Vector Integration
d
dt

F(t)=G(t)
G(t)dt=F(t)+c
If G(t)=ti+t2j+t3k find
 G(t )dt  (ti  t
b
a
b

 G(t)dt=[F(t)] =F(b)-F(a)
a
2
G(t)dt and 1 G(t)dt
t2
t3
t4
i j k
=
=2
3
4 +c
2
2
2
3
4
3
7
17
t
t
t 
1 G(t )dt  2 i  3 j  4 k  2 i  3 j  4 k
1 =
=
2
j  t 3k )dt
2
Problem:If a=xi-x2j+(x-1)k and b=2x2i+6xk find
(axb)dx
Ans
12,
 (a.b)dx,
0
2

0
40
64
k
-24i- 3 j+ 5
ProblemThe acceleration a of a particle at any time t is given by
a= et i  6(t  1) j  3 sin tk . If the velocity v and displacement r are
zero at time t=0 find v and r at any time.
)c
dv
v   adt v  (
at t=0, v=0
dt
v  (1  e t )i  (3t  6t ) j  (3  3 cos t )k
)c
dr
r   v dt r  (
v
at t=0, r=0
dt
a
t
3
c=i+3k
c=-i
r  (t  1  e )i  (t  3t ) j  (3t  3 sin t )k
2
Equation of Curve
In ordinary calculus the equation of the curve is y=f(x),
In vector calculus the equation of the curve is r=F(t)
Example
r=xi+yj x=cost, y=sint
r=costi+sintj is the
equation of the circle.
r=F(t)
Example r=xi+yj x=at2, y=2at
r= at2i+2at j
is the equation of the parabola.
r=F(t)
Three special types of integrals in vector calculus
1. Line integral
2. Surface integral
3. Volume integral
Line integral
The line integrals are
 A.dr
 A x dr,
2.


3.
dr
1.
C
C
C
Explanation
Integration is usually done along x-axis, y-axis but it is also done
along any line and curve
y
y=f(x)
f(a+nh)
b
b
 f ( )df (
)d
f(a)
f(x)
f(a+h)
f(a+2h)
x
x=a
a+h a+2h
2
x
b=a+nh
P2
y
c
s
P1 s
y
o
x
s
x
fig.1
 f ( x)dx  f ( y)dy  f ( x, y)ds
Three types of line integrals in vector are


1. A.Tds
or
A.Tds


P2
C
P1
P2
2.
C
A xTds

or
P1
A xTds
P2

C
3. Tds
or P Tds
where
A(x,y,z)=A1(x,y,z)i+A2(x,y,z)j+A3(x,y,z)k is a vector function
T= unit tangent vector
s=length of the curve
1
s =elementary
length of the curve

r= Lt s0T s
dr= Tds
dr is along the tangent and it is equal to Tds
Lt r 0
3
T =unit tangent
T =unit tangent
ds
Td s
Q
r+  r  r
PQ=  s
r
o
dr
P
s
s
fig.2
we know
a  aaˆ
r  s
Ts
= s

r= Lt s0T s
dr= Tds
dr is along the tangent and it is equal to Tds
Lt r 0
dr
T
=
ds
unit tangent vector
So line integrals are
1.
2.
3.

C

C

C
P2
A.dr
or

A
A.dr
P1

P2
A x dr,


or
P1
P2
dr
or

P1

A x dr
T
A.T  A.1. cos   A cos 
dr
 tan gential component of A
4
 A.Tds=  A.dr is also called the line integral of the tangential
C
C
component of A
dr=dxi+dyj+dzk

C
A.dr=

C
P2
(A1dx+A2dy+A3dz)=

P1
(A1dx+A2dy+A3dz)
Physical meaning of line integral
1.

P2

C
A.Tds
or P A.Tds
If A is the force F on a particle moving along C this line integral
represents the work done by the force F
1
Work=Force x displacement=Fd
d
F
W=Fcos  x d=F.d
Fcos  is the effective force

Fcos 
d
F
fig.3
F=2i+3j+4k constant vector
F=-3x2i+5xyj+2zk variable vector
5
If a body moves along a curve by a variable force how can we
measure the work done.
We consider a curve with elementary length s then
Tds= dr
F
F F
1
T
2
1
n
T
s
T
2
n
s
s
Fig 4
Total work=F1. T1 s + F2. T2 s + F3. T3 s +…………Fn. Tn s
= Lt s0  F.T s
 F.Tds

= F.dr

Hence W= F.dr
=
C
C
C
(4,4)
0 1 2 3 4
6
(4,4)
0 1 2 3 4
1/2 3/2 7/2
5/2
b
Explanation of
Lt
x 0
 f ( x)x   f ( x)dx
a
Area under the curve y=x from x=0 to x=4
0
A=
4
4
 xdx
1
.4.4  8
A= 2
=
 x2 
2
 0 =
8
 f ( x)x  1x1+2x1+3x1+4x1=10 x =1
 f ( x)x  1 x 1  1x 1  3 x 1  2 x 1  5 x 1  3x 1  7 x 1  4 x 1  9
x
1
=2
2 2
2 2 2
2 2 2
2 2 2
2
1 1 2 1 3 1 4 1 5 1 6 1 7 1 8 1
1
 f ( x)x  4 x 4  4 x 4  4 x 4  4 x 4  4 x 4  4 x 4  4 x 4  4 x 4  x = 4
9 1 10 1 11 1 12 1 13 1 14 1 15 1 16 1 16 x17
x  x  x  x  x  x  x  x 
 8.5
4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4
32
x 1
=4
x 1
32x33
 f ( x)x 
8.25
=8
128
x 1
64x65
 f ( x)x 
8.125
= 16
512

x
1
128x129
 f ( x)x 
8.0625
= 32
2048

x

0
 f ( x)x  8
Lt
x 0

4
f ( x)x  8   xdx
0
7
Lt
x 0

b
f ( x)x   f ( x)dx
a
---------------------------------------------------------------------------------
Evaluation of line integral
[open curve problem]

Example 1. Evaluate the line integral C F.dr where
F=-3x2i+5xyj and C is the curve y=2x2 in the xy plane from (0,0) to
(1,2).
OR,
Find the work done in moving a particle in the force field
F=-3x2i+5xyj along the curve y=2x2 in the xy plane from (0, 0) to
(1, 2).
y  2x 2
(1,2)
C
(0,0)
Fig.5
 F.dr

= (-3x2i+5xyj).( dxi+dyj)
C
C
8

= C (-3x2dx+5xydy)
along the curve C, y=2x2 dy=4xdx
1

= 0 (-3x2dx+5x2x24xdx)
1

= 0 (-3x2dx+40x4dx)
1
=  x3  8x5 0
=7
Alternative method
(using dy)
 F.dr

= (-3x2i+5xyj).( dxi+dyj)
C
C
y=2x

2
x2 
= C (-3x2dx+5xydy) along the curve C,
2
y 1
 (3 2 2
=
0
2
x
y
2
y
2
1
y
dy  5
ydy )
2
2 y
3
3
5 2
(
y dy 
y dy )

2
=0 4 2
2
5
 1 32

y  2y2 

 0
=  2 2
=

1
2 2
3
5
22  2 22
=7
Alternative method (using parametric form)
 F.dr

= (-3x2i+5xyj).( dxi+dyj)
C
C
9
dx 
1
1
dy
2 2 y

= C (-3x2dx+5xydy)along the curve C, x=t y=2t2 dx=dt dy=4tdt
1

= 0 (-3t2dt+5t2t24tdt)
1

= 0 (-3t2dt+40t4dt)
1
=  t 3  8t 5 0
=7
[closed curve problem, W  0]

Example 2 Evaluate C F.dr where F=(x-3y)i+(y-2x)j and C is the
closed curve in the xy plane x=2cost, y=3sint from t=0 to t= 2 .
C
fig.6
 F.dr=  [(x-3y)i+(y-2x)j].(dxi+dyj)

= (x-3y) dx+(y-2x) dy along the curve C, x=2cost, y=3sint
C
C
C
2

= 0 (2cost-9sint)(-2sint)dt +(3sint-4cost)(3cost)dt
2

= 0 (-4 sint cost+18sin2t +9sintcost-12cos2t)dt
= 6
[closed curve problem, W=0]
10

Example 3 Evaluate C F.dr where F=xi+yj and C is the closed
curve in the xy plane x=2cost, y=3sint from t=0 to t= 2 .
C
fig.7
 F.dr=  ( xi+yj).(dxi+dyj)

= x dx+y dy along the curve C, x=2cost, y=3sint
C
C
C
2

= 0 (2cost)(-2sint)dt +(3sint)(3cost)dt
2

= 0 (-4 sint cost+9sintcost)dt
2

= 0 5sintcostdt
=0
[different path same result (end points same)]

Example 4. Evaluate C F.dr where F=yi+xj and C is
(a) the arc of y=x2-4 from (2,0) to (4,12) in the xy plane.
(b) the portion of the x-axis from x=2 to x=4 and then the line x=4
from y=0 to y=12.
11
y  x2  4
(4,12)
C
C2
C1
(0,0) (2,0) (4,0)
fig.8
(a)
 F.dr=  ( yi+xj).(dxi+dyj)

= y dx+x dy
C
C
C
4
= 2 ( x2-4)dx+x2xdx along the curve C, y=x2-4 dy=2xdx
=48
(b)
 F.dr=  ( yi+xj).(dxi+dyj)

= y dx+x dy
C
C
C

= C y dx+x dy
along x-axis y=0, dy=0
1

+ C y dx+x dy
2

along x=4, dx=0
12
4
0dx  x.0
=2
=48

+0
y.0  4dy
[different path different results (end points same)
12

Example 5 Evaluate C F.dr where F=xyi+(x2+y2)j and C is
(a)the arc of y=x2-4 from (2,0) to (4,12) in the xy plane.
(b) the portion of the x-axis from x=2 to x=4 and then the line x=4
from y=0 to y=12.
y  x2  4
(4,12)
C
C2
C1
(0,0) (2,0) (4,0)
fig.9
 F.dr=732 or 552 ?

(b) F.dr=768 ok
(a)
C
C



Two other line integrals:
Fx dr,
dr

Ex 6. Evaluate the line integral Fx dr if F=xyi-yj+x2k and C
C
C
C
is the curve x=t , y=2t, z=t from t=0 to t=1.
Along the curve C, F= xyi-yj+x2k=2t4i-2tj+t6k
r=xi+yj+zk= t3i+2tj+t2k, dr=(3t2i+2j+2tk)dt
3
2
Fxdr=(2t4i-2tj+t6k)x(3t2i+2j+2tk)dt
13
i
2t 4
3t 2
j
 2t
2
k
t6
2t
=
dt
2
6
=[i(-4t -2t )+j(3t8-4t5)+k(4t4+6t3)]dt

C
1
1

1


Fxdr= i 0 (-4t2-2t6)dt+j 0 (3t8-4t5)dt+k 0 (4t4+6t3)dt
 34
 1 23
= 21 i+ 3 j+ 10 k

Ex 7. Evaluate the line integral


dr if = xyz and C is the
curve x=t , y=2t, z=t from t=0 to t=1.
 = xyz=2t6, r=xi+yj+zk= t3i+2tj+t2j, dr=(3t2i+2j+2tk)dt
C
3

C

2
1

dr= 0 2t6(3t2i+2j+2tk)dt
1

1
1


=i 0 6t8dt+j 0 4t6dt+k 0 4t7dt
6
4
1
= 9 i+ 7 j+ 2 k
Conservative Force field/ Conservative field
Spring, throw, work done is zero, force is conserved/preserved
within body, called Conservative field of force.
Fig.10
14
Gravitational field of force (g=-gk), electric field of force,
magnetic field of force are all conservative field of force.
g=-gk
fig.11
Field of force-collection of so many vectors
fig.12
Def. Conservative field- For a conservative field,

(i)Work done C F.dr=0 around any closed curve C.
C
fig.13
Ref. For (i) Example 3
15

(ii)Work done C F.dr is independent of the path C joining P1 and P2
. It depends on end points only
C1
P2
C
C3
P1
C= C1  C3
C2
fig. 14
If F is conservative

Now C



C1


C3

C1
 F.dr
=0
C



0
C2
C1


C2
Ref. for (ii) Example 4
Find the condition for a force F to be conservative
If F=  where  is a single valued scalar function
then show that the work done in moving a particle from one
point P(x1, y1, z1) in this field to another point Q (x2, y2, z2) is
independent of the path joining the two points.
Q
W= P F.dr
Q
= P
Q

= P (
Q
= P
.dr
i



 j
k
x
y
z
).(dxi+dyj+dzk)



dx 
dy 
dz

x

y

z
(
)
Exact




dx 
dy 
dz
y
z =d
differential x
16

Q
= P d
=  QP
=  (Q)-  (P)
=  ( x2 , y 2 , z 2 ) -
 ( x1 , y1 , z1 )
Q
Work done P F.dr is independent of the path C joining P and Q.
It depends on end points only
Therefore if F=  the force is conservative.
We know that  x  =0
Force F is conservative if  x F=0 (condition)
Ex.8.(a)Show that F=yi+xj is a conservative force field
(b) Find the work done in moving an object in this field
from (2,0) to (4,12)
(c)Find a Scalar potential 
(a) Curl F=0
i

x
y
j

y
x
k

z
0
Curl F=
=i(
So F is Conservative
)+j(
)+(
)k=0
Q
(b) W= P F.dr
( 4 ,12 )

= ( 2,0) ( yi+xj).(dxi+dyj)
( 4 ,12 )
=

y dx+x dy
[Exact differential]
( 4 ,12 )
xy ( 2,0)

= ( 2,0) d(xy) =
= 48
( 2, 0 )
( 4 ,12 )
(c) As  x F=0, F must be

17
F = 
yi+xj =

x = y

y = x
i


 j
x
y
……. (1)
……….(2)
Integrating (1) with respect to x
 =yx +f(y) ………...(3)

y
f ( y )
=x+ y
……….(4)
f ( y )
y =0
Comparing (2) and (4)
….. (5)
Integrating (5) with respect to y, f(y)= c
hence  =xy+c
Alternative method
 =yx +f(y)
 = xy+g(x)
f(y)=0
g(x)=0
hence  =xy+c
Example 9.
(a) Show that the Force
F=(-4x-3y+4z)i+(-3x+3y+5z)j+(4x+5y+3z)k is conservative.
(b)Find a scalar function  so that F=  .
(a) F is conservative if  xF=0

xF=
i
j
k



x
y
z
- 4x - 3y + 4z - 3x + 3y + 5z 4x + 5y + 3z
(b) As  x F=0, F must be
F = 

18
=0
(-4x-3y+4z)i+(-3x+3y+5z)j+(4x+5y+3z)k =

x = -4x-3y+4z

y = -3x+3y+5z

z =4x+5y+3z
i



 j
k
x
y
z
……. (1)
……….(2)
………….(3)
Integrating (1) with respect to x
 =-2x2-3xy+4xz+f(y,z) …..(4)

y

z
f ( y, z )
=-3x+ y
f ( y, z )
= 4x+ z
……….(5)
…………(6)
f ( y, z )
y =3y+5z
f ( y, z )
z =5y+3z
Comparing (2) and (5)
Comparing (3) and (6)
….. (7)
….. (8)
Integrating (7) with respect to y, f(y, z)=
f ( y, z )
 5 y  g ( z )
z
3
2
2 y +5zy+g(z)
….. (9)
g ( z )  3z
Comparing (8) and (9)
g ( z) 
3 2
z c
2
3
3
2
therefore f(y, z)= 2 y +5yz+ 2 z2+c

3
3
2
2
hence =-2x -3xy+4xz+ 2 y +5yz+ 2 z2+c
Alternative method
 =-2x2-3xy+4xz+


=
-3xy+
=
f(y,z)=
g(z,x)=
+
+
+f(y,z)
3
2
2y +
5yz +
+g(z,x)
4xz+
+5yz +
3
3
2
2
2 y +5yz+ 2 z
3
2
-2x +4xz+ 2 z2
19
3
2
2z
+h(x,y)
h(x,y)= -2x -3xy+
2

hence =-2x
3
2
2y
3
3
2
-3xy+4xz+ 2 y +5yz+ 2 z2+c
2
==============================================
Problem copied from VEC2
Ex. (a)Determine the constant a, b, c so that vector
v=(-4x-3y+az)i+(bx+3y+5z)j+(4x+cy+3z)k is irrotational.
(b)Find a scalar function  so that v=  .
v is irrotational if  xv=0

i
j
k



x
y
z
4x
3y
+
az
bx
+
3y
+
5z
4x
+
cy + 3z
xv=
=(c-5)i-(4-a)j+(b+3)k =0 =0i+0j+0k
a=4,
b=-3, c=5
(b)As  xv=0, v must be
v= 

(-4x-3y+4z)i+(-3x+3y+5z)j+(4x+5y+3z)k =

……. (1)
x = -4x-3y+4z

y = -3x+3y+5z
……….(2)

z =4x+5y+3z ………….(3)
Integrating (1) with respect to x partially
 =-2x2-3xy+4xz+f(y,z) …..(4)

y

z
f ( y, z )
=-3x+ y
f ( y, z )
= 4x+ z
……….(5)
…………(6)
20
i



j
k
x
y
z
f ( y, z )
y =3y+5z
f ( y, z )
z =5y+3z
Comparing (2) and (5)
Comparing (3) and (6)
….. (7)
….. (8)
3
2
2 y +5zy+g(z)
Integrating (7) with respect to y, f(y, z)=
f ( y, z )
 5 y  g ( z )
z
….. (9)
g ( z )  3z
Comparing (8) and (9)
g ( z) 
3 2
z c
2
3
3
2
therefore f(y, z)= 2 y +5yz+ 2 z2+c

3
3
2
2
hence =-2x -3xy+4xz+ 2 y +5yz+ 2 z2+c
Alternative method
v= 
(-4x-3y+4z)i+(-3x+3y+5z)j+(4x+5y+3z)k =
i



j
k
x
y
z

……. (1)
x = -4x-3y+4z

y = -3x+3y+5z
……….(2)

z =4x+5y+3z ………….(3)
 =-2x2-3xy+4xz+f(y,z)


=
=
3
-3xy+ 2 y2+5yz+g(z,x)
3
4xz + 5yz + 2 z2 +h(x,y)
 =-2x2-3xy+4xz+


=
=
-3xy+
+
f(y,z)=
g(z,x)=
+
+
3
2
2y
+
+f(y,z)
+ 5yz +
+g(z,x)
+4xz +
+ 5yz +
3
3
2
2
2 y +5yz+ 2 z
3
2
-2x +4xz+ 2 z2
21
3
2
2z
+h(x,y)
3
2
2y
h(x,y)= -2x -3xy+
2

3
3
2
-3xy+4xz+ 2 y +5yz+ 2 z2+c
hence =-2x
==============================================
2

In fluid mechanics the line integral C A.dr is called the circulation
of A about C where A represents the velocity of a fluid


Circulation= C= C v.dr [Work done=W= C F.dr ]
If no circulation (irrotational)
 v.dr=0
C
curlv=0
Ex 10. Find the circulation of A round the curve C where
A=(2x+y2)i+(3y-4x)j and C is the curve y=x2 from (0,0) to (1,1)
and then y2=x from (1,1) to (0,0).
A.dr = (2x+y2)dx+(3y-4x)dy
Circulation=

C


1
2
A.dr= C A.dr+ C A.dr
y
C
C2: y2=x
B(1,1)
C1: y=x2
A
x
O
22
Fig 15
On C1 , OAB y=x2 so that dy=2xdx

C1

A.dr= C (2x+y2)dx+(3y-4x)dy
1
1
1

= (2x+x4)dx+(3x2-4x)2xdx= 30
0
On C2

C1
BCO
x=y2 so that dx=2ydy

A.dr= C (2x+y2)dx+(3y-4x)dy
1
0
= 1
Circulation=
5
(2y2+y2)2ydy+(3y-4y2)dy=- 3

C


1
2
A.dr= C A.dr+ C
1 5
49
A.dr= 30 - 3 =- 30
Ex 11. Find the work done in moving a particle along the curve C
in the force field F=(2x+y2)i+(3y-4x)j where C is the curve y=x2
from (0,0) to (1,1) and the y2=x from (1,1) to (0,0).
Problem
Determine the work done by the force of gravity F when a mass m
is translated from the point P(a1, b1, c1) to the point Q(a2, b2, c2)
along an arbitrary path C.
y
Q(a2, b2, c2)
C
P(a1, b1, c1)
x
z
23
fig.37
Let F= Xi+Yj+Zk
The projections (components) of the force of gravity F on the
coordinate axes are X=0, Y=0, Z=-mg
Hence, the desired work is

W= C F.dr
Q
= P (Xdx+Ydy+Zdz)
( a 2 , b2 c 2 )

= ( a b , c ) (0dx+0dy +(-mg)dz
=mg(c1-c2)
1
1
1
09 -12-21 Quiz 1 explanation of projection ds-dxdy
equation plane and its projection 1D 2D 3D
Surface integral
Def: Surface integrals are (i)
(iv)
 A.ndS (ii)  dS (iii)
S
S
 AxdS
S
Physical meaning of (i)
 A.ndS
S
Suppose fluid is flowing through the surface
S
S
24
S
 ndS
S
v
n
n
S
v
S
v= vcos0=vcos  =v.n
S
v

S
v
vcos  =v.n
0=


vcos 2 =vcos = v.n
Flow is always along the normal to the surface.
Only component of v along the normal to the surface are taken into
account
vcos 
S
dV=  S vcos 
fig.16
Mass (quantity) of fluid passing through S per unit time
 dV =  v cos S   v.n S where n is outward drawn normal to the
surface S.
25
S
S
S
S
fig.17
Total fluid mass Q passing through S is

v1.n1 S +  v2.n2 S +  v3.n3 S ………………….+  vn.nn S

= Lt s0  v.n S

=

=
S
S


v.ndS
v.dS

Hence fluid flow Q= S
ndS=dS

v.dS
 q.dS

Current flow J.dS
Heat flow
S
S
26

In general Surface integral is S A.ndS
It is a surface integral of the normal component of vector A
 A.ndS is the flux of A over S
S
Flux means flow
So surface integral is also called flux integral or flow integral
Evaluation of Surface integral
To evaluate the surface integral we are to project the surface
on the coordinate planes.
Projection of a surface on a plane becomes a plane.
y
x
z
First we consider the projection on the xy plane
Elementary area in the xy plane is =dxdy
The angle between two planes is equal to the angle between their
normals
n
n
S
dS
k
dS
n
27
S

a
dxdy
dS
Projection of a is

b=acos 
b
Fig. 18
Projection of dS is
dS 
dxdy=dS cos  =dSn.k
dxdy
n.k
Evaluation of surface integral

S

S

S

A.ndS= R

A.ndS= R

A.ndS= R
dxdy
A.n n.k
dydz
A.n n..i
dzdx
A.n n. j
projection on xy plane
projection on yz plane
projection on zx plane
Problem

Example 12.Evaluate S F.ndS where F=10zi+10j+3yk and S is
the part of the plane 2x+3y+6z=12 which is located in the first
octant.
Or,
Find the flux of F=10zi+10j+3yk through the surface S where S is
the part of the plane 2x+3y+6z=12 which is located in the first
octant.
Or,
28
Find the flow of fluid through the surface S where S is the part of
the plane 2x+3y+6z=12 which is located in the first octant, if the
fluid velocity is v=10zi+10j+3yk
Or,
Find the flow of heat through the surface S where S is the part of
the plane 2x+3y+6z=12 which is located in the first octant, if the
heat flux q=10zi+10j+3yk
Or,
Find the flow of current through the surface S where S is the part
of the plane 2x+3y+6z=12 which is located in the first octant, if
the current density J=10zi+10j+3yk
Solution:
Let the projection of the surface S on the xy plane be R

S

F.ndS= R
  i
dxdy
F.n n.k



j
k
x
y
z


n=
normal to the plane 2x+3y+6z=12
2i  3 j  6k
6
n=
n.k= 7
7
20 z  30  18 y 20
(12  2 x  3 y )  30  18 y
7
6
F.n=
=
7
y
4
2
2x+3y+6z=12
x
6
29
70 
=
20
x  8y
3
7
z
x y z
  1
6 4 2
y
4
Area of integration
2x+3y=12
x
6
x y
 1
6 4
 dxdy   dydx   ydx   12  2 x dx
3
R
 dxdy
R
A=
6 y
6
6
=0 0
0
0
12 2 x
6
3
 
=0
0
=12=A
6
12  2 x
dx
3
0
=12
dydx  
projection of the plane on the xy plane will be a triangle


S
R
F.ndS=
F.n
dxdy
n.k
= R
70 
20
x  8y
7
3
dxdy
7
6
1
(210  20 x  24 y )dxdy

= 18 R
1 6 (122 x ) / 3
0(210  20 x  24 y)dydx
18 
x

0
y
=
7 6 210 y  20 xy  12 y 2 (122 x ) / 3 dx
0
4 
= x 0
7 6 
20
4

x(12  2 x)  (12  2 x) 2 dx
70(12  2 x) 

4
3
3

= x 0 
…………………………………………………………


30
Problem 5.21 p-116
  xyz


Example 13. Evaluate
ndS where
and S is the surface
S
of the cylinder x +y =16 included in the first octant between z=0
and z=3.
Fig 5.8 of the book
2

S

ndS= R
  i
n
2
dxdz
n. j



j
k
x
y
z


n=
normal to the surface x2+y2=16 is
2 xi  2 yj
2
2
xi  yj
= 4
y
n.j= 4
n= 4 x 4 y
projection of the surface on the xz plane will will be a rectangle.

S
ndS=

R
4
n
dxdz
n. j
3
3
1
 x3 
i    zdz  j 
3 z 0
= z 0  3  0
3
3

=
4
 xz( xi  yj)dxdz
z 0 x 0
3
4
  x zi  xz
=
2
16 x 2 j )dxdz
z 0 x 0
4
3

2 2
16

x

 zdz

0


3
64i  z 2  64 j  z 2 
  
 
= 3  2 0 3  2 0
= 96i  96 j
Volume Integral


Def: Volume integrals are (i)
dV

 dV
Physical meaning of (i)
V
V
V
V
31
(ii)
 AdV
V
Let  ( x, y, z ) be the density of a body having volume V.
And let  V=  x  y  z be the elementary volume then the mass of
the elementary volume  V is   V
Total Mass M=
1  V
+ 2  V + 3  V+………………..
 
V 0 
V
= Lt

=
V
=

V
n  V


dV
dxdydz


In general this volume integral is
dV
 AdV
The other volume integral is
V
V
Problem


Example 14.Evaluate
dV where
  x2 y
and V is the region
bounded by the planes 4x+2y+z=8, x=0, y=0, z=0.
V
  dV


=
dxdydz
V
V
2
4 2 x 8 4 x  2 y
2
 
= x 0
y 0
x
42 x
  x yz 
2
2
= x 0
2
8 4 x  2 y
0
dydx
y 0
42 x
 
= x 0
ydzdydx
z 0
x 2 y (8  4 x  2 y )dydx
y 0
32
2
42 x
  x (8  4 x) y  2 x y dydx
2
= x 0
2

= x 0
2
2
y 0
42 x
3
 2
y2
2 y 
x
(
8

4
x
)

2
x


2
3 0

dx
3
 2
(4  2 x) 2
2 (4  2 x) 
  x (8  4 x) 2  2 x 3 dx
= x 0 
2
1 2
3
x 0 3 x (4  2 x) dx
=
128
= 45
2
Interpretation: Physically the result can be interpreted as the mass
of the region V in which density  varies according to the formula
  x2 y
Problem 5.26 Do it

Example-15. Evaluate V F dV where F=2xzi-xj+y2k and V is the
region bounded by the surfaces x=0, y=0, y=6, z=x2, z=4.
Since z=x2, z=4,
x2=4,
x=2
 F dV
V
=
2
6
4
0
0
x2
   (2xzi-xj+y2k) dzdydx

Example 16. Evaluate V F dV where F=2x2i-xzj+y2zk and V is
the region bounded by the surfaces x=1, y=0, y=6, z=x2, z=4.
Since z=x2, z=4, x2=4, x=2
33
 F dV
V
2
6
4
1
0
2
6
x2
4
=i 1
0
x2
2
6
=i 1
0
2
6
1
0
=
   (2x2i-xzj+y2zk) dzdydx
2
6
4
0
x2
2
6
4
0
x2
   2x2 dzdydx-j    xz dzdydx+k    y2z dzdydx
 
1
z 
4
x2
2
6
dydx-j 1
0
 z2 
 
x  2 x2
 
2x2
1
1
4
2
6
1
0
2
6
dydx+k 1
0
4
 z2 
 
y2  2  x 2
1 2 6
2  
 
  2x2 (4-x2)dydx-j 2   x(16-x4) dydx+k
=i
 y 60
2

=i 1 2x2 (4-x2)
y
dx-j
6
0
2

=i 1 2x2 (4-x2)
2
 y 60
2
1
2 1
x(16-x4)
dx-j
y
6
0
2
1
2 1
dx+k
x(16-x4)
2
dx+k
1
2
1
2
1
0
dydx
y2(16-x4) dydx
6
2

1
(16-x4)
1
(16-x4)
 y3 
 
 3 0
2
=i 1 12x2 (4-x2) dx-j 1 3x(16-x4) dx+k 1 36 (16-x4) dx
2
2
2
=i 1 (48x2 -12x 4) dx-j 1 (48x -3 x 5 ) dx+k 1 (576-36x4) dx
2
2
 x3
 x2
x5 
x6 
48

12
48

3

5 1 - j  2
6 1
=i  3
188 81
1764
=i 5 -j 2 +k 5
2

x5 
576
x

36

5 1
+k 
34
dx
6
2

 y3 
 
 3 0
dx
Discussion:
1D
o
x
2
1
x
x =0 x=1
y
2D
(0,1)
o
(1,1)
(1,0)
y=1
x y=0
General equation of straight line is Ax+By+C=0
Intercept form of the equation is
x y
 1
a b
Equation of x axis is y=0
Equation of y axis is x=0
x=1 is a straight line parallel to y- axis
y=1 is a straight line parallel to x- axis
35
y
4
2x+3y=12
x
6
x y
 1
6 4
Equation of the circle is x2+y2=16
Equation of the straight line passing through two points
( x2 , y 2 )
is
x  x1
y  y1

x2  x1 y2  y1
y
3D
x
o
z
General equation of plane line is Ax+By+Cz+D=0
Intercept form of the equation is
x y z
  1
a b c
Equation of xy plane is z=0
Equation of yz plane is x=0
Equation of zx plane is y=0
z=1 is a plane parallel to xy plane
36
( x1 , y1 )
and
x=1 is a plane parallel to yz plane
y=1 is a plane parallel to zx plane
y
4
z
2
2x+3y+6z=12
x
6
x y z
  1
6 4 2
Equation of the cylinder is x2+y2=16, z=0, z=3.
Equation of the cylinder is y2+z2=16, x=0, x=3.
Equation of the cylinder is z2+x2=16, y=0, y=3.
Equation of the parabolic sheet is y2=x
Equation of the parabola is y2=x, z=0
General equation of the straight line is
A1x++B1y+C1z+D1=0=A2x +B2y+C2z+D2
Equation of the straight line passing through two points
and
( x2 , y 2 , z 2 )
is
x  x1
y  y1
z  z1


x2  x1 y2  y1 z 2  z1
37
( x1 , y1 , z1 )