel to the axes cut it in no more than two points.Let R be the region enclosed by C.suppose that
M ,N and their first partial derivatives are continuous at every point of some open region containing
C and R.
Let EFGH be the rectangle circumscribing the curve C with sides EF, FG,GH,HE along the lines
x = a, y = p, x = b, y = q respectively. Let the sides EF, FG,GH,HE touch the curve C at points
A,P,B,Q respectively.Now,suppose that the equations of the arcs APB and AQB are y = f1(x)
40
Line Integrals Page 41 Green’s Theorem in Plane
and y = f2(x) respectively.Then we have∫∫R
∂M
∂ydxdy =
∫ b
a
[∫ f2(x)
f1(x)
∂M
∂ydy
]dx
=
∫ b
a
[M(x, y)]f1(x)f2(x)dx
=
∫ b
a
[M(x, f2(x))−M(x, f1(x))]dx
=
∫ b
a
M(x, f2(x))dx−∫ b
a
M(x, f1(x))dx
=−∫ a
b
M(x, f2(x))dx−∫ b
a
M(x, f1(x))dx
=−[∫
arcAPB
Mdx+
∫arcBQA
Mdx
]=−
∮C
Mdx
Therefore
∮C
Mdx = −∫∫
R
∂M
∂ydxdy · · · (1) Next suppose that the equations of the
arcs PAQ and PBQ are x = g1(y) and x = g2(y) respectively.Then we have∫∫R
∂N
∂xdxdy =
∫ q
p
[∫ g2(y)
g1(y)
∂N
∂xdx
]dy
=
∫ q
p
[N(x, y)]g1(y)g2(y)dy
=
∫ q
p
[N(f2(y), y)−N(g1(y), y)]dy
=
∫ q
p
N(g2(y), y)dy −∫ q
p
N(g1(y)), ydy
=
∫ q
p
N(g2(y), y)dy +
∫ p
q
N(g1(y), y)dy
=
[∫arcPBQ
Ndy +
∫arcQAP
Ndy
]=
∮C
Ndy
Therefore
∮C
Ndy =
∫∫R
∂N
∂xdxdy · · · (2)
Adding (1) and (2),we get∮C
(Mdx+Ndy) =
∫∫R
(∂N
∂x− ∂M
∂y
)dxdy
Remark 2.11. Green’s Theorem for Multiply Connected Region
41
Green’s Theorem in Plane Page 42 Line Integrals
We know that a plane region is said to be simply connected if it has no holes and is said to be
multiply connected if it has one or more holes.
We can say that the boundary of the region R has positive orientation if the region R lies on the
left when any portion of the boundary is traversed in the direction of its orientation.This implies
that the outer boundary curve of the region is oriented anticlockwise and the boundary curves
that enclose holes have clockwise orientation.
If all portions of the boundary of a multiply connected region R are oriented in this way,then we
say that the boundary of R has positive orientation. We have proved Green’s theorem for simply
connected region R.We shall now extend it for multiply connected regions which have positive orien-
tations.For simplicity,we will consider a multiply connected region R with one hole bounded by the
simple closed curve C1 and simple closed curve C2 enclosed in C1.
As shown in the Fig.,let us divide R into two regions R′ and R′′ by introducing two cuts in R.The
cuts are shown as line segments.Suppose M and N are continuous functions of x and y having con-
tinuous partial derivatives in R.Note that R′ and R′′ are simply connected regions whose boundaries
are closed smooth curves oriented anticlockwise.Therefore by Green’s,we get∮Boundary of R′
(Mdx+Ndy) =
∫∫R′
(∂N
∂x− ∂M
∂y
)dxdy · · · (1)
42
Line Integrals Page 43 Green’s Theorem in Plane
∮Boundary of R′′
(Mdx+Ndy) =
∫∫R′′
(∂
∂x− ∂M
∂y
)dxdy · · · (2)
Adding (1) and (2),we get∮Boundary of R′
(Mdx+Ndy) +
∮Boundary of R′′
(Mdx+Ndy)
=
∫∫R′
(∂N
∂x− ∂M
∂y
)dxdy +
∫∫R′′
(∂N
∂x− ∂M
∂y
)dxdy
=
∫∫R
(∂N
∂x− ∂M
∂y
)dxdy
The two line integrals are taken in opposite directions along the cuts,
and hence cancel there,leaving only the contributions along C1 and C2.
Thus,we get∮C1
(Mdx+Ndy) +
∮C2
(Mdx+Ndy) =
∫∫R
(∂N
∂x− ∂M
∂y
)dxdy
This is an extension of Green’s theorem to a multiply connected region with one hole. More gener-
ally,if R is a multiply connected region with n holes,then along of (3) involves n + 1 integrals,one
taken anticlockwise around the outer boundary of R and the rest taken clockwise around the holes.
Example 2.15. Evaluate the following line integral using Green’s Theorem and check the answer
by evaluating it directly.
∮C
ydx+ xdy, where C is the unit circle oriented counter clockwise.
Solution. Here M = y, N = x. By Greens theorem∮C
(Mdx+Ndy) =
∫∫R
(∂N
∂x− ∂M
∂y
)dxdy , where R is the region bounded by the circle C.
∮C
(ydx+ xdy) =
∫∫R
(∂
∂x(x)− ∂
∂y(y)
)dxdy
=
∫∫R
0dxdy
= 0
Now we evaluate the line integral using the definition.
The parametric equations of C are x = cos t, y = sin t, (0 ≤ t ≤ 2π).∮C
(ydx+ xdy) =
∫ 2π
0
(− sin2 t+ cos2 t)d
= 0
Therefore Green’s theorem is verified.
43
Green’s Theorem in Plane Page 44 Line Integrals
Example 2.16. Use Green’s theorem to evaluate∮C
(ex + y2)dx+ (ey + x2)dy
, where C is the boundary of the region between y = x2 and y = x.
Solution. Here M = ex + y2, N = ey + x2 and C is the boundary of the region between y = x2
and y = x. By Greens theorem:∮C
(Mdx+Ndy) =
∫∫R
(∂N
∂x− ∂M
∂y
)dxdy , where R is the region bounded by the C.
∮C
((ex + y2)dx+ (ey + x2)dy) =
∫∫R
(∂
∂x(ey + x2)− ∂
∂y(ex + y2)
)dxdy
=
∫ 1
0
[∫ x
x2(2x− 2y)dy
]dx
=
∫ 1
0
[2xy − y2
]xx2dx
=
∫ 1
0
[2x3 − x4 − x2]dx
=− 1
30
Problem Set IV
(1) In each of the following,evaluate the line integral using Green’s Theorem and check the
answer by evaluating it directly.
(a)
∫C
y2dx+x2dy,where C is the square with vertices (0, 0), (1, 0), (1, 1) and (0, 1) oriented
counter clockwise.
(b)
∫C
3xydx+2xydy,where C is the rectangle bounded by x = −2, x = 4, y = 1 and y = 2.
44
Line Integrals Page 45 Illustrative Examples
(c)
∫C
(x2 − y)dx+ xdy,where C is the circle x2 + y2 = 4.
(d)
∫C
ln(1 + y)dx− xy
1 + ydy,where C is the triangle with vertices (0, 0), (2, 0) and (0, 4).
(2) Apply Green’s Theorem to evaluate the integral.
∫C
(xy + y2)dx+ x2dy,where C is the
triangle bounded by x = 0.x+ y = 1, y = 0.
(3) Use Green’s Theorem to find the counter clockwise circulation and outward flux of the field
F = xyi + y2j around and over the boundary of the region enclosed by the curves y = x2
and y = x in the first quadrant.
(4) Use Green’s Theorem to find the work done by F = 2xy3i + 4x2y2j in moving a particle
once counter clockwise around the boundary of the triangular region in the first quadrant
enclosed by the x-axis,the line x = 1 and the curve y = x3..
6. Illustrative Examples
(1) Integrate f(x, y, z) = x+√y − z2 over the path from (0, 0, 0) to (1, 1, 1) given by
C1 : r(t) = tk, 0 ≤ t ≤ 1.
C2 : r(t) = tj, 0 ≤ t ≤ 1.
C3 : r(t) = ti, 0 ≤ t ≤ 1.
Solution. Suppose C is the curve joining C1, C2 and C3, as shown in the figure. By Remark
2.2 (Additivity)∫C
f(x, y, z)ds =
∫C1
f(x, y, z)ds+
∫C2
f(x, y, z)ds+
∫C3
f(x, y, z)ds
Along C1 : x = 0, y = 0, z = t, 0 ≤ t ≤ 1
∴∫C
(x, y, z)ds =
∫ 1
0
−t2dt = −1
3Along C2 : x = 0, y = t, z = 1, 0 ≤ t ≤ 1
∴∫C2
f(x, y, z)ds =
∫ 1
0
(√t− 1)dt = −1
3Along C3 : x = t, y = 1, z = 1, 0 ≤ t ≤ 1
∴∫C3
f(x, y, z)ds =
∫ 1
0
t dt =1
2
∴∫C
f(x, y, z)ds = −1
3− 1
3+
1
2= −1
6.
(2) In each part evaluate the integral∫ydx+ zdy − xdz along the stated curve
(a) The line segment from (0, 0, 0) to (1, 1, 1).
(b) The twisted cubic x = t, y − t2, z = t3 from (0, 0, 0) to (1, 1, 1).
(c) The helix x = cosπt, y = sinπt, z = t from (1, 0, 0) to (−1, 0, 1).
Solution.
45
Illustrative Examples Page 46 Line Integrals
(a) Suppose C1 : The line segment from (0, 0, 0) to (1, 1, 1). The equations of the line
passing through (0, 0, 0) to (1, 1, 1) arex
1=y
1=z
1.
∴ C1 : x = t, t = t, z = t, 0 ≤ t ≤ 1.
∴ dx = dt, dy = dt, dz = dt.
∴∫C1
(ydx+ zdy − xdz) =
∫ 1
0
(t+ t− t)dt =1
2.
(b) C2 : x = t, y = t2, z = t3 ∴ dx = dt, dy = 2tdt, dz = 3t2dt.
∴∫C2
(ydx+ zdy − xdz) =
∫ 1
0
(t2 + t3(2t)− t(3t2))dt
=
∫ 1
0
(t2 − 3t2 + 2t4)dt
=− 1
60
(c) C3 : x = cosπt, y = sinπt, z = t ∴ dx = −π sin πtdt, dy = π cos πtdt, dz = dt.
∴∫C3
(ydx+ zdy − xdz) =
∫ 1
0
[(sinπt)(−π sin πt) + t(π cosπt)− (cosπt)]dt
=
∫ 1
0
(−π sin2 πt+ tπ cosπt− cos πt)dt
=
∫ 1
0
(1− cos 2πt)dt
2+ π
[t sin πt
π
]1
0
−∫ 1
0
sin πt−∫ 1
0
cos πtdt
=− π2 + 4
2π.
(3) Find the mass of a thin wire shaped in the form of the curve x = et cos t, y = et sin t, (0 ≤t ≤ 1), if the density function δ is proportional to the distance from the origin.
Solution. Let (x, y) be a point on the thin wire.
Therefore δ ∝√x2 + y2 ∴ δ = k(
√x2 + y2), where k is the constant of the proportionality.
The mass of thin wire is given by
46
Line Integrals Page 47 Illustrative Examples
M =
∫C
δ(x, y)ds
=
∫ 1
0
k1
√(et cos t)2 + (et sin t)2
√(dx
dt
)2
+
(dy
dt
)2
dt
=
∫ 1
0
k√
2et√
2etdt
=
∫ 1
0
k√
2et√
2etdt
=2k
∫ 1
0
e2tdt
=k(e− 1)
(4) Find the centre of mass of a thin wire lying along the curve
r(t) = ti + 2tj +
(2
3
)t3/2k, 0 ≤ t ≤ 2, if the density is δ = 3
√5 + t.
Solution. Suppose C : r(t) = ti + 2tj +
(2
3
)t3/2k
The coordinates of the centre of mass of a thin wire along the curve C whose density is
δ = 3√
5 + t are given by
x =Myz
M,y =
Mxz
M, z =
Mxy
M.
∴M =
∫C
δ(x, y, z)ds
=
∫ 2
0
3√
5 + t√
(1)t2 + (2)t2 + (t2)2dt
=
∫ 2
0
3√
5 + t√
5 + tdt
=3
∫ 2
0
(5 + t)dt
=3[5t+ t2/2]
=36
47
Illustrative Examples Page 48 Line Integrals
Myz =
∫C
xδ(x, y, z)ds
=
∫ 2
0
3t√
5 + t√
1 + 4 + tdt
=
∫ 2
0
3t(5 + t)dt
=
[15
2t2 + t3
]2
0
Myz =38
Mxy =
∫C
yδ(x, y, z)ds
=
∫ 2
0
6t(5 + t)dt
=[15t2 + 2t3
]Mxy =76
Mxy =
∫C
zδ(x, y, z)ds
=
∫ 2
0
2t3/2(5 + t)dt
=
[4t
5/2 +4
7t7/2
]2
0
Mxy =154
7
√2
x =Myz
M,y =
Mxz
M, z =
Mxy
M
x =19
18,y =
19
9, z =
4√
2
7.
(5) Find the work done by the force field F(x, y, z) = (x + y)i + xyj − z2k on a particle that
moves along the line segment from (1, 3, 1) to (2,−1, 4).
Solution. Suppose A(1, 3, 1) and B(2,−1, 4).
d.r.s of AB are 1,−4, 3. Equations of AB arex− 1
1=y − 3
1=z − 1
1Let C : the line segment AB
∴ C : (1 + t)c + (3 + 4t)j + (1 + 3t)k, 0 ≤ t ≤ 1
48
Line Integrals Page 49 Illustrative Examples
The work done by F along C is given by
W =
∫C
F(x, y, z)dr
=
∫ 1
0
[(4− 3t)c + (3− t− 4t2)j− (1 + 3t2)k
]· [i− 4j + 3k] dt
=
∫ 1
0
[1(4− 3t)− 4(3− t− 4t2)− 3(1 + 3t2)
]dt
=
∫ 1
0
(−9t2 − 17t− 11)dt
=
[−3t3 − 17
2− 11t
]1
0
W =45
2
(6) Find the work done by the gradient of f(x, y) = (x+ y2) counter clockwise around the circle
x2 + y2 = 4 from (2, 0) to itself.
Solution. Suppose
F(x, y) = grad f(x, y) =∂f
∂xi +
∂f
∂yj
=2(x+ y)i + 2(x+ y)j
C : Circle x2 + y2 = 4 from (2, 0) to itself.
∴ C : r(t) = 2 cos ti + 2 sin tj, 0 ≤ t ≤ 2π
The work done by F along C is given by
W =
∫C
F(x, y, z)dr
=
∫ 2π
0
[4(cos t+ sin t)i + 4(cos t+ sin t)j] · (−2 sin i + 2 cos j)dt
=8
∫ 2π
0
(cos2 t− sin2 t)dt
=8
∫ 2π
0
cos 2tdt
=8× 0
W =0
(7) Find the flux of the vector field F = (x + y)i − (x2 + y2)j outward across the triangle with
the vertices (1, 0), (0, 1), (−1, 0).
Solution. Suppose P (1, 0), Q(0, 1) and R(−1, 0)
C : 4PQR
49
Illustrative Examples Page 50 Line Integrals
(Flux of F = M i +N j across C) =
∮C
Mdy −Ndx
∴ Flux of F =
∮C
(x+ y)dy + (x2 + y2)dx
=
∫PQ
(x+ y)dy + (x2 + y2)dx+
∫QR
(x+ y)dy + (x2 + y2)dx+∫RP
(x+ y)dy + (x2 + y2)dx.
Along PQ: The equation of PQ is x = 1− y, dx = −dy and 0 ≤ y ≤ 1.
∴∫PQ
(x+ y)dy + (x2 + y2)dx =
∫ 1
0
(2− 2y + 2y2)dy =5
3.
Similarly
∫QR
(x+ y)dy + (x2 + y2)dx = −2
3and
∫RP
= (x+ y)dy + (x2 + y2)dx =2
3.
Hence Flux of F =5
3− 2
3+
2
3=
5
3.
∂N
∂x= z =
∂M
∂y.
F = yzi + xzj + xyk is conservative.
(8) Show that the vector field F = y sin zi + x sin zj + xy cos zk is conservative.
Solution. Here M = y sin z,N = x sin z, P = xy cos z.∂P
∂y= x cos z =
∂N
∂z,∂M
∂z= y cos z =
∂P
∂x,∂N
∂x= sin z =
∂M
∂y.
Hence F is conservative.
(9) Show that the vector field F = yi + (x+ z)j− yk is not conservative.
Solution. Here M = y,N = x+ z, P = −y.∂P
∂y= −1 ,
∂N
∂z= 1,
∂M
∂z= 0
∂P
∂y6= ∂N
∂z.F is not conservative.
(10) Find a potential function f for the field F.
Solution. Comparing F=(y + z)i + (x+ z)j + (x+ y)k with F= M i +N j + Pk we have
M = y + z,N = x+ z, P = x+ y∂P
∂y= 1 =
∂N
∂z,∂M
∂z= 1 =
∂P
∂x,∂N
∂x= 1 =
∂M
∂yBy Remark 2.9 F is conservative and we can find a potential function f by integrating the
equations∂f
∂x= y + z,
∂f
∂y= x+ z,
∂f
∂z= x+ y
We integrating the first equation with respect to x,keeping y and z constant.
f(x, y, z) = x(y + z) + g(y, z)
We write the constant of integration as a function of y and z.Now we can find∂f
∂yfrom this
equation and equate it with the expression∂f
∂y.
50
Line Integrals Page 51 Illustrative Examples
x+∂g
∂y= x+ z
From the last equation we get,∂g
∂y= z.Therefore g = yz + h(z) .
f(x, y, z) = x(y + z) + yz + h(z)
We now calculate∂f
∂zfrom this equation and
equate it to the expression∂f
∂z.
x+ y +dh
dz= x+ y.It gives
dh
dz= 0.So h(z) = k
Hence f(x, y, z) = xy + xz + yz + k.
(11) Find the work done by
F = eyzi + (xzeyz + z cos y)j + (xyeyz + sin y)k over the following paths from (1, 0, 1) to
(1, π/2, 0)
a. The line segment:x = 1, y = πt/2, z = 1− t ,0 ≤ t ≤ 1.
b. The line segment from (1, 0, 1) to the origin followed by the line segment from the origin
to (1, π/2, 0).
c. The line segment from (1, 0, 1) to (1, 0, 0) followed by the x−axis from (1, 0, 0) to the
origin,followed by the parabola y =πx2
2, z = 2 from there to (1, π/2, 0).
Solution. Comparing F = eyzi+(xzeyz+z cos y)j+(xyeyz+sin y)k with F= M i+N j+Pk
we have
M = eyz, N = xzeyz + z cos y, P = xyeyz + sin y∂P
∂y= xeyz(yz + 1) + cos y =
∂N
∂z,∂M
∂z= yeyz =
∂P
∂x,∂N
∂x= zeyz =
∂M
∂yBy Remark 2.9 F is conservative.Hence the work done by
F =
∫C
F · dr is independent of the path.
Also
∫C
F · dr = f(1, π/2, 0)−f(1, 0, 1),where F = ∇f . Now we can find a potential function
f by integrating the equations∂f
∂x= eyz,
∂f
∂y= xzeyz + z cos y,
∂f
∂z= xyeyz + sin y
We integrating the first equation with respect to x,keeping y and z constant.
f(x, y, z) = xeyz + g(y, z)
We write the constant of integration as a function of y and z.Now we can find∂f
∂yfrom this
equation and equate it with the expression∂f
∂y.
xzeyz +∂g
∂y= xzeyz + z cos y
From the last equation we get,∂g
∂y= z cos y.Therefore g = z sin y + h(z) .
51
Illustrative Examples Page 52 Line Integrals
f(x, y, z) = xeyz + z sin y + h(z)
We now calculate∂f
∂zfrom this equation and
equate it to the expression∂f
∂z.
xyeyz + sin y +dh
dz= xyeyz + sin y.It gives
dh
dz= 0.So h(z) = k
Hence f(x, y, z) = xeyz + z sin y + k.f(1, π/2, 0) = 1 + k, f(1, 0, 1) = 1 + k∫C
F · dr = f(1, π/2, 0)− f(1, 0, 1) = 0
The work done by F along each path is 0.
(12) Show that the integral∫C
(yzdx+ xzdy + yx2)
is not independent of the path.
Solution. It is sufficient to show that F = yzi + xzj + yx2k is not conservative. Here
M = yz,N = xz, P = yx2.∂P
∂y= x2 ,
∂N
∂z= x,
∂M
∂z= y
∂P
∂y6= ∂N
∂z.F is not conservative.Hence the given integral is not independent of the path.
(13) Use Green’s theorem to evaluate the integral
∮C
y tan2 xdx+ tanxdy
,
where C is the circle x2 + (y + 1)2 = 1
Solution. Here M = y tan2 x, N = tanx and C is the boundary of the region between
y = x2 and y = x. By Greens theorem:∮C
(Mdx+Ndy) =
∫∫R
(∂N
∂x− ∂M
∂y
)dxdy , where R is the region bounded by the C.∮
C
(y tan2 xdx+ tanxdy) =
∫∫R
(∂
∂xtanx− ∂
∂yy tan2 x
)dxdy
=
∫∫R
(sec2 x− tan2 x)dxdy
=
∫∫R
1dxdy
=area of the circle
=π
(14) Use Green’s theorem to evaluate the integral∮C
cosx sin ydx+ sinx cos ydy,
where C is the triangle with vertices (0, 0), (3, 3), (0, 3)
52
Line Integrals Page 53 Illustrative Examples
Solution. Here M = cosx sin y, N = sinx cos y. By Greens theorem:∮C
(Mdx+Ndy) =
∫∫R
(∂N
∂x− ∂M
∂y
)dxdy , where R is the region bounded by the C.∮
C
(cosx sin ydx+ sinx cos ydy) =
∫∫R
(∂
∂xsinx cos y − ∂
∂ycosx sin y
)dxdy
=
∫∫R
cosx cos y − cosx cos ydxdy
=0
(15) Use Green’s theorem to find the work done by the force F = xyi + (x2
2+ xy)j, the particle
starts at (5, 0),transverse the upper semi circle x2 + y2 = 25 and returns to its starting point
along the x−axis.
Solution. The work done by the force F is∫C
F · dr =
∮C
Mdx+Ndy,
where M = xy, N = (x2
2+ xy). By Greens theorem:∮
C
(Mdx+Ndy) =
∫∫R
(∂N
∂x− ∂M
∂y
)dxdy , where R is the region bounded by the C.
∮C
(xydx+ (x2
2+ xy)dy) =
∫∫R
(∂
∂x(x2
2+ xy)− ∂
∂yxy
)dxdy
=
∫ 5
−5
(∫ √25−x2
0
ydy
)dx
=
∫ 5
−5
[y2/2
]√25−x2
0dx
=
∫ 5
0
(25− x2)dx
=250
3
(16) Using Green’s theorem,show that the area bounded by a simple closed curve Cis given by1
2
∮C
xdy − ydx.
Hence find the area of the ellipse x2/9 + y2/4 = 1
53
Illustrative Examples Page 54 Line Integrals
Solution. Suppose M = −y,N = x.If R is the region bounded by a simple closed curve
C,then by Green’s theorem we have∮C
(Mdx+Ndy) =
∫∫R
(∂N
∂x− ∂M
∂y
)dxdy
∮C
xdy − ydx =
∫∫R
(∂
∂xx− ∂
∂y(−y)
)dxdy
=
∫∫R
(1 + 1)dxdy
=
∫∫R
2dxdy
=2(area of the regionR)
Hence the area of the region R=1
2
∮C
xdy − ydx
Now we find the area of the ellipse C : x2/9+y2/4 = 1.C : x = 3 cos t, y = 2 sin t; 0 ≤ t ≤ 2π.
area of the ellipse =1
2
∮C
xdy − ydx
=1
2
∫ 2π
0
6 cos2 t+ 6 sin2 tdt
=1
2
∫ 2π
0
6dt
=3π
(17) Evaluate the integral∫C
(e−x + 3y)i + xj · dr
C is the boundary of the region R between the circles x2 + y2 = 16 and x2− 2x+ y2 = 3 and
C is oriented so that the region is on the left when the boundary transverse in the direction
of its orientation.
Solution. Here M = e−x + 3y and N = x.C is the simple closed curve.By Green’s theorem
54
Line Integrals Page 55 Illustrative Examples
∫C
(e−x + 3y)i + xj · dr =
∫∫R
(∂
∂xx− ∂
∂y(e−x + 3y)
)dxdy
=
∫∫R
(1− 3)dxdy
=
∫∫R
(−2)dxdy
=− 2[area of the circle (x2 + y2 = 16)− area of the circle (x2 − 2x+ y2 = 3)]
=− 2(16π − 4π)
=− 24π
55
Exercise Page 56 Line Integrals
7. Exercise
(1) In each part,evaluate the integral∫C
(3x+2y)dx+(2x-y)dy along the stated curves
(a) The line segment from (0, 0) to (1, 1).
(b) The parabolic arc y = x2 from (0, 0) to (1, 1).
(c) The curve y = sin(πx/2) from (0, 0) to (1, 1).
(d) The curve y = x3 from (0, 0) to (1, 1).
(2) Evaluate the line integral∫C
3x2yz ds where C : x = t, y = t2, z =2
3t3.
(3) Evaluate the line integral∫C
(x+y)ds where C is the line segment from (0, 1, 0) to (1, 0, 0).
(4) Integrate f(x, y, z) = (x+ y + z)/(x2 + y2 + z2) over the path
r(t)= ti + tj + tk, 0 ≤ t ≤ 1.
(5) Integrate f(x, y) = x+ y over the curve C : x2 + y2 = 4 in the first quadrant form (2, 0) to
(0, 2).
(6) Find the mass of a thin wire shaped in the form of the helix x = 3 cos t, y = 3 sin t, z = 4t
(0 ≤ t ≤ π/2) if the density function is δ = kx/(1 + y2) k > 0.
(7) Find the mass of a wire that lies along the curve r(t)= (t2 − 1)j + 2tk,
0 ≤ t ≤ 1,if the density is δ = (3/2)t.
(8) Find the centre of mass of a thin wire lying along the curve r(t)= ti+(2√
2/3)t3/2j+t2/2k, 0 ≤t ≤ 2 if the density is δ = 1/(t+ 1).
(9) Find the work done by the force field F(x,y,z)= xyi + yzj + xzk on a particle that moves
along the curve C:r(t)= ti + t2j + t3k, (0 ≤ t ≤ 1).
(10) Suppose that a particle moves through the force field F(x,y)= xyi + (x− y)j from the point
(0, 0) to the (1, 0) along the curve x = t, y = λt(1 − t).For what values of λ will the work
done by the force field be 1?
(11) Find the circulation and flux of the vector fields F(x,y)= −yi + xj around and the closed
semicircular path that consists of the semicircular arch r1(t)= a cos ti + a sin tj, (0 ≤ t ≤ π)
followed by the line segment r2(t)= ti,−a ≤ t ≤ a.
(12) Find the flow of the velocity field F=(x+ y)i− (x2 + y2)j along each of the following paths
from (1, 0) to (−1, 0).
(a) The upper half of the circle x2 + y2 = 1.
(b) The line segment from (1, 0) to (−1, 0).
56
Line Integrals Page 57 Exercise
(c) The line segment from (1, 0) to (0,−1) followed by the line segment from (0,−1) to
(−1, 0).
(13) The field F = xyi + yj − yzk is the velocity field of a flow in space. Find the flow from
(0, 0, 0) to (1, 1, 1) along the curve of intersection of the cylinder y = x2 and the plane z = x.
(14) Show that the line integral∫C
y2dx + 2xydy is independent of the path.Also evaluate the line integral along the line
segment from (−1, 2) to (1, 3).
(15) Show that each of the following integral is independent of the path,and use Remark 2.8 to
find its value.
(a)
∫ (4,0)
(1,2)
3ydx+ 3xdy.
(b)
∫ (3,2)
(0,0)
2xeydx+ x2eydy.
(c)
∫ (−1,0)
(2,−2)
2xy3dx+ 3y2x2dy.
(16) In each of the following find a potential function f for the vector field F
(a) F = 2xi + 3yj + 4zk.
(b) F = ey+2z(i + xj + 2xk).
(c) F = (lnx+ sec2(x+ y))i +
(sec2(x+ y) +
y
y2 + z2
)j +
z
y2 + z2k.
(17) Find a potential function for F =2x
yi +
(1− x2
y2
)j[x2 − 1
y]
(18) Find the work done by F = (x2 +y)i+(y2 +x)j+zezk over the following paths from (1, 0, 0)
to (1, 0, 1).
(a) The line segment x = 1, y = 0, 0 ≤ z ≤ 1.
(b) The helix r(t) = cos ti + sin tj +t
2πk, 0 ≤ t ≤ 2π
(c) The x-axis from (1, 0, 0) to (0, 0, 0) followed by the parabola z = x2, y = 0 from (0, 0, 0)
to (1, 0, 1).
(19) In each of the following,evaluate the line integral using Green’s Theorem and check the
answer by evaluating it directly.
(a)
∫C
x cos ydx − y sinxdy,where C is the square with vertices (0, 0), (π/2, 0), (π/2, π/2)
and (0, π/2).
(b)
∫C
tan−1 ydx− y2x
1 + y2dy,where C is the square with vertices (0, 0), (1, 0), (1, 1) and (0, 1).
(c)
∫C
x2ydx+ (y + xy2)dy,where C is the boundary of the region enclosed by y = x2 and
x = y2.
57
Exercise Page 58 Line Integrals
(20) Use Green’s Theorem to find the work done by the force field F(x, y) = xyi + (1
2x2 + xy)j;
the particle starts at (5, 0),transverse the upper semicircle x2 + y2 = 25 and returns to its
starting point along the x-axis.
(21) Find a simple closed curve C with counter clockwise orientation that maximizes the value
of∫C
1
3y3dx+ (x− 1
3x3)dy and explain your reasoning.
(22) (a) Let C be the line segment from a point (a, b) to a point (c, d).Show that∫C
−ydx+ xdy= ad− bc.
(b) Use the result in part (a) to show that the area A of a triangle with successive vertices
(x1, y1), (x2, y2) and (x3, y3) going counter clockwise is
In the last Chapter, we have seen how to integrate a function over a flat region in a plane. Here
question arises that how to integrate, if the function is defined over a curved surface? That is given
by double integral over a region in a co-ordinate plane.
1. Surface Area and Surface integrals
Surface Area
The surface is defined by the equation f(x, y, z) = C If the surface is smooth, then we can define
and calculate its area as a double integral over region R.
Suppose S is a surface of area, we take the partition of region R into small rectangles ∆Ak. The upper
partition of ∆Ak is ∆σk which may be approximate tangent plane to the partition ∆Pk. We suppose
that ∆Pk is a portion of the plane that is tangent to the surface at that point Tk(xk, yk, zk), which is
upper portion of corner Ck of ∆Ak. If the tangent plane is parallel to R, then ∆Pk will be congruent
to ∆Ak. Let ∆f(xk, yk, zk)k be the gradient at Tk(xk, yk, zk) and a unit vector P that is normal to R.
The angle between ∆f and P is γk. Let uk and vk be the vectors along the edges of portion ∆Pk in
the tangent plane. Thus uk×vk and ∆f both are normal to the tangent plane. Therefore|uk × vk · P |gives the area of the projection of the parallelogram determine by uk and vk onto any plane whose
normal is P . ⇒ |uk × vk.P | = ∆Ak . . . (1) Now, |uk × vk| itself is the area ∆Pk [ by properties of
cross product] so the equation (1) becomes |uk × vk| |P | |cos(angle betweenuk × vkandP )| = ∆Ak (by
definition of dot product) . . . (2) or ∆Pk |cos γk| = ∆Ak.
or ∆Pk =∆Ak|cos γk|
provides cos γk 6= 0. We will have cos γk 6= 0 as long as ∇f is not parallel to the
ground plane and ∇f · P 6= 0 Since the portion ∆Pk approximate the surface portion ∆σk that fit
together to make S, the sum∑
∆Pk =∑ ∆Ak|cos γk|
. . . (3) The sums on the right-hand side of equation
(3) are approximating sums for the double integral
∫∫R
1
cos γdA . . . (4) which gives the area of S to
be the value of this integral whenever it exists that is, for any surface f(x, y, z) = c
We have |∇f · P | = |∇f | |P | |cos γ|,
61
Surface Area and Surface integrals Page 62 Surface and Volume Integrals
So1
cos γ=|∇f ||∇f · P |
Which is equivalent to equation (4)
Therefore the area of the surface f(x, y, z) = c over a closed and bounded plane region R is
surface area =
∫∫R
∇f∇f · P
dA . . . (5),
where P is a unit vector normal to R and ∇f · P 6= 0.
Thus, the area is the double integral over R of the magnitude of ∇f divided by the magnitude of
the scalar component of ∇f normal to R.
We reduced equation (5) under the assumption that ∇f ·P 6= 0 throughout R and ∇f is continuous.
Whenever the integral exists, however, we define its values to be the area of the portion of the surface
f(x, y, z) = c that lies over R.
Example 3.1. Find the area of the surface cut from the bottom of the paraboloid x2 + y2− z = 0
by the plane z = 4.
Solution. The surface S is part of the level surface f(x, y, z) = x2 + y2 − z = 0, and R is the
disk x2 + y2 ≤ 4 in the xy-plane. We take a unit vector normal to the plane of R, is P = k
We have f(x, y, z) = x2 + y2 − z∇f = 2xi+ 2yj − k∴ |∇f | =
√(2x)2 + (2y)2 + (−1)2
62
Surface and Volume Integrals Page 63 Surface Area and Surface integrals
=√
4x2 + 4y2 + 1
|∇f · P | = |∇f · k| = |(−1)| = 1
In the region R, dA = dxdy
∴ Surface area =
∫∫R
∇f∇f · p
dA
=
∫∫x2+y2≤4
√4x2 + 4y2 + 1dxdy
=
∫ 2π
0
∫ 2
0
√4r2 + 1rdrdθ (since x = r cos θ, y = r sin θ and using formula
∫f(x)[f(x)]ndx =
[f(x)]n + 1
n+ 1)
=
∫ 2π
0
[1
12(4r2 + 1)3/2]20dθ
=
∫ 2π
0
1
12(173/2 − 1)dθ
=π
6(17√
17− 1).
Example 3.2. Find the area of the cap cut from the hemisphere x2 + y2 + z2 = 2, z ≥ 0, by the
cylinder x2 + y2 = 1.
Solution. The Cap S is part of the level surface f(x, y, z) = x2 + y2 + z2 = 2. It projects
one-to-one, onto the disk R: x2 + y2 ≤ 1 in the xy−plane. The vector p=k is normal to the plane of
R. At any point on the surface, f(x, y, z) = x2 + y2 + z2
∇f = 2xi+ 2yj + 2zk
∴ |∇f | = 2√
(x)2 + (y)+(z)2 = 2√
2
63
Surface Area and Surface integrals Page 64 Surface and Volume Integrals
|∇f · p| = |∇f · k| = |(2z)| = 2z
∴ Surface area =
∫∫R
|∇f ||∇f · p|
dA
=
∫∫R
2√
2
2zdA =
√2
∫∫R
dA
zNow x2 + y2 + z2 = 2
∴ z =√
2− x2 − y2
∴ surface area =√
2
∫∫R
dA
z=√
2
∫∫x2+y2≤1
dA√2− x2 − y2
=√
2
∫ 2π
0
∫ 1
0
[rdrdθ√2− r2
=√
2
∫ 2π
0
[−(2− r2)1/2]r=1r=0dθ
=√
2
∫ 2π
0
(√
2− 1)dθ.
= 2π(2−√
2)
64
Surface and Volume Integrals Page 65 Surface Area and Surface integrals
Surface integrals
Now we see how to integrate a function over a surface,using the idea of surface area.
Definition 3.1. (Surface Integral):Let R be the shadow region on the ground plane of a surface
S defined by the equation f(x, y, z) = c. If g is a continuous function defined on S,then the integral
of g over S is the integral∫∫R
g(x, y, z)∇f|∇f · p|
dA,
where p is a unit normal to R and ∇f · p 6= 0.
This integral is called the surface integral.
Orientation:We say that a smooth surface S is orientable or two-sided if it is possible to define
a field n of unit normal vectors on S that varies continuously with position.
Spheres and other smooth closed surfaces in space are orientable.The Mobius band is not orientable.
Definition 3.2. (Surface Integral for Flux):The flux of a three-dimensional vector field F across
an oriented surface S in the direction of n is given by
Flux=
∫∫R
F · ndσ
If S is a part of a level surface g(x, y, z) = c,then n may be taken to be one of the two fields
n = ± ∇g|∇g|
Remark1:If we take the projection R of S on xy−plane then∣∣n · k∣∣ ds = dxdy
therefore
∫∫S
F · nds =
∫∫R
F · ndxdyn · k
Remark2 If we take the projection R of S on xz−plane then |n · j| ds = dxdz
therefore
∫∫S
F · nds =
∫∫R
F · ndxdzn · j
Remark3 If we take the projection R of S on yz−plane then |n−.i−| ds = dydz
therefore
∫∫S
F · nds =
∫∫R
F · ndydzn · i
Moments and Masses of Thin Shells:
Thin shells of material like bowls metal drums, and domes are modeled with surfaces. Their moments
and masses are calculated with the formulas, given below
M =
∫∫S
δ(x, y, z)dσ, where δ(x, y, z) = density at (x, y, z), mass per unit area.
First moments about the coordinate planes;
Myz =
∫∫S
xδdσ,Mxz =
∫∫S
yδdσ,Mxy =
∫∫S
zδdσ,.
Coordinates of center of mass:
x =Myz
M, y =
Mxz
M, z =
Mxy
M
65
Surface Area and Surface integrals Page 66 Surface and Volume Integrals
Moments of Inertia:
Ix =
∫∫S
(y2 + z2)δdσ ,Iy =
∫∫S
(x2 + z2)δdσ ,Iz =
∫∫S
(x2 + y2)δdσ,
IL =
∫∫S
r2δdσ, where r(x, y, z) = distnace from point (x, y, z) to line L.
Radius of gyration about a line L:
Rx =√IL/M .
Example 3.3. Integrate g(x, y, z) = xyz over the surface of the cube cut from the first octant by
the plane x = 1, y = 1andz = 1
Solution. We integrate xyz over each of the six sides and add the results since xyz = 0 on
the sides that lie in the coordinate planes, the integral over the surface of the cube reduces to∫∫cubesurface
(xyz)dσ =
∫∫sideA
(xyz)dσ +
∫∫sideB
(xyz)dσ +
∫∫sideC
(xyz)dσ
Side A is the surface f(x, y, z) = z = 1 over the square region Rxy : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, in the
XY−plane.
For this surface and region, p = k,∇f = k, |∇f | = 1, |∇f · p| = |k · k| = 1
dσ =|∇f ||∇f.p|
dA =1
1dxdy = dxdy
xyz = xy(1) = xy
and
∫∫sideA
(xyz)dσ =
∫∫Rxy
(xy)dxdy =
∫ 1
0
∫ 1
0
xydxdy
=
∫ 1
0
(x2
2)10ydy =
∫ 1
0
y
2dy =
1
2(y2
2)10 =
1
4
Similarly, the integral of xyz over sides B and C are1
4and
1
4respectively
∴∫∫
cubesurface
(xyz)dσ =1
4+
1
4+
1
4=
3
4.
66
Surface and Volume Integrals Page 67 Surface Area and Surface integrals
Example 3.4. Evaluate
∫∫S
F·nds over the entire surface S of the region bounded by the cylinder
x2 + z2 = 9, x = 0, y = 0, z = 0 and y = 8, if F = 6zi+ (2x+ y)j − xk
Solution. Since we have to evaluate
∫∫over the entire surface hence we have divide it into
three parts lower and upper circular path and curved part of height clearly surface area of lower
circle is 9π and that of upper circle is 9π.
∴ Total area of circle is 18π . . . (1)
Let us take the projection R on xy−plane.
∴∫∫
S
F · nds =
∫∫F · n∣∣n · k∣∣dxdy . . . (2)
∴ n =∇s|∇s|
=2xi+ 2zk√4x2 + 4z2
=xi+ zk
3n · k =
z
3
F · n = [6zi+ (2x+ y)j − xk] · xi+ zk
3
=1
3(6xz − xz) =
5xz
3
∴F · n∣∣n · k∣∣ =
5xz
3× 1
z/3= 5x.∫∫
F · n∣∣n · k∣∣dxdy =
∫ x=3
x=−3
∫ y=8
y=0
5xdxdy
=
∫ x=3
x=−3
[5xy]80dx = 40
∫ 3
−3
xdx
= 40
(x2
2
)3
−3
= 40(9/2− 9/2) = 0
∴ Required double integral from (1) and (2)∫∫SF · nds = 18π.
Example 3.5. Evaluate
∫∫S
F · nds where F = zi+xj− 3y2zk and S the surface of the cylinder
x2 + y2 = 16 included in the first octant between z = 0 and z = 5.
Solution. Let us take the projection R of S on yz−plane
∴∫∫
S
F · nds =
∫∫R
F · n dydz|n · i|
n =∇s|∇s|
=2xi+ 2yj√4x2 + 4y2
=xi+ yj
4
67
Surface Area and Surface integrals Page 68 Surface and Volume Integrals
n · i =x
4
F · n = [zi+ xj − 3y2zk] · xi+ yj
4
=1
4(xz + xy) =
5xz
3
∴F · n|n · i|
=1
4× xz + xy
x/4= y + z.∫∫
R
F · n|n · i|
dydz =
∫ y=4
y=0
∫ z=5
z=0
(y + z)dzdy ( by putting x = 0 in x2 + y2 = 16 so y = ±4 )
=
∫ 4
0
[yz +z2
2]50dx =
∫ 4
0
(5y +25
2)dy
=
(5y2
2+
25y
2
)4
0
= 40 + 50 = 90
Example 3.6. Calculate flux of F where F = xi+ yj + 2zk over the surface x2 + y2 + z2 = a2.
Solution. Since we have calculate flux over the entire surface but entire surface is 8 time the
surface in first octant
∴∫∫
S
F · nds =
∫∫S
F · nds. . . (1)
where S is surface in first octant. Let us consider the projection of R of S on XY plane.
∴∫∫
S
F · nds =
∫∫R
F · n∣∣n · k∣∣dxdyNow n =
∇s|∇s|
=2xi+ 2yj + 2zk√
4x2 + 4y2 + 4z2
∴ n =xi+ yj + zk
an · k =
z
a
F · n =x2 + y2 + 2z2
a
∴F · n∣∣n · k∣∣ =
x2 + y2 + 2z2
z.∫∫
R
F · n∣∣n.k∣∣dxdy =
∫∫R
x2 + y2 + 2z2
zdxdy
=
∫∫R
x2 + y2 + 2(a2 − x2 − y2)√a2 − x2 − y2
dxdy
We have to find double integral over R, integrate over an elementary strip parallel over an elementary
srtip parallel to Y-axis from y = 0 to y =√a2 − x2
(Since plane is XY, so z = 0)
Again keeping x fixed and then move the strip from x = 0 to x = a (on X-axis, y = 0 and z = 0)∫ x=a
x=0
∫ y=√a2−x2
y=0
2a2 − x2 − y2√a2 − x2 − y2
dxdy
68
Surface and Volume Integrals Page 69 Surface Area and Surface integrals
=
∫ x=a
x=0
∫ y=√a2−x2
y=0
√a2 − x2 − y2dxdy +
∫ x=a
x=0
∫ y=√a2−x2
y=0
a2√a2 − x2 − y2
dxdy
=
∫ x=a
x=0
[y
2
√a2 − x2 − y2 +
a2 − x2
2sin−1 y√
a2 − x2
]dx+ a2
∫ a
0
[sin−1 y√
a2 − x2
]√a2−x20
dx
=
∫ a
0
π
2
(a2 − x2)
2dx+ a2
∫ a
0
π
2dx
=π
4
∫ a
0
[a2 − x2]dx+πa2
2
∫ a
0
dx
=π
4[a2x− x3
3]a0 +
πa2
2[x]a0
=π
4[a3 − a3
3] +
πa3
2
=π8
12a3 =
π2a3
3
∴∫∫
S
F.nds =π8
12a3 =
π2a3
3
∴ by (1)
∫∫S
F · nds =π16
3a3 .
Example 3.7. Find the center of mass of a thin hemispherical shell of radius a and constant
density δ.
Solution. We model the shell with the hemisphere f(x, y, z) = x2 + y2 + z2 = a2, z ≥ 0.
The symmetry of the surface abput the Z-axis tells us that x = y = 0. It remains only to find z from
the formula z = Mxy/M .
The mass of the shell is M =
∫∫S
δ(x, y, z)dσ
= δ
∫∫S
dσ = δ(areaofS)
= δ(2πa2)
69
Surface Integral for parametrized surface Page 70 Surface and Volume Integrals
= 2πa2δ.
To evaluate the integral for Mxy we take p=k and calculate
|∇f | = |2xi+ 2yj + 2zk| = 2√x2 + y2 + z2 = 2a
|∇f · p| = |∇f · k| = |2z| = 2z
dσ =|∇f | dA|∇f.k|
=2a
2zdA =
a
zdA
Then using the formula for Mxy =
∫∫S
zδdσ
= δ
∫∫S
z × a
zdA = δa
∫∫S
dA = δa(πa2) = πδa3
∴ Using z =Mxy
M=
πδa3
2πa2δ=a
2∴ The Shell’s center of mass is the point (0, 0, a/2).
Problem Set I
(1) Find the area of the band cut from the paraboloid x2 + y2 − z = 0 by the plane z = 6.
(2) Find the area of the region cut from the plane X + 2y + 2z = 5 by the cylinder where wells
are x = y2 and x = 2− y2.
(3) Integrate g(x, y, z) = (xyz) over the surface of the rectangular solid cut from the first octant
by the planes x = a, y = b and z = c.
(4) Integrate g(x, y, z) = x+ y + z over the surface of the cube cut from the first octant by the
planes x = a, y = a, z = a.
(5) Evaluate
∫∫S
F.nds where F = (x + y2)i + 2xj + 2yzk and S is the part of the plane
2x+ y + 2z = 6 in the first octant.
(6) If F = 4xzi− y2j + yzk, evaluate
∫∫S
F.nds where S is the surface of the cube bounded by
x = 0, x = 1, y = 0, y = 1, z = 0, z = 1.
(7) Calculate the Flux of F where F = −i + 2j + 3k and S is the part of rectangular surface
z = 0, 0 ≤ x ≤ 2, 0 ≤ y ≤ 3 by taking projection xy−plane.
(8) Calculate the Flux of F where F = xi + yj + 3k over the surface x2 + y2 + z2 = a2 in the
first octant.
2. Surface Integral for parametrized surface
Parametrization of Surface
We know that the equations of the form z = f(x, y)(explicit form) and F (x, y, z) = 0(implicit form)
represent surfaces in space.
There is also a parametric form of a surface,which can be expressed in terms of two parameters. Let
r(u, v) = f(u, v)i+ g(u, v)j + h(u, v)k. . . (1)
70
Surface and Volume Integrals Page 71 Surface Integral for parametrized surface
be the continuous vector function that is defined on a region R in the uv−plane and one-to-one
on the interior of R. We call the range of r the surface S.The equation (1) together with the
domain R represent a parametrization of the surface. The variables u and v are the parameters,
and R is parameter domain.The equation (1) is the vector equivalent of three parametric equations:
x = f(u, v), y = g(u, v), z = h(u, v)
A parametrized surface r(u, v) = f(u, v)i + g(u, v)j + h(u, v)k is smooth if ru =∂r
∂uand rv =
∂r
∂v
are continuous; and∂r
∂u× ∂r
∂vis never zero on the parameter domain.
The area of a smooth surface r(u, v) = f(u, v)i+ g(u, v)j+h(u, v)k;a ≤ u ≤ b, c ≤ v ≤ d is given by
A =
∫ d
c
∫ b
a
|ru × rv|dudvSurface Integral Using the Parametrized form
If S is a smooth surface defined parametrically as r(u, v) = f(u, v)i + g(u, v)j + h(u, v)k, a ≤ u ≤b, c ≤ v ≤ d and G(x, y, z) is a continuous function defined on S, then the integral of G over S is∫∫
S
G(x, y, z)dσ =
∫ d
c
∫ b
a
G (f(u, v), g(u, v), h(u, v)) |ru × rv| dudv.
Example 3.8. Find a parametrization of the cone z =√x2 + y2, 0 ≤ z ≤ 1.
Solution. Here cylindrical co-ordinates provide everything we need, a point (x, y, z) on the cone
has x = r cos θ, y = r sin θ and z =√x2 + y2 = r with 0 ≤ z ≤ 1 and 0 ≤ θ ≤ 2π.
Taking u = r, v = θ in the equation r(u, v) = f(u, v)i + g(u, v)j + h(u, v)k r(r, θ)=(r cos θ)i +
(r sin θ)j + rk
0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π.
This is required parametrization.
Example 3.9. Find a parametrization of the paraboloid z = x2 + y2, z ≤ 4.
71
Surface Integral for parametrized surface Page 72 Surface and Volume Integrals
Solution. In cylindrical co-ordinates, a point (x, y, z) has x = r cos θ, y = r sin θ, z = z
For points on the paraboloid : z = x2 + y2, x = r cos θ, y = r sin θ, z = r2 0 ≤ r ≤ 2, 0 ≤ θ ≤ 2π
Taking u = r and v = θ.
r(r, θ)=(r cos θ)i+ (r sin θ)j + r2k
0 ≤ r ≤ 2, 0 ≤ θ ≤ 2π.
This is required parametrization.
Example 3.10. Find a parametization of the sphere x2 + y2 + z2 = a2.
Solution. In spherical co-ordinate, point (x, y, z) has x = a sinφ cos θ, y = a sinφ sin θ and
z = a cosφ with 0 ≤ φ ≤ π and 0 ≤ θ ≤ 2π.
Taking u = φ and v = θ in equation r(u, v) = f(u, v)i+ g(u, v)j + h(u, v)k r(φ, θ)=(a sinφ cos θ)i+
(a sinφ sin θ)j + a cosφk
Which gives the required parametrization.
Example 3.11. Find a parametization of the cylinder x2 + (y − 3)2 = 9, 0 ≤ z ≤ 5.
Solution. In cylindrical co-ordinate, point (x, y, z) has x = r cos θ, y = r sin θ and z = z .For
points on the cylinder x2 + (y − 3)2 = 9 with r = 6 sin θ and 0 ≤ θ ≤ π.
∴ x = r cos θ = 6sinθ cos θ = 3sin2θ, y = r sin θ = 6sin2θ, Taking u = θ and v = z in equation