This document discusses various vector integration topics: 1. It defines line, surface, and volume integrals and provides examples of evaluating each. Line integrals deal with vector fields along paths, surface integrals deal with vector fields over surfaces, and volume integrals deal with vector fields throughout a volume. 2. Green's theorem, Stokes' theorem, and Gauss's theorem are introduced as relationships between these types of integrals but their proofs are not shown. 3. Examples are provided to demonstrate evaluating line integrals of conservative and non-conservative vector fields, as well as a surface integral over a spherical surface.

1. Unit-5 VECTOR INTEGRATION
RAI UNIVERSITY, AHMEDABAD 1
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

2. Unit-5 VECTOR INTEGRATION
RAI UNIVERSITY, AHMEDABAD 2
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

3. Unit-5 VECTOR INTEGRATION
RAI UNIVERSITY, AHMEDABAD 3
∴ 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 ]

4. Unit-5 VECTOR INTEGRATION
RAI UNIVERSITY, AHMEDABAD 4
= ∫ (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.

5. Unit-5 VECTOR INTEGRATION
RAI UNIVERSITY, AHMEDABAD 5
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.

6. Unit-5 VECTOR INTEGRATION
RAI UNIVERSITY, AHMEDABAD 6
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
=
𝑥𝑖̂+ 𝑦𝑗̂+ 𝑧𝑘̂
𝑎
[∵ 𝑥2
+ 𝑦2
+ 𝑧2
= 𝑎2]
Here, 𝐹 = 𝑦𝑧𝑖̂ + 𝑧𝑥𝑗̂ + 𝑥𝑦𝑘̂
𝐹. 𝑛̂ = (𝑦𝑧𝑖̂ + 𝑧𝑥𝑗̂ + 𝑥𝑦𝑘̂).(
𝑥𝑖̂+ 𝑦𝑗̂+ 𝑧𝑘̂
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

7. Unit-5 VECTOR INTEGRATION
RAI UNIVERSITY, AHMEDABAD 7
= 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

8. Unit-5 VECTOR INTEGRATION
RAI UNIVERSITY, AHMEDABAD 8
=
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

9. Unit-5 VECTOR INTEGRATION
RAI UNIVERSITY, AHMEDABAD 9
= (
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.

10. Unit-5 VECTOR INTEGRATION
RAI UNIVERSITY, AHMEDABAD 10
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

11. Unit-5 VECTOR INTEGRATION
RAI UNIVERSITY, AHMEDABAD 11
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
]

12. Unit-5 VECTOR INTEGRATION
RAI UNIVERSITY, AHMEDABAD 12
= −
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

13. Unit-5 VECTOR INTEGRATION
RAI UNIVERSITY, AHMEDABAD 13
= [
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

14. Unit-5 VECTOR INTEGRATION
RAI UNIVERSITY, AHMEDABAD 14
= ∭ (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

15. Unit-5 VECTOR INTEGRATION
RAI UNIVERSITY, AHMEDABAD 15
= 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
.

16. Unit-5 VECTOR INTEGRATION
RAI UNIVERSITY, AHMEDABAD 16
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