b.tech ii unit-5 material vector integration

Unit-5 VECTOR INTEGRATION

Unit-V: VECTOR INTEGRATION

Sr.

No.

Name of the Topic Page

No.1 Line Integral 2

2 Surface integral 5

3 Volume Integral 6

4 Green’s theorem (without proof) 8

5 Stoke’s theorem (without proof) 10

6 Gauss’s theorem of divergence (without proof) 13

7 Reference book 16

RAI UNIVERSITY, AHMEDABAD 1


Vector integration

1.1 LINE INTEGRAL:

Line integral ¿∫c(F . drds )ds=∫c F .dr

Note:1) Work: If F represents the variable force acting on a particle along arc AB,

then the total work done ¿∫A

B

F .dr

2) Circulation: If V represents the velocity of a liquid then ∮cV .dr is called the

circulation of V round the closed curve c .If the circulation of V round every closed curve is zero then V is said to be irrotational there.

3) When the path of integration is a closed curve then notation of integration is ∮ in place of∫ .

Note: If ∫A

B

F .dr is to be proved to be independent of path, then F=∇∅

here F is called Conservative (irrotational) vector field and ∅ is called the Scalar potential. And ∇× F=∇×∇∅=0

Example 1: Evaluate ∫cF .dr where F=x2 i+xy j and C is the boundary of the square

in the plane z=0 and bounded by the linesx=0 , y=0 , x=a∧¿

y=a.

Solution: ∫cF .dr=∫

OAF .dr+∫

ABF .dr+∫

BCF .dr+∫

COF .dr

Here r=x i+ y j , dr=dx i+dy j , F=x2 i+xy j



F .dr=x2dx+xydy _______ (i)

On OA , y=0 ∴F .dr=x2dx (From (i))

∫OAF .dr=∫

0

a

x2dx=[ x3

3 ]0

3

=a3

3 _______ (ii)

On AB , x=a ∴dx=0

∴F .dr=ay dy (From (i))

∫ABF .dr=∫

0

a

ay dy=a[ y2

2 ]0

a

=a3

2 _______ (iii)

On BC , y=a ∴dy=0

∴F .dr=x2dx (From (i))

∫BCF .dr=∫

a

0

x2dx=[ x3

3 ]a

0

=−a3

3 _______ (iv)

On CO, x=0

∴F .dr=0 (From (i))

∫CO

F .dr=0 _______ (v)

On adding (ii), (iii), (iv) and (v), we get

∫CF .dr=a3

3+ a3

2−a3

3+0=a3

2 ________ Ans.

Example 2: A vector field is given by

F=(2 y+3 ) i+( xz ) j+( yz−x) k . Evaluate ∫CF .dr along the path c is

x=2 t , y=t , z=t3 ¿ t=0¿ t=1.

Solution:



∫CF .dr=∫

C(2 y+3 ) dx+ (xz ) dy+( yz−x )dz

[ since x=2 t y=t z=t3

∴ dxdt

=2 dydt

=1 dzdt

=3 t 2] ¿∫

0

1

(2 t+3 ) (2dt )+ (2t ) ( t3 )dt+(t 4−2 t )(3 t 2dt)

¿∫0

1

( 4 t+6+2t 4+3 t 6−6 t 3 )dt

¿ [4 t2

2 +6 t+ 25 t

5+37 t7−6

4 t 4]0

1

¿ [2 t 2+6 t+ 25t 5+ 3

7t7−3

2t 4]0

1

¿2+6+ 25+3

7−3

2

¿7.32857 _________ Ans.

Example 3: Suppose F ( x , y , z )=x3 i+ y j+z k is the force field. Find the work done by F along the line from the (1, 2, 3) to (3, 5, 7).

Solution: Work done ¿∫cF .dr

¿ ∫(1,2,3)

(3,5,7)

(x3 i+ y j+z k ) . d ( i dx+ j dy+k dz )

¿ ∫(1,2,3)

(3,5,7)

(x3dx+ ydy+ zdz )

¿∫1

3

x3dx+∫2

5

y dy+∫3

7

z dz

¿ [ x4

4 ]1

3

+[ y2

2 ]2

5

+[ z2

2 ]3

7



¿ [ 814

−14 ]+[25

2− 4

2 ]+[492

−92 ]

¿ 804

+ 212

+ 402

¿ 2024

¿50.5 units _______ Ans.

1.2Exercise:

1) If a force F=2 x2 y i+3 xy j displaces a particle in the xy-plane from (0, 0) to (1, 4) along a curvey=4 x2. Find the work done.

2) If A=(3 x2+6 y ) i−14 yz j+20 x z2 k, evaluate the line integral ∮ A dr from (0, 0, 0) to (1, 1, 1) along the curveC.

3) Show that the integral ∫(1,2)

(3,4)

( x y2+ y3 )dx+(x2 y+3 x y2)dy is independent of

the path joining the points (1, 2) and (3, 4). Hence, evaluate the integral.

2.1 SURFACE INTEGRAL:

Let F be a vector function and S be the given surface.Surface integral of a vector function F over the surface S is defined as the integral of the components of F along the normal to the surface.

Component of F along the normal¿F . nWhere n = unit normal vector to an element ds and

n= grad f|grad f|

ds=dx dy(n . k )



Surface integral of F over S

¿∑ F . n ¿∬S

(F .n )ds

Note:

1) Flux ¿∬S

(F .n )ds where, F represents the velocity of a liquid.

If∬S

(F . n )ds=0, then F is said to be a Solenoidal vector point function.

3.1 VOLUME INTEGRAL:

Let F be a vector point function and volume V enclosed by a closed surface.

The volume integral ¿∭V

Fdv

Example 1: Evaluate ∬S

( yz i+zx j+xy k ) . ds where S the surface of the sphere

is x2+ y2+z2=a2 in the first octant.

Solution: Here, ∅=x2+ y2+ z2−a2

Vector normal to the surface ¿∇∅

¿ i ∂∅∂x + j ∂∅∂ y

+k ∂∅∂z

¿( i ∂∂x + j ∂∂ y

+k ∂∂ z ) (x2+ y2+ z2−a2 )



¿2 x i+2 y j+2 z k

n= ∇∅|∇∅|=

2x i+2 y j+2 z k√4 x2+4 y2+4 z2

¿ x i+ y j+ z k√ x2+ y2+z2

¿ x i+ y j+ z ka

[∵ x2+ y2+ z2=a2 ]

Here, F= yz i+zx j+xy k

F . n=( yz i+zx j+xy k ) .( x i+ y j+z ka )=3 xyza

Now, ∬SF . n ds=∬

S

( F .n ) dx dy|k . n|

¿∫0a

∫0

√a2−x2

3xyz dx dy

a( za )

¿3∫0

a

∫0

√a2−x2

xy dy dx

¿3∫0

a

x ( y2

2 )0

√a2− x2

dx

¿ 32∫0

a

x (a2−x2)dx

¿ 32 ( a2 x2

2− x4

4 )0

a

¿ 32 ( a4

2−a4

4 ) ¿ 3a4

8 ________ Ans.



Example 2: IfF=2 z i−x j+ y k, evaluate ∭V

F dv where, v is the region bounded by

the surfacesx=0 , y=0 , x=2 , y=4 , z=x2 , z=2.

Solution: ∭V

F dv=∭ (2 z i−x j+ y k )dx dydz

¿∫0

2

dx∫0

4

dy∫x2

2

(2 z i−x j+ y k ) dz

¿∫0

2

dx∫0

4

dy [ z2 i−xz j+ yz k ]x2

2

¿∫0

2

dx∫0

4

dy [ 4 i−2 x j+2 y k−x4 i+x3 j−x2 y k ]

¿∫0

2

dx [4 y i−2 xy j+ y2 k−x4 y i+x3 y j− x2 y2

2 k ]0

4

¿∫0

2

(16 i−8 x j+16 k−4 x4 i+4 x3 j−8x2 k )dx

¿ [16 x i−4 x2 j+16 x k−4 x5

5 i+x4 j−8 x3

3 k ]0

2

¿32 i−16 j+32 k−1285

i+16 j−643k

¿ 32 i5

+ 32 k3

¿ 3215

(3 i+5 k ) _________ Ans.

3.2 Exercise:

1) Evaluate∬S

(F . n ) ds, where, F=18 z i−12 j+3 y k and S is the surface of the

plane 2 x+3 y+6 z=12 in the first octant.

2) IfF=(2 x2−3 z) i−2 xy j−4 x k, then evaluate∭V∇ F dv, where V is bounded

by the plane x=0 , y=0 , z=0 and2 x+2 y+z=4.



4.1 GREEN’S THEOREM: (Without proof)

If ∅ (x , y ) ,Ψ ( x , y ) ,

∂ϕ∂ y

∧∂Ψ

∂ x be continuous functions over a region R bounded

by simple closed curve C in x− y plane, then

∮C

(ϕdx+Ψ dy )=∬R

( ∂Ψ∂ x − ∂ϕ∂ y )dxdy

Note: Green’s theorem in vector form

∫cF .dr=∬

R(∇×F ) . k dR

Where, F=∅ i+Ψ j , r=x i+ y j , k is a unit vector along z-axis and dR=dx dy.

Example 1: Using green’s theorem, evaluate∫c(x2 y dx+x2dy ), where c is the

boundary described counter clockwise of the triangle with vertices(0,0 ) , (1,0 ) ,(1,1).Solution: By green’s theorem, we have

∮C

(ϕdx+Ψ dy )=∬R

( ∂Ψ∂ x − ∂ϕ∂ y )dxdy

∫c

(x2 y dx+x2dy )=∬R

(2x−x2 )dx dy

¿∫0

1

(2 x−x2 )dx∫0

x

dy

¿∫0

1

(2 x−x2 )dx [ y ]0x¿¿

¿∫0

1

(2 x2−x3)dx

¿( 2 x3

3− x4

4 )0

1

¿( 23−1

4 ) ¿ 5

12 _______ Ans.



Example 2: Use green’s theorem to evaluate

∫c

(x2+xy )dx+(x2+ y2)dy, where c is the square formed by the lines

y=±1 , x=±1.

Solution: By green’s theorem, we have

∮C

(ϕdx+Ψ dy )=∬R

( ∂Ψ∂ x −∂ϕ∂ y )dxdy

¿∫−1

1

∫−1

1

[ ∂∂x

(x2+ y2 )− ∂∂ y

( x2+xy )]dxdy¿∫

−1

1

∫−1

1

(2x−x )dxdy

¿∫−1

1

∫−1

1

x dxdy

¿∫−1

1

x dx∫−1

1

dy

¿∫−1

1

x dx ( y)−1

1¿ ¿

¿∫−1

1

x dx (1+1)

¿∫−1

1

2 xdx

¿(x2)−1

1¿¿

¿1−1

¿0 ________ Ans.



4.2 Exercise:

1) Apply Green’s theorem to evaluate

∫C

[ (2x2− y2 )dx+(x2+ y2)dy ], where C is the boundary of the area enclosed by

the x-axis and the upper half of circlex2+ y2=a2.

2) A vector field F is given byF=sin y i+x (1+cos y) j.

Evaluate the line integral ∫CF .dr where C is the circular path given by

x2+ y2=a2.

5.1 STOKE’S THEOREM: (Relation between Line integral and Surface integral) (Without Proof)

Surface integral of the component of curl F along the normal to the surfaceS, taken over the surface S bounded by curve C is equal to the line integral of the vector point function F taken along the closed curveC.

Mathematically

∮F .dr=∬Scurl F .n ds

Where n=cos∝ i+cos β j+cos γ k

is a unit external normal to any surface ds.

OR

The circulation of vector F around a closed curve C is equal to the flux of the curve of the vector through the surface S bounded by the curveC.

∮F .dr=∬ScurlF .n ds=∬

Scurl F . d S



Example 1: Apply Stoke’s theorem to find the value of

∫c( y dx+ zdy+x dz)

Where c is the curve of intersection of x2+ y2+z2=a2 andx+z=a.

Solution: ∫c( y dx+ zdy+x dz)

¿∫c

( y i+z j+x k ) .(i dx+ j dy+ k dz)

¿∫c

( y i+z j+x k ) . d r

¿∬Scurl ( y i+z j+x k ). n ds (By Stoke’s theorem)

¿∬S

(i ∂∂ x + j ∂∂ y

+ k ∂∂ z )× ( y i+z j+x k ) . n ds

¿∬S– (i+ j+ k) . n ds _______ (i)

Where S is the circle formed by the integration of x2+ y2+z2=a2 and

x+z=a.

n=∇∅

|∇∅|

¿(i ∂∂ x

+ j ∂∂ y

+ k ∂∂ z )(x+z−a)

|∇∅|

¿ i+ k√1+1



¿ i√2

+ k√2

Putting the value of n in (i), we have

¿∬S– (i+ j+ k) .( i

√2+ k

√2 )ds

¿∬S

−¿( 1√2

+ 1√2 )ds ¿ [Use r2=R2−p2=a2−

a2

2=a2

2 ] ¿− 2

√2∬Sds=−2

√2π ( a

√2 )2

=−π a2

√2 ______ Ans.

Example 2: Evaluate ∮CF .dr by stoke’s theorem, where

F= y2 i+x2 j−(x+z ) k and C is the boundary of triangle with vertices at (0,0,0 ) ,(1,0,0) and (1,1,0).

Solution: We have, curl F=∇× F

¿| i j k∂∂ x

∂∂ y

∂∂z

y2 x2 −(x+z )|

¿0. i+ j+2(x− y ) k

We observe that z co-ordinate of each vertex of the triangle is zero.

Therefore, the triangle lies in the xy-plane.

∴ n=k

∴ curlF .n=[ j+2 (x− y) k ] . k=2 ( x− y ) .

In the figure, only xy-plane is considered.

The equation of the line OB is y=x

By Stoke’s theorem, we have



∮CF .dr=∬

S(curl F . n)ds

¿ ∫x=0

1

∫y=0

x

2 ( x− y )dxdy

¿2∫0

1 [ x2− x2

2 ]dx ¿2∫

0

1 x2

2dx

¿∫0

1

x2dx

¿ [ x3

3 ]0

1

¿ 13 ________ Ans.

5.2 Exercise:

1) Use the Stoke’s theorem to evaluate ∫C

[ ( x+2 y ) dx+( x−z )dy+( y−z )dz ]

where C is the boundary of the triangle with vertices (2,0,0 ) , (0,3,0 )∧(0,0,6) oriented in the anti-clockwise direction.

2) Apply Stoke’s theorem to calculate ∫c

4 y dx+2 z dy+6 y dz

Where c is the curve of intersection of x2+ y2+z2=6 z and z=x+3

3) Use the Stoke’s theorem to evaluate∫Cy2dx+xy dy+ xzdz, where C is

the bounding curve of the hemispherex2+ y2+z2=1 , z≥0, oriented in the positive direction.

6.1 GAUSS’S THEOREM OF DIVERGENCE: (Without Proof)



The surface integral of the normal component of a vector function F taken around a closed surface S is equal to the integral of the divergence of F taken over the volume V enclosed by the surfaceS.Mathematically

∬SF . n ds=∭

V¿ F dv

Example 1: Evaluate ∬SF . n ds where F=4 xz i− y2 j+ yz k and S is the surface

of the cube bounded byx=0 , x=1 , y=0 , y=1 , z=0 , z=1.Solution: By Gauss’s divergence theorem,

∬SF . n ds=∭

V

(∇ . F )dv

¿∭v

(i ∂∂ x

+ j ∂∂ y

+ k ∂∂ z ) . (4 xz i− y2 j+ yz k )dv

¿∭v

[ ∂∂ x

(4 xz )+ ∂∂ y

(− y2)+ ∂∂z

( yz )]dxdy dz ¿∭

v(4 z−2 y+ y )dx dy dz

¿∭v

(4 z− y )dxdy dz

¿∫0

1

∫0

1

( 4 z2

2− yz )

0

1¿ dx dy¿

¿∫0

1

∫0

1

(2 z2− yz )0

1¿ dxdy ¿

¿∫0

1

∫0

1

(2− y )dx dy

¿∫0

1

(2 y− y2

2 )0

1

dx

¿ 32∫0

1

dx

¿ 32

[ x ]01¿¿

¿ 32(1)



¿ 32 ________ Ans.

Example 2: Evaluate surface integral ∬ F . n ds , where F=(x2+ y2+ z2 ) ( i+ j+ k ) , S is the surface of the tetrahedron x=0 , y=0 , z=0 , x+ y+z=2 and n is the unit normal in the outward direction to the closed surfaceS.

Solution: By gauss’s divergence theorem,

∬SF . n ds=∭

V¿ F . dv

Where S is the surface of tetrahedron x=0 , y=0 , z=0 , x+ y+z=2

¿∭V

(i ∂∂ x

+ j ∂∂ y

+ k ∂∂ z ) . (x2+ y2+z2 ) ( i+ j+k )dv

¿∭V

(2 x+2 y+2 z)dv

¿2∭V

(x+ y+z )dx dydz

¿2∫0

2

dx∫0

2− x

dy ∫0

2− x− y

( x+ y+z )dz

¿2∫0

2

dx∫0

2− x

dy (xz+ yz+ z2

2 )0

2− x− y¿ ¿

¿2∫0

2

dx∫0

2− x

dy [2x−x2−xy+2 y−xy− y2+ (2−x− y )2

2 ] ¿2∫

0

2

dx [2xy−x2 y−x y2+ y2− y3

3−

(2−x− y)3

6 ]0

2− x¿

¿

¿2∫0

2

dx [2x (2−x )−x2 (2−x )−x (2−x )2+(2−x )2−(2−x )3

3+(2−x)3

6 ] ¿2∫

0

2 [4 x−2x2−2 x2+x3−4 x+4 x2− x3+(2−x )2−(2−x )3

3+(2−x)3

6 ]



¿2[2x2− 4 x3

3+ x4

4−2x2+ 4 x3

3− x4

4−

(2−x)3

3+(2−x )4

12−

(2−x)4

24 ]0

2

¿2[−(2−x)3

3+(2−x)4

12−

(2−x )4

24 ]0

2

¿2[ 83−16

12+ 16

24 ]¿4 ________ Ans.

6.2 Exercise:

1) Evaluate ∬SF . n ds where S is the surface of the sphere x2+ y2+z2=16

andF=3 x i+4 y j+5 z k.

2) Find∬SF . n ds, where F=(2 x+3 z ) i−( xz+ y ) j+( y2+2 z) k and S is the

surface of the sphere having centre (3,-1, 2) and radius 3.

3) Use divergence theorem to evaluate∬SA . ds, where A=x3 i+ y3 j+ z3 k

and S is the surface of the spherex2+ y2+z2=a2.

4) Use divergence theorem to show that∬S∇ (x2+ y2+z2 ). ds=6V , where S

is any closed surface enclosing volumeV .

7.1 REFERECE BOOKS:

1) Introduction to Engineering MathematicsBy H. K. DASS. & Dr. RAMA VERMA S. CHAND

2) Higher Engineering MathematicsBy B.V. RAMANAMc Graw Hill Education

3) Higher Engineering MathematicsBy Dr. B.S. GREWAL



KHANNA PUBLISHERS4) http://mecmath.net/calc3book.pdf


http://mecmath.net/calc3book.pdf