Btech_II_ engineering mathematics_unit5

Unit-5 VECTOR INTEGRATION
RAI UNIVERSITY, AHMEDABAD 1
Course: B.Tech- II
Subject: Engineering Mathematics II
Unit-5
RAI UNIVERSITY, AHMEDABAD

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

Vector integration
1.1 LINE INTEGRAL:
Line integral = ∫ (𝐹̅.
𝑑𝑟⃗⃗⃗⃗⃗
𝑑𝑠
)𝑐
𝑑𝑠 = ∫ 𝐹̅. 𝑑𝑟̅̅̅
𝑐
Note:
1) Work: If 𝐹̅ represents the variable force acting on a particle along arc AB,
then the total work done = ∫ 𝐹̅. 𝑑𝑟̅̅̅𝐵
𝐴
2) Circulation: If 𝑉̅ represents the velocity of a liquid then ∮ 𝑉̅. 𝑑𝑟̅̅̅
𝑐
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 ∫ 𝐹̅. 𝑑𝑟̅̅̅𝐵
𝐴
is to be proved to be independent of path, then 𝐹̅ = ∇∅
here 𝐹 is called Conservative (irrotational) vector field and ∅ is called the
Scalarpotential. And ∇ × 𝐹̅ = ∇ × ∇∅ = 0
Example 1: Evaluate ∫ 𝐹̅. 𝑑𝑟̅̅̅
𝑐
where 𝐹̅ = 𝑥2
𝑖̂ + 𝑥𝑦𝑗̂ and 𝐶 is the boundary of the
square in the plane 𝑧 = 0 and bounded by the lines 𝑥 = 0, 𝑦 = 0, 𝑥 = 𝑎 𝑎𝑛𝑑
𝑦 = 𝑎.
Solution: ∫ 𝐹̅. 𝑑𝑟̅̅̅
𝑐
= ∫ 𝐹̅. 𝑑𝑟̅̅̅
𝑂𝐴
+ ∫ 𝐹̅. 𝑑𝑟̅̅̅
𝐴𝐵
+ ∫ 𝐹̅. 𝑑𝑟̅̅̅
𝐵𝐶
+ ∫ 𝐹̅. 𝑑𝑟̅̅̅
𝐶𝑂
Here 𝑟̅ = 𝑥𝑖̂ + 𝑦𝑗̂, 𝑑𝑟̅̅̅ = 𝑑𝑥𝑖̂ + 𝑑𝑦𝑗̂, 𝐹̅ = 𝑥2
𝑖̂ + 𝑥𝑦𝑗̂
𝐹̅. 𝑑𝑟̅̅̅ = 𝑥2
𝑑𝑥 + 𝑥𝑦𝑑𝑦 _______(i)
 On 𝑂𝐴, 𝑦 = 0

∴ 𝐹̅. 𝑑𝑟̅̅̅ = 𝑥2
𝑑𝑥 (From (i))
∫ 𝐹̅. 𝑑𝑟̅̅̅
𝑂𝐴
= ∫ 𝑥2
𝑑𝑥 = [
𝑥3
3
]
0
3
=
𝑎3
3
𝑎
0
_______ (ii)
 On 𝐴𝐵, 𝑥 = 𝑎
∴ 𝑑𝑥 = 0
∴ 𝐹̅. 𝑑𝑟̅̅̅ = 𝑎𝑦 𝑑𝑦 (From (i))
∫ 𝐹̅. 𝑑𝑟̅̅̅
𝐴𝐵
= ∫ 𝑎𝑦 𝑑𝑦 = 𝑎 [
𝑦2
2
]
0
𝑎
=
𝑎3
2
𝑎
0
_______(iii)
 On 𝐵𝐶, 𝑦 = 𝑎
∴ 𝑑𝑦 = 0
∴ 𝐹̅. 𝑑𝑟̅̅̅ = 𝑥2
𝑑𝑥 (From (i))
∫ 𝐹̅. 𝑑𝑟̅̅̅
𝐵𝐶
= ∫ 𝑥2
𝑑𝑥 = [
𝑥3
3
]
𝑎
0
= −
𝑎3
3
0
𝑎
_______(iv)
 On 𝐶𝑂, 𝑥 = 0
∴ 𝐹̅. 𝑑𝑟̅̅̅ = 0 (From (i))
∫ 𝐹̅. 𝑑𝑟̅̅̅
𝐶𝑂
= 0 _______(v)
On adding (ii), (iii), (iv) and (v), we get
∫ 𝐹̅. 𝑑𝑟̅̅̅
𝐶
=
𝑎3
3
+
𝑎3
2
−
𝑎3
3
+ 0 =
𝑎3
2
________ Ans.
Example 2: A vector field is given by
𝐹̅ = (2𝑦 + 3) 𝑖̂ + ( 𝑥𝑧) 𝑗̂ + (𝑦𝑧− 𝑥)𝑘̂. Evaluate ∫ 𝐹̅. 𝑑𝑟̅̅̅
𝐶
along the path 𝑐 is 𝑥 =
2𝑡, 𝑦 = 𝑡, 𝑧 = 𝑡3
𝑓𝑟𝑜𝑚 𝑡 = 0 𝑡𝑜 𝑡 = 1.
Solution:
∫ 𝐹̅. 𝑑𝑟̅̅̅
𝐶
= ∫ (2𝑦 + 3) 𝑑𝑥 + ( 𝑥𝑧) 𝑑𝑦 + (𝑦𝑧− 𝑥)𝑑𝑧𝐶
[
𝑠𝑖𝑛𝑐𝑒 𝑥 = 2𝑡 𝑦 = 𝑡 𝑧 = 𝑡3
∴
𝑑𝑥
𝑑𝑡
= 2
𝑑𝑦
𝑑𝑡
= 1
𝑑𝑧
𝑑𝑡
= 3𝑡2 ]

= ∫ (2𝑡 + 3)(2 𝑑𝑡) + (2𝑡)( 𝑡3) 𝑑𝑡 + ( 𝑡4
− 2𝑡)(3𝑡2
𝑑𝑡)
1
0
= ∫ (4𝑡 + 6 + 2𝑡4
+ 3𝑡6
− 6𝑡3) 𝑑𝑡
1
0
= [4
𝑡2
2
+ 6𝑡 +
2
5
𝑡5
+
3
7
𝑡7
−
6
4
𝑡4
]
0
1
= [2𝑡2
+ 6𝑡 +
2
5
𝑡5
+
3
7
𝑡7
−
3
2
𝑡4
]
0
1
= 2 + 6 +
2
5
+
3
7
−
3
2
= 7.32857 _________Ans.
Example 3: Suppose 𝐹( 𝑥, 𝑦, 𝑧) = 𝑥3
𝑖̂ + 𝑦𝑗̂ + 𝑧𝑘̂ 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= ∫ 𝐹̅. 𝑑𝑟̅̅̅
𝑐
= ∫ (𝑥3
𝑖̂ + 𝑦𝑗̂ + 𝑧𝑘̂). 𝑑(𝑖̂ 𝑑𝑥 + 𝑗̂ 𝑑𝑦 + 𝑘̂ 𝑑𝑧)
(3,5,7)
(1,2,3)
= ∫ ( 𝑥3
𝑑𝑥 + 𝑦𝑑𝑦 + 𝑧𝑑𝑧)
(3,5,7)
(1,2,3)
= ∫ 𝑥3
𝑑𝑥 + ∫ 𝑦 𝑑𝑦 + ∫ 𝑧 𝑑𝑧
7
3
5
2
3
1
= [
𝑥4
4
]
1
3
+ [
𝑦2
2
]
2
5
+ [
𝑧2
2
]
3
7
= [
81
4
−
1
4
] + [
25
2
−
4
2
] + [
49
2
−
9
2
]
=
80
4
+
21
2
+
40
2
=
202
4
= 50.5 units _______Ans.

1.2Exercise:
1) If a force 𝐹̅ = 2𝑥2
𝑦𝑖̂ + 3𝑥𝑦𝑗̂ displaces a particle in the 𝑥𝑦-plane from
(0, 0) to (1, 4) along a curve 𝑦 = 4𝑥2
. Find the work done.
2) If 𝐴 = (3𝑥2
+ 6𝑦) 𝑖̂ − 14𝑦𝑧𝑗̂ + 20𝑥𝑧2
𝑘̂, evaluate the line integral
∮ 𝐴 𝑑𝑟⃗⃗⃗⃗ from (0, 0, 0) to (1, 1, 1) along the curve 𝐶.
3) Show that the integral ∫ ( 𝑥𝑦2
+ 𝑦3) 𝑑𝑥 + (𝑥2
𝑦 + 3𝑥𝑦2
)𝑑𝑦
(3,4)
(1,2)
is
independent of the path joining the points (1, 2) and (3, 4). Hence,
evaluate the integral.
2.1 SURFACE INTEGRAL:
Let 𝐹̅ be a vector function and 𝑆 be the given surface.
Surface integral of a vector function 𝐹̅ over the surface 𝑆 is defined as the
integral of the components of 𝐹̅ along the normal to the surface.
Component of 𝐹̅ along the normal= 𝐹̅. 𝑛̂
Where n = unit normal vector to an element 𝑑𝑠 and
𝑛̂ =
𝑔𝑟𝑎𝑑 𝑓
| 𝑔𝑟𝑎𝑑 𝑓|
𝑑𝑠 =
𝑑𝑥 𝑑𝑦
( 𝑛̂.𝑘̂)
Surface integral of F over S
= ∑ 𝐹̅. 𝑛̂ = ∬ ( 𝐹̅. 𝑛̂) 𝑑𝑠𝑆
Note:
1) Flux = ∬ ( 𝐹̅. 𝑛̂) 𝑑𝑠𝑆
where, 𝐹̅ represents the velocity of a liquid.
If ∬ ( 𝐹̅. 𝑛̂) 𝑑𝑠𝑆
= 0, then 𝐹̅ is said to be a Solenoidal vector point function.

3.1 VOLUME INTEGRAL:
Let 𝐹̅ be a vector point function and volume 𝑉 enclosed by a closed surface.
The volume integral = ∭ 𝐹̅ 𝑑𝑣𝑉
Example 1: Evaluate ∬ (𝑦𝑧𝑖̂ + 𝑧𝑥𝑗̂ + 𝑥𝑦𝑘̂). 𝑑𝑠𝑆
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 = ∇∅
= 𝑖̂
𝜕∅
𝜕𝑥
+ 𝑗̂
𝜕∅
𝜕𝑦
+ 𝑘̂ 𝜕∅
𝜕𝑧
= (𝑖̂
𝜕
𝜕𝑥
+ 𝑗̂
𝜕
𝜕𝑦
+ 𝑘̂ 𝜕
𝜕𝑧
)( 𝑥2
+ 𝑦2
+ 𝑧2
− 𝑎2)
= 2𝑥𝑖̂ + 2𝑦𝑗̂ + 2𝑧𝑘̂
𝑛̂ =
∇∅
|∇∅|
=
2𝑥𝑖̂+2𝑦𝑗̂+2𝑧𝑘̂
√4𝑥2+4𝑦2+4𝑧2
=
𝑥𝑖̂+ 𝑦𝑗̂+ 𝑧𝑘̂
√𝑥2+𝑦2+𝑧2
=
𝑎
[∵ 𝑥2
+ 𝑦2
+ 𝑧2
= 𝑎2]
Here, 𝐹 = 𝑦𝑧𝑖̂ + 𝑧𝑥𝑗̂ + 𝑥𝑦𝑘̂
𝐹. 𝑛̂ = (𝑦𝑧𝑖̂ + 𝑧𝑥𝑗̂ + 𝑥𝑦𝑘̂).(
𝑎
) =
3𝑥𝑦𝑧
𝑎
Now, ∬ 𝐹. 𝑛̂𝑆
𝑑𝑠 = ∬ (𝐹. 𝑛̂)𝑆
𝑑𝑥 𝑑𝑦
| 𝑘̂.𝑛̂|
= ∫ ∫
3𝑥𝑦𝑧 𝑑𝑥 𝑑𝑦
𝑎 (
𝑧
𝑎
)
√𝑎2−𝑥2
0
𝑎
0

= 3∫ ∫ 𝑥𝑦 𝑑𝑦 𝑑𝑥
√𝑎2−𝑥2
0
𝑎
0
= 3∫ 𝑥 (
𝑦2
2
)
0
√𝑎2−𝑥2
𝑑𝑥
𝑎
0
=
3
2
∫ 𝑥 (𝑎2
− 𝑥2
)𝑑𝑥
𝑎
0
=
3
2
(
𝑎2 𝑥2
2
−
𝑥4
4
)
0
𝑎
=
3
2
(
𝑎4
2
−
𝑎4
4
)
=
3𝑎4
8
________Ans.
Example 2: If 𝐹̅ = 2𝑧𝑖̂ − 𝑥𝑗̂ + 𝑦𝑘̂, evaluate ∭ 𝐹̅ 𝑑𝑣𝑉
where, 𝑣 is the region
bounded by the surfaces 𝑥 = 0, 𝑦 = 0, 𝑥 = 2, 𝑦 = 4, 𝑧 = 𝑥2
, 𝑧 = 2.
Solution: ∭ 𝐹̅ 𝑑𝑣𝑉
= ∭(2𝑧𝑖̂ − 𝑥𝑗̂ + 𝑦𝑘̂) 𝑑𝑥 𝑑𝑦 𝑑𝑧
= ∫ 𝑑𝑥 ∫ 𝑑𝑦 ∫ (2𝑧𝑖̂ − 𝑥𝑗̂ + 𝑦𝑘̂) 𝑑𝑧
2
𝑥2
4
0
2
0
= ∫ 𝑑𝑥 ∫ 𝑑𝑦 [𝑧2
𝑖̂ − 𝑥𝑧𝑗̂ + 𝑦𝑧𝑘̂] 𝑥2
24
0
2
0
= ∫ 𝑑𝑥 ∫ 𝑑𝑦 [4𝑖̂ − 2𝑥𝑗̂ + 2𝑦𝑘̂ − 𝑥4
𝑖̂ + 𝑥3
𝑗̂ − 𝑥2
𝑦𝑘̂]
4
0
2
0
= ∫ 𝑑𝑥 [4𝑦𝑖̂ − 2𝑥𝑦𝑗̂ + 𝑦2
𝑘̂ − 𝑥4
𝑦𝑖̂ + 𝑥3
𝑦𝑗̂ −
𝑥2 𝑦2
2
𝑘̂]
0
42
0
= ∫ (16𝑖̂ − 8𝑥𝑗̂ + 16𝑘̂ − 4𝑥4
𝑖̂ + 4𝑥3
𝑗̂ − 8𝑥2
𝑘̂)
2
0
𝑑𝑥
= [16𝑥𝑖̂ − 4𝑥2
𝑗̂ + 16𝑥𝑘̂ −
4𝑥5
5
𝑖̂ + 𝑥4
𝑗̂ −
8𝑥3
3
𝑘̂]
0
2
= 32𝑖̂ − 16𝑗̂ + 32𝑘̂ −
128
5
𝑖̂ + 16𝑗̂ −
64
3
𝑘̂
=
32𝑖̂
5
+
32𝑘̂
3

=
32
15
(3𝑖̂ + 5𝑘̂) _________ Ans.
3.2 Exercise:
1) Evaluate ∬ ( 𝐹̅. 𝑛̂) 𝑑𝑠𝑆
, where, 𝐹 = 18𝑧𝑖̂ − 12𝑗̂ + 3𝑦𝑘̂ and 𝑆 is the surface
of the plane 2𝑥 + 3𝑦 + 6𝑧 = 12 in the first octant.
2) If 𝐹 = (2𝑥2
− 3𝑧) 𝑖̂ − 2𝑥𝑦𝑗̂ − 4𝑥𝑘̂, then evaluate ∭ ∇𝑉
𝐹 𝑑𝑣, where 𝑉 is
bounded by the plane 𝑥 = 0, 𝑦 = 0, 𝑧 = 0 and 2𝑥 + 2𝑦 + 𝑧 = 4.
4.1 GREEN’S THEOREM:(Without proof)
If ∅( 𝑥, 𝑦),Ψ( 𝑥, 𝑦),
𝜕𝜙
𝜕𝑦
𝑎𝑛𝑑
𝜕Ψ
𝜕𝑥
be continuous functions over a region R
bounded by simple closed curve 𝐶 in 𝑥 − 𝑦 plane, then
∮( 𝜙 𝑑𝑥 + Ψ 𝑑𝑦) = ∬(
𝜕Ψ
𝜕𝑥
−
𝜕𝜙
𝜕𝑦
) 𝑑𝑥 𝑑𝑦
𝑅𝐶
Note:Green’s theorem in vector form
∫ 𝐹̅. 𝑑𝑟̅̅̅ = ∬(∇ × 𝐹̅). 𝑘̂ 𝑑𝑅
𝑅𝑐
Where, 𝐹̅ = ∅𝑖̂ + Ψĵ, r̅ = 𝑥𝑖̂ + 𝑦𝑗̂, 𝑘̂ is a unit vector along 𝑧-axis and 𝑑𝑅 =
𝑑𝑥 𝑑𝑦.
Example 1: Using green’s theorem, evaluate ∫(𝑥2
𝑦 𝑑𝑥 + 𝑥2
𝑑𝑦)𝑐
, 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
∮ ( 𝜙 𝑑𝑥 + Ψ 𝑑𝑦) = ∬ (
𝜕Ψ
𝜕𝑥
−
𝜕𝜙
𝜕𝑦
) 𝑑𝑥 𝑑𝑦𝑅𝐶
∫ ( 𝑥2
𝑦 𝑑𝑥 + 𝑥2
𝑑𝑦) = ∬ (2𝑥 − 𝑥2) 𝑑𝑥 𝑑𝑦𝑅𝑐
= ∫ (2𝑥 − 𝑥2) 𝑑𝑥∫ 𝑑𝑦
𝑥
0
1
0
= ∫ (2𝑥 − 𝑥2) 𝑑𝑥 [ 𝑦]0
𝑥1
0
= ∫ (2𝑥2
− 𝑥3
)𝑑𝑥
1
0

= (
2𝑥3
3
−
𝑥4
4
)
0
1
= (
2
3
−
1
4
)
=
5
12
_______Ans.
Example 2: Use green’s theorem to evaluate
∫ ( 𝑥2
+ 𝑥𝑦) 𝑑𝑥 + (𝑥2
+ 𝑦2
)𝑑𝑦𝑐
, where c is the square formed by the lines
𝑦 = ±1, 𝑥 = ±1.
Solution: By green’s theorem, we have
∮ ( 𝜙 𝑑𝑥 + Ψ 𝑑𝑦) = ∬ (
𝜕Ψ
𝜕𝑥
−
𝜕𝜙
𝜕𝑦
) 𝑑𝑥 𝑑𝑦𝑅𝐶
= ∫ ∫ [
𝜕
𝜕𝑥
( 𝑥2
+ 𝑦2) −
𝜕
𝜕𝑦
(𝑥2
+ 𝑥𝑦)] 𝑑𝑥 𝑑𝑦
1
−1
1
−1
= ∫ ∫ (2𝑥 − 𝑥) 𝑑𝑥𝑑𝑦
1
−1
1
−1
= ∫ ∫ 𝑥 𝑑𝑥𝑑𝑦
1
−1
1
−1
= ∫ 𝑥 𝑑𝑥 ∫ 𝑑𝑦
1
−1
1
−1
= ∫ 𝑥 𝑑𝑥 (𝑦)−1
11
−1
= ∫ 𝑥 𝑑𝑥 (1 + 1)
1
−1
= ∫ 2𝑥 𝑑𝑥
1
−1
= (𝑥2
)−1
1
= 1 − 1
= 0 ________Ans.

4.2 Exercise:
1) Apply Green’s theorem to evaluate
∫ [(2𝑥2
− 𝑦2) 𝑑𝑥 + (𝑥2
+ 𝑦2
)𝑑𝑦]𝐶
, where 𝐶 is the boundary of the area
enclosed by the 𝑥-axis and the upper half of circle 𝑥2
+ 𝑦2
= 𝑎2
.
2) A vector field 𝐹̅ is given by 𝐹̅ = sin 𝑦 𝑖̂ + 𝑥 (1 + cos 𝑦) 𝑗̂.
Evaluate the line integral ∫ 𝐹̅. 𝑑𝑟̅̅̅
𝐶
where 𝐶 is the circular path given
by 𝑥2
+ 𝑦2
= 𝑎2
.
5.1 STOKE’S THEOREM:(RelationbetweenLine integral and Surface
integral) (Without Proof)
Surface integral of the component of curl 𝐹̅ along the normal to the
surface 𝑆, taken over the surface 𝑆 bounded by curve 𝐶 is equal to the line
integral of the vector point function 𝐹̅ taken along the closed curve 𝐶.
Mathematically
∮ 𝐹̅. 𝑑𝑟̅̅̅ = ∬ 𝑐𝑢𝑟𝑙𝑆
𝐹̅. 𝑛̂ 𝑑𝑠
Where 𝑛̂ = cos ∝ 𝑖̂ + cos 𝛽 𝑗̂ + cos 𝛾 𝑘̂
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 curve 𝐶.
∮ 𝐹̅. 𝑑𝑟̅̅̅ = ∬ 𝑐𝑢𝑟𝑙
𝑆
𝐹̅. 𝑛̂ 𝑑𝑠 = ∬ 𝑐𝑢𝑟𝑙
𝑆
𝐹̅. 𝑑𝑆̅

Example 1: Apply Stoke’s theorem to find the value of
∫ (𝑦 𝑑𝑥 + 𝑧 𝑑𝑦 + 𝑥 𝑑𝑧)𝑐
Where 𝑐 is the curve of intersection of 𝑥2
+ 𝑦2
+ 𝑧2
= 𝑎2
and 𝑥 + 𝑧 = 𝑎.
Solution: ∫ (𝑦 𝑑𝑥 + 𝑧 𝑑𝑦 + 𝑥 𝑑𝑧)𝑐
= ∫ (𝑦 𝑖̂ + 𝑧 𝑗̂ + 𝑥 𝑘̂).(𝑖̂ 𝑑𝑥 + 𝑗̂ 𝑑𝑦 + 𝑘̂ 𝑑𝑧)𝑐
= ∫ (𝑦 𝑖̂ + 𝑧 𝑗̂ + 𝑥 𝑘̂). 𝑑𝑟̅𝑐
= ∬ 𝑐𝑢𝑟𝑙 (𝑦 𝑖̂ + 𝑧 𝑗̂ + 𝑥 𝑘̂). 𝑛̂ 𝑑𝑠𝑆
(By Stoke’s theorem)
= ∬ (𝑖̂
𝜕
𝜕𝑥
+ 𝑗̂
𝜕
𝜕𝑦
+ 𝑘̂ 𝜕
𝜕𝑧
)×𝑆
(𝑦 𝑖̂ + 𝑧 𝑗̂ + 𝑥 𝑘̂). 𝑛̂ 𝑑𝑠
= ∬ – ( 𝑖̂ + 𝑗̂ + 𝑘̂). 𝑛̂ 𝑑𝑠𝑆
_______(i)
Where 𝑆 is the circle formed by the integration of 𝑥2
+ 𝑦2
+ 𝑧2
= 𝑎2
and
𝑥 + 𝑧 = 𝑎.
𝑛̂ =
∇∅
|∇∅|
=
( 𝑖̂
𝜕
𝜕𝑥
+𝑗̂
𝜕
𝜕𝑦
+𝑘̂ 𝜕
𝜕𝑧
)(𝑥+𝑧−𝑎)
|∇∅|
=
𝑖̂+ 𝑘̂
√1+1
=
𝑖̂
√2
+
𝑘̂
√2
Putting the value of 𝑛̂ in (i), we have
= ∬ –( 𝑖̂ + 𝑗̂ + 𝑘̂). (
𝑖̂
√2
+
𝑘̂
√2
)𝑆
𝑑𝑠
= ∬ −𝑆
(
1
√2
+
1
√2
)𝑑𝑠 [𝑈𝑠𝑒 𝑟2
= 𝑅2
− 𝑝2
= 𝑎2
−
𝑎2
2
=
𝑎2
2
]

= −
2
√2
∬ 𝑑𝑠 = −
2
√2
𝜋 (
𝑎
√2
)
2
= −
𝜋𝑎2
√2𝑆
______Ans.
Example 2: Evaluate ∮ 𝐹̅. 𝑑𝑟̅̅̅
𝐶
by stoke’s theorem, where
𝐹̅ = 𝑦2
𝑖̂ + 𝑥2
𝑗̂ − (𝑥 + 𝑧)𝑘̂ and 𝐶 is the boundary of triangle with vertices at
(0,0,0),(1,0,0) and (1,1,0).
Solution: We have, curl 𝐹̅ = ∇ × 𝐹̅
= ||
𝑖̂ 𝑗̂ 𝑘̂
𝜕
𝜕𝑥
𝜕
𝜕𝑦
𝜕
𝜕𝑧
𝑦2
𝑥2
−(𝑥 + 𝑧)
||
= 0. 𝑖̂ + 𝑗̂ + 2(𝑥 − 𝑦)𝑘̂
We observethat z co-ordinate of each vertex of the triangle is zero.
Therefore, the triangle lies in the 𝑥𝑦-plane.
∴ 𝑛̂ = 𝑘̂
∴ 𝑐𝑢𝑟𝑙 𝐹̅. 𝑛̂ = [𝑗̂ + 2(𝑥 − 𝑦)𝑘̂]. 𝑘̂ = 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
∮ 𝐹̅. 𝑑𝑟̅̅̅ = ∬ (𝑐𝑢𝑟𝑙 𝐹̅. 𝑛̂)𝑑𝑠𝑆𝐶
= ∫ ∫ 2( 𝑥 − 𝑦) 𝑑𝑥𝑑𝑦
𝑥
𝑦=0
1
𝑥=0
= 2∫ [𝑥2
−
𝑥2
2
]
1
0
𝑑𝑥
= 2 ∫
𝑥2
2
1
0
𝑑𝑥
= ∫ 𝑥2
𝑑𝑥
1
0

= [
𝑥3
3
]
0
1
=
1
3
________ Ans.
5.2 Exercise:
1) Use the Stoke’s theorem to evaluate ∫ [( 𝑥 + 2𝑦) 𝑑𝑥 + ( 𝑥 − 𝑧) 𝑑𝑦 +𝐶
(𝑦 − 𝑧)𝑑𝑧] where 𝐶 is the boundary of the triangle with vertices
(2,0,0),(0,3,0) 𝑎𝑛𝑑 (0,0,6) oriented in the anti-clockwise direction.
2) Apply Stoke’s theorem to calculate ∫ 4 𝑦 𝑑𝑥 + 2𝑧 𝑑𝑦 + 6𝑦 𝑑𝑧𝑐
Where 𝑐 is the curve of intersection of 𝑥2
+ 𝑦2
+ 𝑧2
= 6𝑧 and 𝑧 =
𝑥 + 3
3) Use the Stoke’s theorem to evaluate ∫ 𝑦2
𝑑𝑥 + 𝑥𝑦 𝑑𝑦 + 𝑥𝑧 𝑑𝑧𝐶
,
where 𝐶 is the bounding curve of the hemisphere 𝑥2
+ 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 𝑉enclosed by the surface 𝑆.
Mathematically
∬ 𝐹. 𝑛̂ 𝑑𝑠 = ∭ 𝑑𝑖𝑣 𝐹 𝑑𝑣
𝑉𝑆
Example 1: Evaluate ∬ 𝐹. 𝑛̂ 𝑑𝑠𝑆
where 𝐹 = 4𝑥𝑧𝑖̂ − 𝑦2
𝑗̂ + 𝑦𝑧𝑘̂ and 𝑆 is the
surface of the cube bounded by 𝑥 = 0, 𝑥 = 1, 𝑦 = 0, 𝑦 = 1, 𝑧 = 0, 𝑧 = 1.
Solution: By Gauss’s divergence theorem,
∬ 𝐹. 𝑛̂ 𝑑𝑠 = ∭ (∇. 𝐹) 𝑑𝑣𝑉𝑆
= ∭ (𝑖̂
𝜕
𝜕𝑥
+ 𝑗̂
𝜕
𝜕𝑦
+ 𝑘̂ 𝜕
𝜕𝑧
).𝑣
(4𝑥𝑧𝑖̂ − 𝑦2
𝑗̂ + 𝑦𝑧𝑘̂) 𝑑𝑣
= ∭ [
𝜕
𝜕𝑥
(4𝑥𝑧) +
𝜕
𝜕𝑦
(−𝑦2
) +
𝜕
𝜕𝑧
(𝑦𝑧)] 𝑑𝑥 𝑑𝑦 𝑑𝑧𝑣
= ∭ (4𝑧− 2𝑦 + 𝑦) 𝑑𝑥 𝑑𝑦 𝑑𝑧𝑣

= ∭ (4𝑧− 𝑦)𝑣
𝑑𝑥 𝑑𝑦 𝑑𝑧
= ∫ ∫ (
4𝑧2
2
− 𝑦𝑧)
0
1
1
0
1
0
𝑑𝑥 𝑑𝑦
= ∫ ∫ (2𝑧2
− 𝑦𝑧)0
11
0
𝑑𝑥 𝑑𝑦
1
0
= ∫ ∫ (2 − 𝑦)
1
0
𝑑𝑥 𝑑𝑦
1
0
= ∫ (2𝑦 −
𝑦2
2
)
0
1
𝑑𝑥
1
0
=
3
2
∫ 𝑑𝑥
1
0
=
3
2
[ 𝑥]0
1
=
3
2
(1)
=
3
2
________ Ans.
Example 2: Evaluate surface integral ∬ 𝐹. 𝑛̂ 𝑑𝑠, where 𝐹 = ( 𝑥2
+ 𝑦2
+
𝑧2)(𝑖̂ + 𝑗̂ + 𝑘̂), 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
surface 𝑆.
Solution: By gauss’s divergence theorem,
∬ 𝐹. 𝑛̂ 𝑑𝑠 = ∭ 𝑑𝑖𝑣 𝐹. 𝑑𝑣
𝑉𝑆
Where 𝑆 is the surface of tetrahedron 𝑥 = 0, 𝑦 = 0, 𝑧 = 0, 𝑥 + 𝑦 + 𝑧 = 2
= ∭ (𝑖̂
𝜕
𝜕𝑥
+ 𝑗̂
𝜕
𝜕𝑦
+ 𝑘̂ 𝜕
𝜕𝑧
). ( 𝑥2
+ 𝑦2
+ 𝑧2)(𝑖̂ + 𝑗̂ + 𝑘̂)𝑑𝑣𝑉
= ∭ (2𝑥 + 2𝑦 + 2𝑧)𝑑𝑣𝑉
= 2∭ ( 𝑥 + 𝑦 + 𝑧) 𝑑𝑥 𝑑𝑦 𝑑𝑧𝑉
= 2∫ 𝑑𝑥 ∫ 𝑑𝑦
2−𝑥
0
∫ ( 𝑥 + 𝑦 + 𝑧) 𝑑𝑧
2−𝑥−𝑦
0
2
0

= 2∫ 𝑑𝑥 ∫ 𝑑𝑦
2−𝑥
0
(𝑥𝑧 + 𝑦𝑧 +
𝑧2
2
)
0
2−𝑥−𝑦
2
0
= 2∫ 𝑑𝑥 ∫ 𝑑𝑦
2−𝑥
0
[2𝑥 − 𝑥2
− 𝑥𝑦 + 2𝑦 − 𝑥𝑦 − 𝑦2
+
(2−𝑥−𝑦)2
2
]
2
0
= 2∫ 𝑑𝑥 [2𝑥𝑦 − 𝑥2
𝑦 − 𝑥𝑦2
+ 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
0
(2−𝑥)3
6
]
= 2∫ [4𝑥 − 2𝑥2
− 2𝑥2
+ 𝑥3
− 4𝑥 + 4𝑥2
− 𝑥3
+ (2 − 𝑥)2
−
(2−𝑥)3
3
+
2
0
(2−𝑥)3
6
]
= 2[2𝑥2
−
4𝑥3
3
+
𝑥4
4
− 2𝑥2
+
4𝑥3
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[
8
3
−
16
12
+
16
24
]
= 4 ________Ans.
6.2 Exercise:
1) Evaluate ∬ 𝐹. 𝑛̂ 𝑑𝑠𝑆
where 𝑆 is the surface of the sphere 𝑥2
+ 𝑦2
+
𝑧2
= 16 and 𝐹 = 3𝑥𝑖̂ + 4𝑦𝑗̂ + 5𝑧𝑘̂.
2) Find ∬ 𝐹. 𝑛̂ 𝑑𝑠𝑆
, where 𝐹 = (2𝑥 + 3𝑧) 𝑖̂ − ( 𝑥𝑧 + 𝑦) 𝑗̂ + (𝑦2
+ 2𝑧)𝑘̂
and 𝑆 is the surface of the sphere having centre (3,-1, 2) and radius 3.
3) Use divergence theorem to evaluate ∬ 𝐴. 𝑑𝑠⃗⃗⃗⃗
𝑆
, where 𝐴 = 𝑥3
𝑖̂ +
𝑦3
𝑗̂ + 𝑧3
𝑘̂ and 𝑆 is the surface of the sphere 𝑥2
+ 𝑦2
+ 𝑧2
= 𝑎2
.

4) Use divergence theorem to show that∬ ∇𝑆
( 𝑥2
+ 𝑦2
+ 𝑧2). 𝑑𝑠 = 6𝑉,
where 𝑆 is any closed surface enclosing volume 𝑉.
7.1 REFERECE BOOKS:
1) Introduction to Engineering Mathematics
By H. K. DASS. & Dr. RAMA VERMA
S. CHAND
2) Higher Engineering Mathematics
By B.V. RAMANA
Mc Graw Hill Education
3) Higher Engineering Mathematics
By Dr. B.S. GREWAL
KHANNA PUBLISHERS
4) http://mecmath.net/calc3book.pdf

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