VECTOR INTEGRATION AND ORTHOGONAL CURVILINEAR COORDINATES

173

4.3 INTEGRAL THEOREM

4.3.1 Greens Theorem in a Plane
This theorem gives the relation between the plane, surface and the line integrals. Statement. If R is a closed region in the xy-plane bounded by a simple closed curve C and M (x, y) and N (x, y) are continuous functions having the partial derivatives in R then

z
C

M dx + N dy =

zz FGH
R

N M dx dy. y x

IJ K

4.3.2 Surface integral and Volume integral

Surface Integral An integral evaluated over a surface is called a surface integral. Consider a surface S and a point P on it. Let A be a vector function of x, y, z defined and continuous over S. ^ In n is the unit outward normal to the surface S and P then the integral of the normal component of A ^ at P (i.e., A n ) over the surface S is called the surface integral written as
Z n s k
^ ^

ds

zz
S

 ds An
R X dx dy

where ds is the small element area. To evaluate integral we have to find the double integral over the orthogonal projection of the surface on one of the coordinate planes.

Fig. 4.1

^ Suppose R is the orthogonal projection of S on the XOY plane and n is the unit outward normal ^ ^ ^ to S then it should be noted that n k ds (k being the unit vector along z-axis) is the projection of ^ the vectorial area element n ds on the XOY plane and this projection is equal to dx dy which being ^ ^ the area element in the XOY plane. That is to say that n k ds = dx dy. Similarly, we can argue ^ ^ ^ ^ to state that n j ds = dz dx and n i ds = dy dz. All these three results hold good if we write ^ n ds = dy dz i + dz dx j + dx dy k.

Sometimes we also write

 ds = dy dz i ds = n
Volume Integral If V is the volume bounded by a surface and if F (x, y, z) is a single valued function defined over V then the volume integral of F (x, y, z) over V is given by

zzz
V

F dv . If the volume is divided

into sub-elements having sides dx, dy, dz then the volume integral is given by the triple integral

zzz

F x , y , z dx dy dz which can be evaluated by choosing appropriate limits for x, y, z.

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ENGINEERING MATHEMATICSII

4.3.3 Stokes Theorem

Statement. If S is a surface bounded by a simple closed curve C and if F is any continuously differentiable vector function then

z
C

F dr =

zz
S
V

 ds = Curl F n

z z FH
S

 ds F n

IK

4.3.4 Gauss Divergence Theorem

Statement. If V is the volume bounded by a surface S and F is a continuously differentiable vector function then

zzz

div F dV =

zz
S

 dS F n

^ where n is the positive unit vector outward drawn normal to S.

WORKED OUT EXAMPLES

1. Verify Greens theorem in the plane for

zd
C

3x 2 8y 2 dx + 4y 6xy dy where C is the

boundary for the region enclosed by the parabola y2 = x and x2 = y. Solution. We shall find the points of intersection of the parabolas y 2 = x and x2 = y i.e., Equating both, we get y =
x and y = x2
(0,1) O (0,0) (1,0) y=x
2

Y y=x
2

x or x x4 x (1 x3) x and hence y = 0, 1 the points of Let M

= x2 x = x4 = 0 = 0 = 0, 1 intersection are (0, 0) and (1, 1). = 3x2 8y2, N = 4y 6xy = 16y
N = 6y x

A(1,1) x

M y
By Greens theorem,

Fig. 4.2

z
C

Mdx + Ndy

L.H.S = =

OA

zz FGH z zb
R C

N M dx dy y x

IJ K

Mdx + Ndy M dx + N dy +

g b M dx + N dyg = I
AO

+ I2

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175

Along OA:

y = x2 dy = 2xdx, x varies from 0 to 1 I1 =

x =0 1

= = Along AO:

x=0

zd zd
1 0

3x 2 8 x 4 dx + 4 x 2 6x 3 2 x dx

3x 2 + 8 x 3 20 x 4 dx
1 0

x 3 + 2x 4 4x5

= 1

y 2 = x dx = 2y dy, y varies from 1 to 0 I2 =

zd zd
y =1 0 1

3 y 4 8 y 2 2 y dy + 4 y 6 y 3 dy

4 y 22 y 3 + 6 y 5 dy 11 4 y + y6 2

= Hence, Also

LM2 y N

OP Q

=
1

5 2

5 3 L.H.S. = I1 + I2 = 1 + = 2 2
R.H.S. =

= 5

zz FGH IJK z zb z z z LMN OPQ zd i

R 1 x x = 0 y = x2 1 x x = 0 y = x2 1 x =0 1

N M dx dy y x 6 y + 16 y dy dx

10 y dy dx
x

10 y 2 2

dx

y = x2

x x 4 dx
5 1

x=0

= =

Lx x O 5M P N2 5Q L 1 1O 3 5M P = N 2 5Q 2
2

x=0

L.H.S. = R.H.S. =

3 . 2

Hence verified.

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ENGINEERING MATHEMATICSII

2. Verify Greens theorem in the plane for of the region bounded by y = x and y = x2.

z {d
C

xy + y 2 dx + x 2 dy where C is the closed curve

Solution. We shall find the points of intersection of y = x and y = x2. Equating the R.H.S. x = x2 x x2 = 0 x (1 x) = 0 x = 0, 1 y = 0, 1 and hence (0, 0), (1, 1) are the points of intersection. We have Greens theorem in a plane,
(0, 1)

(1, 1)
2

x
y=x

o (0, 0)

x (1, 0)

z
C

M dx + N dy

The line integral,

zd
C

xy + y 2 dx + x 2 dy

OA

zz FGH z {d
R x =0 1

N M dx dy x y

IJ K

Fig. 4.3

xy + y 2 dx + x 2 dy +

AO

z {d

xy + y 2 dx + x 2 dy

Along OA, we have y =

x2,

= I1 + I2 dy = 2x dx and x varies from 0 to 1.

I1 =

x =0

zd zd
1

x x 2 + x 4 dx + x 2 2 x dx

3x 3 + x 4 dx

LM 3x N4

x5 + 5

OP Q

=
0

3 1 19 + = 4 5 20

Along AO, we have y = x dy = dx x varies from 1 to 0 I2 =

zd z
0 1 0 1 R

x x + x 2 dx + x 2 dx
0 1

3x 2 dx = x 3

= 1

Hence, Also

L.H.S. = I1 + I2

19 1 1= = 20 20

R.H.S. =

zz FGH

N M dx dy y x

IJ K

VECTOR INTEGRATION AND ORTHOGONAL CURVILINEAR COORDINATES

177

where

N = x2

= xy + y2

N = 2x x R is the region bounded by y = x2 and y = x

M y = x + 2y

zz
R

FG N M IJ dx dy H x y K

x = 0 y = x2 1 x

x = 0 y = x2 1

x =0 1

x =0 1

x=0

z zb z zb z zd zd
1 x 5

2 x x 2 y dy dx

x 2 y dy dx
x y = x2

xy y 2

dx

x 2 x 2 x 3 x 4 dx

i d

x 4 x 3 dx
4 1

LM x x OP N5 4Q

=
0

1 1 1 = 5 4 20
Hence verified.

L.H.S. = R.H.S. =

1 . 20

3. Apply Greens theorem in the plane to evaluate

zd
C

2x 2 y 2 dx + x 2 + y 2 dy where C is

the curve enclosed by the x-axis and the semicircle x2 + y2 = 1. Solution. The region of integration is bounded by AB and the semicircle as shown in the figure. By Greens theorem,

z
C

Mdx + Ndy =

zz FGH
R

N M dx dy x y

IJ K

...(1)

Given where

zd
C

2 x 2 y 2 dx + x 2 + y 2 dy
N = x2 + y2

M = 2x2 y2,

M y

= 2y

N x

Fig. 4.4

= 2x

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ENGINEERING MATHEMATICSII

From the equation (1),

zd
C 1

2 x 2 y 2 dx + x 2 + y 2 dy =

zz b
R

2 x + 2 y dx dy

In the region, x varies from 1 to 1 and y varies from 0 to = 2

1 x2

x = 1 1

= 2

x = 1 1

= 2

x = 1

z zb LM zN z LMN
y=0

1 x2

x + y dy dx

y2 xy + 2

OP Q

1 x2

dx
y=0

x 1 x2 +

1 1 x 2 2

iOPQ dx

Since, x 1 x 2 is odd and (1 x2) is even function = 0+2

zd
1 0

1 x 2 dx

L xO 2 Mx P N 3Q
4 . 3

3 1

= 4. Evaluate

zd
C

xy x 2 dx + x 2 y dy where C is the closed curve formed by y = 0, x = 1 and

y = x (i) directly as a line integral (ii) by employing Greens theorem. Solution (i) Let M = xy x2, N = x2y

Fig. 4.5

z
C

Mdx + Ndy =

OA

zb

Mdx + Ndy +

AB

zb

Mdx + Ndy +

BO

zb

Mdx + Ndy

VECTOR INTEGRATION AND ORTHOGONAL CURVILINEAR COORDINATES

179

(a) Along OA: (b) Along AB: (c) Along BO:

y = 0 x = 1 y = x =

dy = 0 and x varies from 0 to 1. dx = 0 and y varies from 0 to 1. dy = dx and x varies from 1 to 0.

z
C

b M dx + N dyg

x =0 3 1

z
1

x 2 dx +

y=0 2 1 4

z
1

y dy +

x =1 0

z
0

x 3 dx

LM x OP + LM y OP + LM x OP N3Q N2Q N4Q

0 0

1 1 1 1 = + = 3 2 4 12
Thus (ii) We have Greens theorem,

zb
C

Mdx + Ndy

R.H.S. =

zd i zz FGH zz b g z zb z z
C

xy x 2 dx + x 2 ydy =

1 12

N M dx dy x y

IJ K

2 xy x dx dy
x

2 xy x dy dx (from the figure)

x y =0

x=0 y=0 1

xy 2 xy

x =0 1

x 3 x 2 dx x3 3

x =0

LM x N4

OP Q

= R.H.S. =

1 1 1 = 4 3 12

1 . 12

5. Verify Stokes theorem for the vector F = (x2 + y2) i 2xyj taken round the rectangle bounded by x = 0, x = a, y = 0, y = b. Solution By Stokes theorem :

z
C

F dr =

zz FH
S

F n dS

IK

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ENGINEERING MATHEMATICSII

F dr F dr

= (x2 + y2) i 2xyj = dxi + dyj = (x2 + y2) dx 2xy dy

y=b R Q Y

(1) Along OP: y = 0, dy = 0, x varies from 0 to a

OP

(2) Along PQ: x = a, dx = 0 ; y varies from 0 to b

b

PQ

z z

F dr

F dr

z z
a 0 0 0 a

a3 x dx = 3
2

x=0

x=a

2ay dy = ab 2

y=0

(3) Along QR: y = b, dy = 0; x varies from a to 0

Fig. 4.6
0

QR

(4) Along RO: x = 0, dx = 0 ; x varies from b to 0

RO

L.H.S. =

z z z
C

F dr

F dr

zd zb

x 2 b 2 dx =

LM x b xOP N3 Q
3 2

= ab 2
a

a3 3

0 0 dy = 0

F dr

a3 a3 + ab 2 + ab 2 +0 3 3 = 2ab2
=
i x x2 y2 j y 2 xy k = 4y k z 0

Now,

curl F

For the surface, S n = k R.H.S. =

 curl F n

= 4y

zz
S

 dS = curl F n

2 = 2b

zz z LMN z
a b 0 0 a

4 y dy dx

y2 2

OP Q

dx
0

dx

= 2ab2 L.H.S. = R.H.S. Hence, the Stokes theorem is verified.

VECTOR INTEGRATION AND ORTHOGONAL CURVILINEAR COORDINATES

181

6. Verify Stokes theorem for the vector field F = 2x y i yz 2 j y 2 zk over the upper half surface of x2 + y2 + z2 = 1 bounded by its projection on the xy-plane.

Solution C is the circle: i.e.,

z
C

F dr =

zz
S

 dS curl F n

(Stokes theorem)

x2 + y2 = 1, z = 0 (xy-plane) x = cos t, y = sin t, z = 0 r = xi + y j where 0 2

dr = dxi + dy j
where, L.H.S. =

F =

b2 x yg i yz

j y 2 zk
( z = 0)

F dr = (2x y) dx

z
C

F dr =

zb zb zd zd z RST b
C 0 0 0 0

2 x y dx

2 cos t sin t sin t dt

gb

sin 2 t 2 cos t sin t dt

sin 2 t sin 2 t dt 1 1 cos 2t sin 2t dt 2

U V W

= = Hence,

LM t sin 2t + cos 2t OP N2 4 2 Q FG 1 1 IJ + b 0g = H 2 2K

2 0

F dr =
i curl F = F = x 2x y j y yz 2 k z y2z

...(1)

Also,

= i 2 yz + 2 yz j 0 + k 0 +1

g bg b g

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= k

 dS = ndS = dydz i + dzdx j + dxdy k

R.H.S. =

Hence,

zz
S

 = curl F ndS

zz

dx dy
...(2)

zz

dx dy represents the area of the circle x2 + y2 = 1 which is .

Thus, from (1) and (2) we conclude that the theorem is verified. 7. If F = 3yi xz j + yz 2 k and S is the surface of the paraboloid 2z = x2 + y2 bounded by z = 2, show by using Stokes theorem that

zz
S

 ds = 20 . curl F n

Solution. If z = 0 then the given surface becomes x2 + y2 = 4. Hence, C is the circle x2 + y2 = 4 in the plane z = 2 i.e., x = 2 cos t, y = 2 sin t, 0 t 2 Hence by Stokes theorem, we have

z
C

F dr =

zz
S

 dS curl F n

L.H.S. put where

F = 3 yi xz j + yz 2 k , dr = dxi + dy j + dzk

z z
C C

F dr =

z = 2, dz = 0

F dr =

zd zb
C C

3 y dx xz dy + yz 2 dz

3 y dx 2 x dy

g
dx = 2 sin t dt dy = 2 cos t dt

x = 2 cos t y = 2 sin t

z
C

F dr =

z
0

6 sin t 2 sin t dt 4 cos t 2 cos t dt

Since, the surface S lies below the curve C =

2 2

zd
0

12 sin 2 t + 8 cos2 t dt

zd
0

12 sin 2 t + 8 cos 2 t dt

VECTOR INTEGRATION AND ORTHOGONAL CURVILINEAR COORDINATES

2 2

183

= = =

48

z
0

sin 2 t dt + 32 cos2 t dt
0

zz
S

48 + 32 = 20 4 4
 dS = 20 curl F n

Hence proved. 8. Using divergence theorem, evaluate

zz
S

 where F = 4xzi y 2 j + yzk and S is the F ndS

surface of the cube bounded by x = 0, x = 1, y = 0, y = 1, z = 0, z = 1. Solution. We have divergence theorem:

zzz
V

div F dV =

zz
S

 dS F n

Now

div F = F = =

FGi + j + k IJ d4 xzi y H x y z K 4 xz g + y i + b b yzg d x y z

2

j + yzk

= 4 z 2y + y = 4z y Hence, by divergence theorem, we have

zz
S

 dS = F n

zzz z z zb zz z zb g z LMN OPQ

div F dV
1 1 V 1 x =0 y=0z =0 1 1 x=0 y=0 1 1 x=0 y=0 1 x =0

4 z y dz dy dx

2 z 2 yz

1 z=0

dy dx

2 y dy dx
1

y2 2y 2

dx
0

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ENGINEERING MATHEMATICSII

x =0 1

x =0

z LMN z
1

1 dx 2

OP Q

3 dx 2
1 0

3 x 2

3 2

9. Using divergence theorem, evaluate

zz
S

 dS where F = x 2 i + y 2 j + z 2 k and S is the F n

surface of the solid cut off by the plane x + y + z = a from the first octant.

2 2 2 x + y + z x y z = 2x + 2y + 2z = 2 (x + y + z) Hence, by divergence theorem, we have

Solution. Now

div F = F =

d i

d i

d i

zz
S

 dS = F n
=

= 2

= 2

= 2

1 = 3x

zzz zzz b g z z zb g z z LMNb g OPQ zz b g LM b g OP z MN PQ i zd

div F dV
V

2 x + y + z dV
a x a x y

x + y + z dz dy dx
a x y

x=0 y=0 a a x

z=0

x+ y z+

x=0 y =0 a a x

1 2 z 2

dy dx
z=0

x=0 y =0

1 2 a x+y 2 x+ y 3
3

dy dx

ax

a y

dx

x =0

y=0

2a 3 3a 2 x + x 3 dx

=0

1 x2 x4 2a 3 x 3a 2 + = 3 2 4

LM N

OP Q

zz
S

 dS = 1 a 4 . Fn 4

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185

10. Using divergence theorem, evaluate surface of the sphere x2 + y2 + z2 = a2. Solution.

zz
S

 dS where F = x3i + y3j + z3k and S is the F n

div F =

3 x3 + y3 + z x y z = 3x2 + 3y2 + 3z2 = 3 (x2 + y2 + z2)

d i

d i

d i

by divergence theorem, we get

zz
S

d S = F n

zzz zzz
V V

div F dV 3 x 2 + y 2 + z 2 dx dy dz

...(1)

Since, V is the volume of the sphere we transform the above triple integral into spherical polar coordinates (r, , ). For the spherical polar coordinates (r, , ), we have x2 + y2 + z2 = r2 and dx dy dz = dV dV = r2 sin dr d d Also, 0 r a, 0 and 0 2 Therefore (1) reduces to,

zz
S

 dS = 3 F n

r =0 =0 =0 a

= 3

r=0

z z zd i z z
a 2

r 2 r 2 sin dr d d
2

r 4 dr
5 a

sin d

=0

=0 =0 2 =0

= = 11. Evaluate in the first octant.

L r O cos 3 M P N5Q 3a b cos + 1g 2 5

r=0 5

12 a 5 5

zz b
S

 dS where S is the surface of the sphere x2 + y2 + z2 = a2 yzi + zxj + xyk n

Solution. The given surface is x2 + y2 + z2 = a2, we know that is a vector normal to the surface (x, y, z) = c. Taking (x, y, z) = x2 + y2 + z2

VECTOR INTEGRATION AND ORTHOGONAL CURVILINEAR COORDINATES

187

3 2

z
a 0

x a 2 x 2 dx
a

3 2

LM a x x OP N 2 4Q a O 3 La P M 2 N2 4Q
2 2 4 4 4

3 a4 8 3 a4 . 8

Thus

zz
S

 b yzi + zxj + xyk g ndS

12. Evaluate

zz b
S

 d S where S is the surface of the sphere x2 + y2 + z2 = 1. axi + byj + czk n F = axi + byj + czk

Solution. Let we have

zz
S

 dS = F n

zzz
V

div F dV

div F = F =

FGi + j + k IJ baxi + byj + czk g H x y z K baxg + y bbyg + z bczg x

= (a + b + c)

zz
S

 dS = F n

zzz b
V

a + b + c dV
...(1)

= (a + b + c) V where V is the volume of the sphere with unit radius and V = Here, since we have r = 1, Thus, V =

4 3 r for a sphere of radius r. 3

4 3 4 a +b+ c . 3

zz
S

 dS = F n