Notesale: Turn your study into money

Document Preview

Extracts from the notes are below, to see the PDF you'll receive please use the links above

---

KIET GROUP OF INSTITUTIONS, DELHI NCR, GHAZIABAD
B
...
(I Sem
...
Vector Integration: Line
integral, Surface integral, Volume integral, Gauss’s Divergence theorem, Green’s theorem,
Stoke’s theorem (without proof) and their applications
...
Also for solving line, surface and volume integrals
...
It is used extensively in physics
and engineering, especially in the description of electromagnetic fields, gravitational
fields and fluid flow
...
Vector differentiation
1
...

(3-22)
3

1
...
1 Geometrical Interpretation of Gradient

4

1
...
2 Properties of Gradient

5

1
...
3 Divergence of a vector point function

11

1
...
1 Physical Interpretation of Divergence
1
...
4
...
5 Tangent and Normal Planes
2
...
1 Line Integral

12
13
13
19
(23-69)
23

2
...
1 Work done by a force

23

2
...
2 Circulation

26

2
...
3 Volume integral

34

2
...
4
...
5 Green’s theorem (without proof)
2
...
1 Application of Green’s theorem
2
...
6
...
E-resources

41
45
50
56
65
70

2|P a ge

1
...
It is used extensively in physics and engineering, especially in the
description of electromagnetic fields, gravitational fields and fluid flow
...


Scalar Point Function: If to each point P (x, y, z) of a region R in space there corresponds a
unique scalar f(P), then f is called a scalar point function
...

(i) Temperature distribution in a heated body,
(ii) Density of a body & (iii) Potential due to gravity
...


Examples
...


1
...

The value of the gradient at a point is a tangent vector
...


Thus, 𝑔𝑟𝑎𝑑 ∅ = 𝛻∅

1
...
1 Geometrical Interpretation of Gradient: If a surface ∅(𝑥, 𝑦, 𝑧) = 𝑐
passes through a point P
...

Then such a surface is called a level surface through P
...
If ø(x,y,z) represent potential at the P
...


Note: Two level surfaces can’t intersect
...


Consider another level surface passing through Q, Where the value of function ø + dø
...
(𝑖̂𝑑𝑥 + 𝑗̂𝑑𝑦 + 𝑘̂ 𝑑𝑧)

=

𝜕𝜙
𝜕𝑥

𝜕𝜙

𝑑𝑥 + 𝜕𝑦 𝑑𝑦 +

𝜕𝜙
𝜕𝑧

𝑑𝑧 = 𝑑𝜙-------- (1)

If Q lies on the level surface of P, then 𝑑𝜙 = 0
From equation (1), we get
∇𝜙
...
1
...

(c) ∇(𝜙1 𝜙2 ) = 𝜙1 ∇𝜙2 + 𝜙2 ∇𝜙1
𝜙

𝜙2 ∇𝜙1−𝜙1 ∇𝜙2

𝜙2

𝜙22

(d) ∇ ( 1 ) =

, 𝜙2 ≠ 0

Example 1
...

Solution
...
What is the greatest rate of increase of 𝑢 = 𝑥𝑦𝑧 2 at the point (1,0,3) ?
Solution
...
Find a unit vector normal to the surface 𝑦𝑥 2 + 2𝑥𝑧 = 4 at the point (2, −2,3)
...
Let ∅ = 𝑦𝑥 2 + 2𝑥𝑧 − 4
𝑔𝑟𝑎𝑑 ∅ = 𝛻∅ = (𝑖̂

𝜕
𝜕
𝜕
+ 𝑗̂ + 𝑘̂ ) (𝑦𝑥 2 + 2𝑥𝑧 − 4)
𝜕𝑥
𝜕𝑦
𝜕𝑧

= (2𝑥𝑦 + 2𝑧)𝑖̂ + 𝑥 2 𝑗̂ + 2𝑥𝑘̂
Now,

𝑔𝑟𝑎𝑑 ∅](2,−2,3) = −2𝑖̂ + 4𝑗̂ + 4𝑘̂ and |𝑔𝑟𝑎𝑑 ∅| = √4 + 16 + 16 = 6

5|P a ge

𝑔𝑟𝑎𝑑 ∅

The unit vector normal to the surface ∅ = |𝑔𝑟𝑎𝑑 ∅| =

̂
−2𝑖̂ +4𝑗̂ +4𝑘
6

=

̂
−𝑖̂ +2𝑗̂ +2𝑘
3

Example 4
...

Solution
...
𝑑𝑟⃗ [taking dot product with 𝑑𝑟⃗ ]
𝜕∅
𝜕∅
𝜕∅
𝜕∅
𝜕∅
𝐹⃗
...
𝑑𝑟⃗ = 𝑑∅

𝜕∅

[𝑎𝑠

𝜕𝑥

𝜕∅

𝑑𝑥 + 𝜕𝑦 𝑑𝑦+

𝜕∅
𝜕𝑧

𝜕∅
𝜕𝑧

𝑑𝑧

𝑑𝑧 = 𝑑∅]

Or 𝑑∅ = 𝐹⃗
...
(𝑑𝑥𝑖̂ + 𝑑𝑦𝑗̂ + 𝑑𝑧𝑘̂)
= (𝑦 2 − 2𝑥𝑦𝑧 3 )𝑑𝑥 + (3 + 2𝑥𝑦 − 𝑥 2 𝑧 3 )𝑑𝑦 + (6𝑧 3 − 3𝑥 2 𝑦𝑧 2 )𝑑𝑧 ----------------- (1)
= 3𝑑𝑦 + 6𝑧 3 𝑑𝑧 + 𝑦 2 𝑑𝑥 + 2𝑥𝑦𝑑𝑦 − 2𝑥𝑦𝑧 3 𝑑𝑥 − 𝑥 2 𝑧 3 𝑑𝑦 − 3𝑥 2 𝑦𝑧 2 𝑑𝑧
𝑑∅ = 3𝑑𝑦 + 6𝑧 3 𝑑𝑧 + 𝑑(𝑥𝑦 2 ) − 𝑑(𝑥 2 𝑦𝑧 3 )
3

Integrate, we get ∅ = 3𝑦 + 2 𝑧 4 + 𝑥𝑦 2 − 𝑥 2 𝑦𝑧 3 + 𝐶

Example 5
...


Solution
...
Then
𝑛
...
4+(−2)
...
(−1)
√16+4+16 √16+4+1

=

16
6√21

=

8
3√21

𝟖

∴ 𝜽 = 𝐜𝐨𝐬−𝟏 (𝟑√𝟐𝟏)

Example 6
...


Solution
...
𝑔𝑟𝑎𝑑 𝑣 × 𝑔𝑟𝑎𝑑 𝑤 = |
𝑧+𝑦

1
2𝑦
𝑧+𝑥

1
2𝑧 | = 0
𝑦+𝑥

Hence, 𝑔𝑟𝑎𝑑 𝑢, 𝑔𝑟𝑎𝑑 𝑣 𝑎𝑛𝑑 𝑔𝑟𝑎𝑑 𝑤 𝑎𝑟𝑒 𝑐𝑜𝑝𝑙𝑎𝑛𝑎𝑟 𝑣𝑒𝑐𝑡𝑜𝑟𝑠
...
Show that ∇(𝑎⃗
...

2
...

3
...

4
...

5
...


Answers
1

1

2
...
− √11 (𝑖̂ + 3𝑗̂ − 𝑘̂)

4
...
cos −1 (− √22)

1
...
𝑎̂
1

Example 1: Find the directional derivative of ∅ = (𝑥 2 + 𝑦 2 + 𝑧 2 )−2 at the point 𝑃(3,1,2) in the
direction of the vector 𝑦𝑧𝑖̂ + 𝑧𝑥𝑗̂ + 𝑥𝑦𝑘̂
...
Here, ∅ = (𝑥 2 + 𝑦 2 + 𝑧 2 )−2 ⇒ 𝑔𝑟𝑎𝑑 ∅ = −
Let 𝑎̂ be the unit in the given direction, then 𝑎̂ =
𝑔𝑟𝑎𝑑 ∅](3,1,2) = −

̂
3𝑖̂ +𝑗̂ +2𝑘

and

14√14

Directional derivative =

𝑑∅
𝑑𝑠

̂
𝑥𝑖̂ +𝑦𝑗̂ +𝑧𝑘
3

(𝑥 2+𝑦 2 +𝑧 2 )2

̂
𝑦𝑧𝑖̂ +𝑧𝑥𝑗̂ +𝑥𝑦𝑘
√𝑥 2+𝑦 2 +𝑧 2

𝑎̂](3,1,2) =

̂
2𝑖̂ +6𝑗̂ +3𝑘

6+6+6

7
𝟗

= (𝑔𝑟𝑎𝑑 ∅)
...


7|P a ge

5

Solution
...
𝑎̂ =

25
3

𝑦−3
−2

𝑧

= 1, then

̂
2𝑖̂ −2𝑗̂ +𝑘

+

3
10
3

=

35
6

Example 3: Find the directional derivative of ∅ = 𝑥 2 − 𝑦 2 + 2𝑧 2 at the point 𝑃(1,2,3) in the
direction of the line 𝑃𝑄 where 𝑄 is the point (5,0,4)
...


Solution
...
𝑎̂ =

8+8+12
√21

=

28
√21

Directional derivative will be maximum in the direction of normal to the given surface i
...
, in the
direction of 𝛻∅
...


Solution
...
𝑎̂
= (𝑖̂ − 3𝑗̂ − 3𝑘̂ )
...


Solution
...
𝑎̂
Required Directional derivative = (𝛻𝑉
= (324𝑖̂ + 432𝑘̂)
...
(𝛻∅) at a point (1, −2,1) in the direction of the
normal to the surface 𝑥𝑦 2 𝑧 = 3𝑥 + 𝑧 2 , where ∅ = 2𝑥 3 𝑦 2 𝑧 4
...
Here, ∅ = 2𝑥 3 𝑦 2 𝑧 4 ⇒ 𝛻∅ = 6𝑥 2 𝑦 2 𝑧 4 𝑖̂ + 4𝑥 3 𝑦𝑧 4 𝑗̂ + 8𝑥 3 𝑦 2 𝑧 3 𝑘̂
𝛻
...
(𝛻∅)} = (12𝑦 2 𝑧 4 + 12𝑥 2 𝑧 4 + 72𝑥 2 𝑦 2 𝑧 2 )𝑖̂ + (24𝑥𝑦𝑧 4 + 48𝑥 3 𝑦𝑧 2 )𝑗̂
+(48𝑥𝑦 2 𝑧 3 + 16𝑥 3 𝑧 3 + 48𝑥 3 𝑦 2 𝑧)𝑘̂
𝛻{𝛻
...

̂

𝑖̂ −4𝑗̂ +2𝑘
1724
)=
Required Directional derivative = (348𝑖̂ − 144𝑗̂ + 400𝑘̂)
...
If the directional derivative of ∅ = 𝑎𝑥 2 𝑦 + 𝑏𝑦 2 𝑧 + 𝑐𝑧 2 𝑥 at the point (1,1,1) has maximum
magnitude 15 in the direction parallel to the line

𝑥−1
2

=

𝑦−3
−2

𝑧

= 1 find the values of 𝑎, 𝑏 𝑎𝑛𝑑 𝑐
...
For the function ∅ = 𝑥 2+𝑦2, find the magnitude of the directional derivative making an angle
30° with the positive 𝑥 − 𝑎𝑥𝑖𝑠 at the point (0,1)
...
Find the values of constants 𝑎, 𝑏, 𝑐 so that the maximum value of the directional derivative of
∅ = 𝑎𝑥𝑦 2 + 𝑏𝑦𝑧 + 𝑐𝑧 2 𝑥 3 at (1,2, −1) has a magnitude 64 in the direction parallel to 𝑧 −
𝑎𝑥𝑖𝑠
...
Find the directional derivative of 𝑓 (𝑥, 𝑦, 𝑧) = 2𝑥 2 + 3𝑦 2 + 𝑧 2 at the point 𝑃(2,1,3) in the
direction of the vector 𝑎⃗ = 𝑖̂ − 2𝑘̂
...
Find the directional derivative of Ψ(𝑥, 𝑦, 𝑧) = 4𝑒 𝑥+5𝑦−13𝑧 at the point (1,2,3) in the direction
towards the point (−3,5,7)
...
𝑎 = ±

20
9

,𝑏 = ∓

55
9

,𝑐 = ±

50
9

1

2
...
𝑎 = 6, 𝑏 = 24, 𝑐 = −8 4
...
−4√41𝑒 −28

Gradient in Polar Form
Sometimes the surface is given in the form of ∅ = 𝑓(𝑟, 𝜃)
...

Let 𝑢𝑟 𝑎𝑛𝑑 𝑢𝜃 be the unit vectors along and perpendicular to 𝑟⃗
...
𝑢𝑟
Thus

𝜕∅
𝜕𝑠

𝜕∅

= 𝜕𝑟 = 𝛻∅
...
𝑢𝜃
Thus

𝜕∅
𝜕𝑠

𝜕∅

= 𝑟𝑑𝜃 = 𝛻∅
...
𝑢𝜃 )𝑢𝜃
10 | P a g e

Or

𝜕∅

𝜕∅

𝛻∅ = ( 𝜕𝑟 ) 𝑢𝑟 + (𝑟𝑑𝜃 ) 𝑢𝜃

Example 1
...
(i) 𝛻 (

𝑒 𝜇𝑟
𝑟

(ii) 𝛻 |𝑟⃗|2 = 𝛻𝑟 2 =

)=

𝜕
𝜕𝑟

𝑒 𝜇𝑟
𝑟
𝜕

(ii) 𝛻 |𝑟⃗|2

)

(

𝜕𝑟

[ from equation (1) and (2)]

𝑒 𝜇𝑟
𝑟

) 𝑟̂ = [

(iii) 𝛻 log 𝑟 𝑛

𝑟𝜇𝑒 𝜇𝑟−𝑒 𝜇𝑟 𝑟⃗
𝑟2

] =
𝑟

(iv) 𝑔𝑟𝑎𝑑𝑓 ′ (𝑟) × 𝑟⃗

𝑒 𝜇𝑟(𝜇𝑟−1)𝑟⃗
𝑟3

𝑟⃗

(𝑟 2 )𝑟̂ = 2𝑟 = 2𝑟⃗
𝑟

𝜕

1

𝑟⃗

(iii) 𝛻 log 𝑟 𝑛 = 𝜕𝑟 (log 𝑟 𝑛 )𝑟̂ = (𝑟𝑛 𝑛𝑟 𝑛−1 ) 𝑟 =

𝑛𝑟⃗
𝑟2

𝜕

𝑟⃗

⃗⃗
(iv) 𝑔𝑟𝑎𝑑𝑓 ′ (𝑟) × 𝑟⃗ = [𝜕𝑟 {𝑓 ′ (𝑟)}𝑟̂ ] × 𝑟⃗ = 𝑓 ′′ (𝑟) 𝑟 × 𝑟⃗ = 0
⃗⃗ (𝑘
⃗⃗
...
If 𝑉(𝑥, 𝑦) = 2 log(𝑥 2 + 𝑦 2 ) prove that 𝑔𝑟𝑎𝑑𝑉 = {𝑟⃗−𝑘⃗⃗(𝑘⃗⃗
...
Here, 𝑉 = 2 log(𝑥 2 + 𝑦 2 ) == 2 log 𝑟 2 = log 𝑟
𝜕

1 𝑟⃗

𝑟⃗

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

⃗⃗ (𝑘
⃗⃗
...
{𝑟⃗−𝑘
⃗⃗ (𝑘
⃗⃗
...
{𝑟⃗−0}

𝑟⃗

[ as 𝑘⃗⃗
...
𝑟⃗ = 𝑟 2

1
...


Definition: The divergence of a differentiable vector point function 𝑉⃗⃗ is denoted by 𝑑𝑖𝑣𝑉⃗⃗ and
⃗⃗ = 𝛻
...
𝑉
⃗⃗ = (𝑖̂ 𝜕𝑉⃗⃗ + 𝑗̂ 𝜕𝑉⃗⃗ + 𝑘̂ 𝜕𝑉⃗⃗)
defined as: 𝑑𝑖𝑣𝑉
𝜕𝑥
𝜕𝑦
𝜕𝑧
𝜕𝑥
𝜕𝑦
𝜕𝑧

11 | P a g e

1
...
1 Physical Interpretation of Divergence
⃗⃗ are the density and velocity at a point 𝑃(𝑥, 𝑦, 𝑧) at any time 𝑡 respectively, of a
Let 𝜌 𝑎𝑛𝑑 𝑉
moving fluid in a rectangular parallelepiped
...

Let us consider a small parallelepiped with edges 𝛿𝑥, 𝛿𝑦, 𝛿𝑧 parallel to the coordinate axes
...
(𝑖̂𝑉𝑥 + 𝑗̂𝑉𝑦 + 𝑘̂ 𝑉𝑧 )

⃗⃗ = 𝑑𝑖𝑣 𝑉
⃗⃗
= 𝛻
...

Hence, 𝑑𝑖𝑣 𝑉
⃗⃗ = 0, then (i) 𝑉
⃗⃗ is called solenoidal vector (ii) fluid is called compressible
Note: If 𝑑𝑖𝑣 𝑉

1
...


Definition: The curl of a differentiable vector point function 𝐹⃗ is denoted by curl 𝐹⃗ and defined
as:
𝜕
𝜕
𝜕
𝑐𝑢𝑟𝑙 𝐹⃗ = 𝛻 × 𝐹⃗ = (𝑖̂ 𝜕𝑥 + 𝑗̂ 𝜕𝑦 + 𝑘̂ 𝜕𝑧 ) × 𝐹⃗

1
...
1 Physical Interpretation of Curl
Let us consider a rigid body rotating about a fixed axis through 𝑂 with uniform angular velocity
𝜔
⃗⃗ = 𝜔1 𝑖̂ + 𝜔2 𝑗̂+ 𝜔3 𝑘̂
...


𝑖̂
∴ 𝑣⃗ = 𝜔
⃗⃗ × 𝑟⃗ = |𝜔1
𝑥
𝑖̂
No𝑐𝑢𝑟𝑙 𝑣⃗ = |

𝑗̂

𝑘̂

𝜕

𝜕

𝜕

𝜕𝑥

𝜕𝑦

𝜕𝑧

𝜔2 𝑧 − 𝜔3 𝑦
= 2𝜔
⃗⃗

𝑘̂
𝜔3 | = (𝜔2 𝑧 − 𝜔3 𝑦)𝑖̂ + (𝜔3 𝑥 − 𝜔1 𝑧)𝑗̂ + (𝜔1 𝑦 − 𝜔2 𝑥)𝑘̂
𝑧

𝑗̂
𝜔2
𝑦

| = 2𝜔1 𝑖̂ + 2𝜔2 𝑗̂ + 2𝜔3 𝑘̂

𝜔3 𝑥 − 𝜔1 𝑧 𝜔1 𝑦 − 𝜔2 𝑥
1

⇒ 𝜔
⃗⃗ = 2 𝑐𝑢𝑟𝑙 𝑣⃗

Thus, the angular velocity at any point is equal to the half of the 𝑐𝑢𝑟𝑙 of the linear velocity at that
point of the body
...


13 | P a g e

Example 1
...

Solution
...
(𝑥𝑖̂ + 𝑦𝑗̂ + 𝑧𝑘̂)

= 1+1+1 = 𝟑
𝑐𝑢𝑟𝑙 𝑟⃗ = 𝛻 × 𝑟⃗ = (𝑖̂

=|

𝜕
𝜕𝑥

𝜕𝑦

+ 𝑘̂

𝜕
𝜕𝑧

) × (𝑥𝑖̂ + 𝑦𝑗̂ + 𝑧𝑘̂)

𝑘̂

𝑖̂

𝑗̂

𝜕

𝜕

𝜕

𝜕𝑥

𝜕𝑦

𝜕𝑧

𝑥

𝜕

+ 𝑗̂

𝑦

⃗⃗
| = (0 − 0)𝑖̂ + (0 − 0)𝑗̂ + (0 − 0)𝑘̂ = 𝟎

𝑧

Example 2
...

𝑅⃗⃗ = (𝑥 2 + 𝑦𝑧)𝑖̂ + (𝑦 2 + 𝑧𝑥)𝑗̂ + (𝑧 2 + 𝑥𝑦)𝑘̂

Solution
...
[(𝑥 2 + 𝑦𝑧)𝑖̂ + (𝑦 2 + 𝑧𝑥)𝑗̂ + (𝑧 2 + 𝑥𝑦)𝑘̂ ]

= 2𝑥 + 2𝑦 + 2𝑧 = 𝟐(𝒙 + 𝒚 + 𝒛)
𝑐𝑢𝑟𝑙 𝐹⃗ = 𝛻 × 𝑅⃗⃗
= (𝑖̂

𝜕
𝜕𝑥

+ 𝑗̂

𝜕
𝜕𝑦

𝑖̂
=|

+ 𝑘̂

𝜕
𝜕𝑧

) × [(𝑥 2 + 𝑦𝑧)𝑖̂ + (𝑦 2 + 𝑧𝑥)𝑗̂ + (𝑧 2 + 𝑥𝑦)𝑘̂ ]
𝑘̂

𝑗̂

𝜕

𝜕

𝜕

𝜕𝑥

𝜕𝑦

𝜕𝑧

𝑥 2 + 𝑦𝑧 𝑦 2 + 𝑧𝑥

⃗⃗
| = (𝑥 − 𝑥)𝑖̂ + (𝑦 − 𝑦)𝑗̂ + (𝑧 − 𝑧)𝑘̂ = 𝟎

𝑧 2 + 𝑥𝑦

Example 3
...


Solution
...
𝑉
⃗⃗ = −2 + 2𝑥 − 2𝑥 + 2 = 0
𝑑𝑖𝑣 𝑉

⃗⃗ is solenoidal
∴𝑉
𝑘̂

𝑖̂

𝑗̂

𝜕

𝜕

𝜕

𝜕𝑥

𝜕𝑦

𝜕𝑧

𝑦 2 − 𝑧 2 + 3𝑦𝑧 − 2𝑥

3𝑥𝑧 + 2𝑥𝑦

3𝑥𝑦 − 2𝑥𝑧 + 2𝑧

⃗⃗ = 𝛻 × 𝑉
⃗⃗ = |
𝑐𝑢𝑟𝑙 𝑉

|

= (3𝑥 − 3𝑥)𝑖̂ + (−2𝑧 + 3𝑦 − 3𝑦 + 2𝑧)𝑗̂ + (3𝑧 + 2𝑦 − 2𝑦 − 3𝑧)𝑘̂
= ⃗𝟎⃗

⃗⃗ is irrotational
...
Show that the vector field 𝐴⃗, where 𝐴⃗ = (𝑥 2 − 𝑦 2 + 𝑥)𝑖̂ − (2𝑥𝑦 + 𝑦)𝑗̂ is
irrotational
...


14 | P a g e

Solution
...


0

Let 𝐴⃗ = 𝛻∅ , where ∅ is scalar potential
...
𝑑𝑟⃗
⇒ [(𝑥 2 − 𝑦 2 + 𝑥)𝑖̂ − (2𝑥𝑦 + 𝑦)𝑗̂]
...
Find the constants 𝑎, 𝑏, 𝑐 so that
̂ is irrotational
...


Solution
...
𝑑𝑟⃗
̂ ]
...


Vector Identities
If 𝑎⃗, 𝑏⃗⃗ are vector functions and ∅ is a scalar function, then

(1) 𝑑𝑖𝑣(∅𝑎⃗) = ∅𝑑𝑖𝑣𝑎⃗ + (𝑔𝑟𝑎𝑑∅)
...
(∅ 𝜕𝑥 + 𝜕𝑥 𝑎⃗)
𝜕𝑎⃗⃗

𝜕∅

= ∅ ∑ 𝑖̂
...
𝑐𝑢𝑟𝑙 (∅𝑎⃗) = ∑ 𝑖̂ × 𝜕𝑥 (∅𝑎⃗) = ∑ 𝑖̂ × (∅ 𝜕𝑥 + 𝜕𝑥 𝑎⃗)
𝜕𝑎⃗⃗

𝜕∅

= ∅ ∑ 𝑖̂ × 𝜕𝑥 + (∑ 𝑖̂ 𝜕𝑥 ) × 𝑎⃗ = ∅𝑐𝑢𝑟𝑙𝑎⃗ + (𝑔𝑟𝑎𝑑∅) × 𝑎⃗

(3) 𝑑𝑖𝑣(𝑎⃗ × 𝑏⃗⃗) = 𝑏⃗⃗
...
(𝑎⃗ × 𝜕𝑥 + 𝜕𝑥 × 𝑏⃗⃗)
⃗⃗

𝜕𝑏
𝜕𝑎⃗⃗
= ∑ 𝑖̂
...
(𝜕𝑥 × 𝑎⃗)
⃗⃗
𝜕𝑏

𝜕𝑎⃗⃗

= ∑ (𝑖̂ × 𝜕𝑥 )
...
(𝑐𝑢𝑟𝑙𝑏⃗⃗)

(4) 𝑐𝑢𝑟𝑙(𝑎⃗ × 𝑏⃗⃗) = 𝑎⃗𝑑𝑖𝑣𝑏⃗⃗ − 𝑏⃗⃗𝑑𝑖𝑣𝑎⃗ + (𝑏⃗⃗
...
𝑐𝑢𝑟𝑙(𝑎⃗ × 𝑏⃗⃗) = ∑ 𝑖̂ × 𝜕𝑥 (𝑎⃗ × 𝑏⃗⃗) = ∑ 𝑖̂ × (𝑎⃗ × 𝜕𝑥 + 𝜕𝑥 × 𝑏⃗⃗)
⃗⃗
𝜕𝑏

𝜕𝑎⃗⃗

= ∑ 𝑖̂ × (𝑎⃗ × 𝜕𝑥 ) + ∑ 𝑖̂ × ( 𝜕𝑥 × 𝑏⃗⃗)
= ∑ [(𝑖̂
...


⃗⃗
𝜕𝑏
𝜕𝑥
⃗⃗
𝜕𝑏
𝜕𝑥

) 𝑎⃗ − (𝑖̂
...
𝑖̂)

] + ∑ [(𝑖̂
...


𝜕𝑎⃗⃗
𝜕𝑥

) 𝑏⃗⃗]

𝜕𝑎⃗⃗
𝜕𝑎⃗⃗
+ ∑(𝑏⃗⃗
...
∑ 𝑖̂ 𝜕𝑥 ) 𝑎⃗ − (𝑑𝑖𝑣𝑎⃗)𝑏⃗⃗
= 𝑎⃗𝑑𝑖𝑣𝑏⃗⃗ − 𝑏⃗⃗𝑑𝑖𝑣𝑎⃗ + (𝑏⃗⃗
...
(𝑖̂ 𝜕𝑥 + 𝑗̂ 𝜕𝑦 + 𝑘̂ 𝜕𝑧 )
𝜕

𝜕∅

𝜕

𝜕∅

𝜕

𝜕2 ∅

𝜕∅

= 𝜕𝑥 (𝜕𝑥 ) + 𝜕𝑦 (𝜕𝑦 ) + 𝜕𝑧 ( 𝜕𝑧 ) = 𝜕𝑥 2 + 𝜕𝑦2 + 𝜕𝑧 2
𝜕2

𝜕2

𝜕2

= (𝜕𝑥 2 + 𝜕𝑦2 + 𝜕𝑧 2 ) ∅ = 𝛻 2 ∅; 𝛻 2 is known as Laplacian operator

(6) 𝑐𝑢𝑟𝑙 (𝑔𝑟𝑎𝑑∅) = 𝛻 × (𝛻∅) = ⃗0⃗
𝜕

𝜕

𝜕

𝜕∅

𝜕∅

𝜕∅

Proof
...
Let 𝑉⃗⃗ = 𝑉1 𝑖̂ + 𝑉2 𝑗̂ + 𝑉3 𝑘̂ ⇒ 𝑐𝑢𝑟𝑙 𝑉⃗⃗ = |
𝜕𝑉

= ( 𝜕𝑦3 −
𝜕

𝜕𝑉2
𝜕𝑧

) 𝑖̂ + (

𝜕𝑉

𝜕𝑉2

⃗⃗) = ( 3 −
Now, 𝑑𝑖𝑣(𝑐𝑢𝑟𝑙 𝑉
𝜕𝑥 𝜕𝑦
𝜕2 𝑉

𝜕𝑧

𝜕𝑉1
𝜕𝑧

)+

𝜕2 𝑉

−

𝜕𝑉3

𝜕
𝜕𝑦

𝜕𝑥

(

𝑘̂

𝑖̂
𝜕

𝜕𝑉1
𝜕𝑧

𝜕2 𝑉

−

𝜕𝑥

𝜕𝑉3
𝜕𝑥

−

)+

𝜕2 𝑉

|

𝜕𝑉1
𝜕𝑦

𝜕
𝜕𝑧

𝜕2 𝑉

(

) 𝑘̂

𝜕𝑉2
𝜕𝑥

−

𝜕𝑉1
𝜕𝑦

)

𝜕2 𝑉

= (𝜕𝑥𝜕𝑦3 − 𝜕𝑥𝜕𝑧2 ) + (𝜕𝑦𝜕𝑧1 − 𝜕𝑦𝜕𝑥3 ) + (𝜕𝑧𝜕𝑥2 − 𝜕𝑧𝜕𝑦1 ) = 0

Example 1
...
𝑉⃗⃗
𝜕𝑢

𝜕𝑢

𝜕𝑢

⃗⃗
= 3𝑢 + (𝑖̂ 𝜕𝑥 + 𝑗̂ 𝜕𝑦 + 𝑘̂ 𝜕𝑧 )
...
(𝑥𝑖̂ + 𝑦𝑗̂ + 𝑧𝑘̂ )
= 3𝑢 + (2𝑥 2 + 2𝑦 2 + 2𝑧 2 )
= 3𝑢 + 2(𝑥 2 + 𝑦 2 + 𝑧 2 )
= 3𝑢 + 2𝑢 = 𝟓𝒖

Example 2
...
(i) 𝑐𝑢𝑟𝑙 (𝑟 𝑛 𝑟⃗) = 𝑟 𝑛 𝑐𝑢𝑟𝑙 𝑟⃗ + (𝑔𝑟𝑎𝑑 𝑟 𝑛 ) × 𝑟⃗
𝑟⃗

⃗⃗ + (𝑛𝑟 𝑛−1 𝑟̂ ) × 𝑟⃗ = 𝑛𝑟 𝑛−1 × 𝑟⃗ = 0
⃗⃗
=0
𝑟
(ii) 𝛻 2 (𝑟 𝑛 𝑟⃗) = 𝛻[𝛻
...
𝑟⃗]
𝑟⃗

= 𝑔𝑟𝑎𝑑[3𝑟 𝑛 + (𝑛𝑟 𝑛−1 𝑟̂ )
...
𝑟⃗]
= 𝑔𝑟𝑎𝑑 [3𝑟 𝑛 + 𝑛𝑟 𝑛−1

𝑟2
𝑟

] = 𝑔𝑟𝑎𝑑[(𝑛 + 3)𝑟 𝑛 ]

𝑟⃗

= (𝑛 + 3)𝑛𝑟 𝑛−1 𝑟 = 𝑛(𝑛 + 3)𝑟 𝑛−2 𝑟⃗

17 | P a g e

Example 3
...

Solution
...
𝑟⃗ = 0
𝑟⃗

⇒ 3𝑓 (𝑟) + [𝑓 ′ (𝑟)𝑟̂ ]
...
𝑟⃗ = 0
⇒ 3𝑓 (𝑟) + 𝑓 ′ (𝑟)

𝑟2
𝑟

𝑓 ′ (𝑟)

=0⇒

𝑓(𝑟)

3

= −𝑟

---------------------- (1)

Integrate eqn (1), we get
log[𝑓 (𝑟)] = −3 log 𝑟 + log 𝑐
𝑐

⇒ 𝑓(𝑟) = 𝑟3

Example 4
...

Solution
...

2

Example 5
...

1

Hence evaluate 𝛻 2 (log 𝑟) if 𝑟 = (𝑥 2 + 𝑦 2 + 𝑧 2 )2

Solution
...
𝛻𝑓 (𝑟) = 𝑑𝑖𝑣[𝛻𝑓(𝑟)] = 𝑑𝑖𝑣[𝑔𝑟𝑎𝑑 𝑓(𝑟)]
𝑟⃗

𝑓 ′(𝑟)

𝑟

𝑟

= 𝑑𝑖𝑣[𝑓 ′ (𝑟)𝑟̂ ] = 𝑑𝑖𝑣 [𝑓 ′ (𝑟) ] = 𝑑𝑖𝑣 [{
=

𝑓 ′ (𝑟)
𝑟

𝑑𝑖𝑣𝑟⃗ + [𝑔𝑟𝑎𝑑 {

3

= 𝑓 ′ (𝑟) + [
3

𝑟

}]
...
𝑟⃗ = 𝑓 ′ (𝑟) +
𝑟

𝑟

𝑟𝑓 ′′ (𝑟)−𝑓 ′ (𝑟)
𝑟2

𝑟̂ ]
...
𝑟⃗
𝑟2

( )
𝑟

2

= 𝑓 ′′ (𝑟) + 𝑟 𝑓 ′ (𝑟)

Put 𝑓(𝑟) = log 𝑟, we get
𝛻 2 (log 𝑟) = −

1
𝑟2

+

21
𝑟𝑟

=

𝟏
𝒓𝟐

18 | P a g e

1
...
Let 𝑟 = 𝑥𝑖̂ + 𝑦𝑗̂ + 𝑧𝑘̂ be the position vector
of any point 𝑃(𝑥, 𝑦, 𝑧) on this surface
...


i
...
𝛻∅ is perpendicular to the tangent plane at 𝑃
...

⃗⃗⃗⃗⃗⃗ = 𝑅 − 𝑟 = (𝑋 − 𝑥)𝑖̂ + (𝑌 − 𝑦)𝑗̂ + (𝑍 − 𝑧)𝑘̂
∴ 𝑃𝑄
and ⃗⃗⃗⃗⃗⃗
𝑃𝑄 lies in the tangent plane at 𝑃
...

∴ ⃗⃗⃗⃗⃗⃗
𝑃𝑄
...
[
𝜕∅

𝜕∅

𝜕∅
𝜕𝑥

𝑖̂ +

𝜕∅
𝜕𝑦

𝑗̂ +

𝜕∅
𝜕𝑧

𝑘̂] = 0

𝜕∅

⇒ (𝑋 − 𝑥) 𝜕𝑥 + (𝑌 − 𝑦) 𝜕𝑦 + (𝑍 − 𝑧) 𝜕𝑧 = 0
This is the equation of tangent plane
...
Let 𝑟 = 𝑥𝑖̂ + 𝑦𝑗̂ + 𝑧𝑘̂ be the position vector
of any point 𝑃(𝑥, 𝑦, 𝑧) on this surface
...


i
...
𝛻∅ is perpendicular to the tangent plane at 𝑃
...

∴ ⃗⃗⃗⃗⃗⃗
𝑃𝑄 = 𝑅 − 𝑟 = (𝑋 − 𝑥)𝑖̂ + (𝑌 − 𝑦)𝑗̂ + (𝑍 − 𝑧)𝑘̂
⃗⃗⃗⃗⃗⃗ lies along the normal of 𝑃 to the surface ∅
...

and 𝑃𝑄
∴ ⃗⃗⃗⃗⃗⃗
𝑃𝑄 × 𝛻∅ = ⃗0⃗ ⇒ (𝑅 − 𝑟) × 𝛻∅ = ⃗0⃗ [vector form of normal]

Cartesian form of Normal:
𝜕∅
𝜕∅
𝜕∅
The vectors (𝑋 − 𝑥)𝑖̂ + (𝑌 − 𝑦)𝑗̂ + (𝑍 − 𝑧)𝑘̂ and 𝛻∅ = 𝜕𝑥 𝑖̂ + 𝜕𝑦 𝑗̂ + 𝜕𝑧 𝑘̂ will be parallel, if
𝜕∅
𝜕∅
𝜕∅
(𝑋 − 𝑥)𝑖̂ + (𝑌 − 𝑦)𝑗̂ + (𝑍 − 𝑧)𝑘̂ = 𝜆 [ 𝑖̂ + 𝑗̂ + 𝑘̂ ]
𝜕𝑥

𝜕𝑦

𝜕𝑧

Equating the coefficient of 𝑖̂ , 𝑗̂ and 𝑘̂, we get
𝑋−𝑥=𝜆
OR

𝑋−𝑥
𝜕∅
𝜕𝑥

=

𝜕∅
𝜕𝑥

𝑌−𝑦
𝜕∅
𝜕𝑦

,𝑌−𝑦=𝜆

=

𝑍−𝑧
𝜕∅
𝜕𝑧

𝜕∅
𝜕𝑦

,𝑍−𝑧 =𝜆

𝜕∅
𝜕𝑧

, which are equations of Normal
...
Find the equations of the tangent plane and normal to the surface
...


Solution
...
(7𝑖̂ − 3𝑗̂ + 8𝑘̂ ) = 0
Or 7𝑋 − 3𝑌 + 8𝑍 = 26
Equations of the normal to the given surface at the point (1, −1,2) are:
𝑋−𝑥
𝜕∅
𝜕𝑥

=

𝑌−𝑦
𝜕∅
𝜕𝑦

=

𝑍−𝑧
𝜕∅
𝜕𝑧

i
...


𝑋−1
7

=

𝑌+1
−3

=

𝑍+2
8

Example 2
...


Solution
...
(4𝑖̂ + 2𝑗̂ + 2 𝑘̂ ) = 0
Or 4𝑋 + 2𝑌 + 2𝑍 = 12 Or 2𝑍 + 𝑌 + 𝑍 = 6
Equations of the normal to the given surface at the point (1, −1,2) are:
𝑋−𝑥
𝜕∅
𝜕𝑥

=

𝑌−𝑦
𝜕∅
𝜕𝑦

=

𝑍−𝑧
𝜕∅
𝜕𝑧

i
...


𝑋−1
4

=

𝑌−2
2

=

𝑍−2
2

⇒

𝑋−1
2

=

𝑌−2
1

=

𝑍−2
1

Example 3
...


Solution
...
(−3𝑖̂ + 3𝑗̂ − 2𝑘̂ ) = 0
Or −3𝑋 + 3𝑌 − 2𝑍 + 10 = 0 ⇒ 3𝑋 − 3𝑌 + 2𝑍 = 10
Equations of the normal to the given surface at the point (1, −1,2) are:
𝑋−𝑥
𝜕∅
𝜕𝑥

=

𝑌−𝑦
𝜕∅
𝜕𝑦

=

𝑍−𝑧
𝜕∅
𝜕𝑧

𝑋−1

i
...


−3

=

𝑌+1
3

=

𝑍+2
−2

Example 4
...


Solution
...
(−4𝑖̂ + 2𝑗̂ + 𝑘̂ ) = 0
Or −4𝑋 + 2𝑌 + 𝑍 + 5 = 0 ⇒ 4𝑋 − 2𝑌 − 𝑍 = 5
Equations of the normal to the given surface at the point (1, −1,2) are:
𝑋−𝑥
𝜕∅
𝜕𝑥

=

𝑌−𝑦
𝜕∅
𝜕𝑦

=

𝑍−𝑧
𝜕∅
𝜕𝑧

i
...


𝑋−2
−4

=

𝑌+1
2

=

𝑍−5
1

Exercise
1
...

2
...

3
...

(ii) Is the motion possible for an incompressible fluid?
21 | P a g e

4
...
𝑉

2
√𝑥 2+𝑦 2+𝑧 2

⃗⃗ = 0
⃗⃗
and ∇ × 𝑉

5
...
Find the velocity
potential ∅ such that 𝐴⃗ = ∇∅
...
A fluid motion is given by 𝑣⃗ = (𝑦 sin 𝑧 − sin 𝑥 )𝑖̂ + (𝑥 sin 𝑧 + 2𝑦𝑧)𝑗̂ + (𝑥𝑦 cos 𝑧 + 𝑦 2 )𝑘̂
...

⃗⃗ are irrotational, prove that 𝐸⃗⃗ × 𝐻
⃗⃗ is irrotational
...
If 𝐸⃗⃗ 𝑎𝑛𝑑 𝐻

8
...
𝑐𝑢𝑟𝑙 𝐹⃗ = 0
...
If 𝑟⃗ = 𝑥𝑖̂ + 𝑦𝑗̂ + 𝑧𝑘̂ and 𝑟 = |𝑟⃗| , show that
(i) 𝑑𝑖𝑣(𝑟⃗)∅ = 3∅ + 𝑟⃗
...
Prove that 𝑑𝑖𝑣(𝑔𝑟𝑎𝑑 𝑟 𝑛 ) = ∇2 𝑟 𝑛 = 𝑛(𝑛 + 1)𝑟 𝑛−2 , where 𝑟⃗ = 𝑥𝑖̂ + 𝑦𝑗̂ + 𝑧𝑘̂
...
Hence or otherwise evaluate ∇ × (𝑟2 )
...
𝑑𝑖𝑣 𝐹⃗ = 𝑧 3 − 2𝑥 2 𝑧 + 8𝑦𝑧 3 , 𝑐𝑢𝑟𝑙 𝐹⃗ = 2(𝑥 2 𝑦 + 𝑧 4 )𝑖̂ + 3𝑥𝑧 2 𝑗̂ − 4𝑥𝑦𝑧𝑘̂
⃗⃗ = 2𝑥𝑦 2 + 2𝑥, 𝑐𝑢𝑟𝑙 𝐹⃗ = (2𝑦 − 𝑥)𝑖̂ + 𝑦𝑗̂ + 2𝑦(1 − 𝑥 2 )𝑘̂
2
...
(i) Yes; velocity potential ∅ = 𝑥𝑦 + 𝑦𝑧 + 𝑧𝑥 + 𝑐 (ii) Yes
5
...
Yes; ∅ = 𝑥𝑦 sin 𝑧 + cos 𝑥 + 𝑦 2 𝑧 + 𝑐

⃗⃗
10
...
Vector Integration
Introduction: The process of integration to compute the integrals of vector functions of a real
variable is known vector integration
...
e
...


2
...
The
terms path integral, curve integral, and curvilinear integral are also used for line integral
...


2
...
1 Work done by a force: Let 𝐹⃗ be the force acting on a moving particle along the path 𝐶,
then the total work done (𝑊 ) by 𝐹⃗ during displacement from 𝐴 𝑡𝑜 𝐵 on 𝐶 is given by
𝐵
𝑊 = ∫𝐴 𝐹⃗
...
If 𝐴⃗ = (𝑥 − 𝑦)𝑖̂ + (𝑥 + 𝑦)𝑗̂, evaluate ∮𝐶 𝐴⃗
...

Solution
...
𝑑𝑟⃗ ; 𝐶: 𝑦 = 𝑥 2 and 𝑥 = 𝑦 2

On 𝐶1 ; 𝑦 = 𝑥 2 ⇒ 𝑑𝑦 = 2𝑥𝑑𝑥 and 𝑥 → 0 𝑡𝑜 1
On 𝐶2 ; 𝑥 = 𝑦 2 ⇒ 𝑑𝑥 = 2𝑦𝑑𝑦 and 𝑦 → 0 𝑡𝑜1
𝐴⃗
...
𝑑𝑟⃗ = ∮𝐶1 𝐴⃗
...
𝑑𝑟⃗
1

0

= ∫0 [(𝑥 − 𝑥 2 )𝑑𝑥 + (𝑥 + 𝑥 2 )2𝑥𝑑𝑥] + ∫1 [(𝑦 2 − 𝑦)2𝑦𝑑𝑦 + (𝑦 2 + 𝑦)𝑑𝑦]
1

0

= ∫0 (2𝑥 3 + 𝑥 2 + 𝑥 )𝑑𝑥 + ∫1 (2𝑦 3 − 𝑦 2 + 𝑦)𝑑𝑦
2

1

1

2

1

1

𝟐

=4+3+2−4+3−2= 𝟑
Example 2
...


23 | P a g e

Solution
...
Find the work done by force field 𝐹⃗ = 𝑦𝑧𝑖̂ + (𝑥𝑧 + 1)𝑗̂ + 𝑥𝑦𝑘̂ in moving a particle
in from (1,0,0) 𝑡𝑜 (2,1,4)
...
Here, 𝐹⃗ = 𝑦𝑧𝑖̂ + (𝑥𝑧 + 1)𝑗̂ + 𝑥𝑦𝑘̂ , 𝑑𝑟⃗ = 𝑖̂𝑑𝑥 + 𝑗̂𝑑𝑦 + 𝑘̂𝑑𝑧
𝐹⃗
...
𝑑𝑟⃗ = ∫𝐶 [𝑦𝑧𝑑𝑥 + (𝑥𝑧 + 1)𝑑𝑦 + 𝑥𝑦𝑑𝑧]
1

= ∫0 [4𝑡 2 𝑑𝑡 + (4𝑡 2 + 4𝑡 + 1)𝑑𝑡 + (𝑡 2 + 𝑡)4𝑑𝑡]
1

= ∫0 (12𝑡 2 + 8𝑡 + 1)𝑑𝑡 = [4𝑡 3 + 4𝑡 2 + 𝑡]10 = 𝟗
Example 4
...


Solution
...
𝑑𝑟⃗ = 3𝑥 2 𝑑𝑥 + (2𝑥𝑧 − 𝑦)𝑑𝑦 + 𝑧𝑑𝑧
Here, 𝑥 2 = 4𝑦 and 3𝑥 3 = 8𝑧
Or 𝑦 =

𝑥2
4

𝑥

3𝑥 3

2

8

⇒ 𝑑𝑦 = 𝑑𝑥 and 𝑧 =

⇒ 𝑑𝑧 =

9𝑥 2
8

𝑑𝑥 and 𝑥 → 0 𝑡𝑜 2

Required work done (𝑊) = ∫𝐶 𝐹⃗
...
Evaluate ∫𝐶 𝐹⃗
...


Solution
...
𝑑𝑟⃗ = ∫𝐶 [(𝑦𝑧 + 2𝑥 )𝑑𝑥 + 𝑥𝑧𝑑𝑦 + (𝑥𝑦 + 2𝑧)𝑑𝑧]

25 | P a g e

𝑥𝑑𝑥

On 𝐶; 𝑥 2 + 𝑦 2 = 1 ⇒ 𝑦 = √1 − 𝑥 2 ⇒ 𝑑𝑦 = − √1−𝑥 2
and 𝑧 = 1 ⇒ 𝑑𝑧 = 0

and 𝑥 → 0 𝑡𝑜1
𝑥2

1

∴ 𝐼 = ∫0 [(√1 − 𝑥 2 + 2𝑥) − √1−𝑥 2] 𝑑𝑥
1

1

= ∫0 [2√1 − 𝑥 2 + 2𝑥 − √1−𝑥 2] 𝑑𝑥
1

= [𝑥√1 − 𝑥 2 + sin−1 𝑥 + 𝑥 2 − sin−1 𝑥]0
1

= [𝑥√1 − 𝑥 2 + 𝑥 2 ]0 = 𝟏

2
...
2 Circulation: If 𝑉⃗⃗ represents the velocity of a fluid particle and 𝐶 is the closed curve, then
⃗⃗
...

the integral ∮𝐶 𝑉

Example 1
...


Solution
...
𝑑𝑟⃗ = 𝑒 𝑥 sin 𝑦 𝑑𝑥 + 𝑒 𝑥 cos 𝑦 𝑑𝑦

26 | P a g e

On 𝑂𝐴; 𝑦 = 0, 𝑑𝑦 = 0, 𝑥 → 0 𝑡𝑜 1
𝝅

On 𝐴𝐵; 𝑥 = 1, 𝑑𝑥 = 0, 𝑦 → 0 𝑡𝑜 𝟐
𝝅

On 𝐵𝐶; 𝑦 = , 𝑑𝑦 = 0, 𝑥 → 1 𝑡𝑜 0
𝟐

On 𝐶𝑂; 𝑥 = 0, 𝑑𝑥 = 0, 𝑦 →

𝝅
𝟐

𝑡𝑜 0

Circulation = ∫𝐶 𝐹⃗
...
𝑑𝑟⃗
𝜋

0

0

= 0 + ∫02 𝑒 cos 𝑦 𝑑𝑦 + ∫1 𝑒 𝑥 𝑑𝑥 + ∫𝜋 cos 𝑦 𝑑𝑦
2

𝜋
2

= 𝑒[sin 𝑦]0 + [𝑒 𝑥 ]10 + [sin 𝑦]0𝜋 = 𝑒 + (1 − 𝑒) + (0 − 1) = 𝟎
2

Exercise
1
...


2
...
𝑑𝑟⃗, where 𝐶 is the arc of the parabola 𝑦 = 2𝑥 2 from (0,0)
to (1,2)
...
A vector field is given by 𝐹⃗ = (sin 𝑦)𝑖̂ + 𝑥 (1 + cos 𝑦)𝑗̂
...


4
...
𝑑𝑟⃗ from
(0,0,0) 𝑡𝑜 (1,1,1) along the following paths 𝐶:
(i) 𝑥 = 𝑡, 𝑦 = 𝑡 2 , 𝑧 = 𝑡 3
(ii) the straight line joining (0,0,0) 𝑡𝑜 (1,1,1)
(iii) the straight line joining (0,0,0) 𝑡𝑜 (1,0,0) then to (1,1,0) and then to (1,1,1)
...
Evaluate ∫𝐶 𝐹⃗
...


Answers

1
...
− 6

3
...
(i) 5

(ii)

13
3

(iii)

23
3

5
...
2 Surface Integral
Any integral which is to be evaluated over a surface is called a surface integral
...

Flux: Suppose 𝑆 is a piecewise smooth surface and 𝐹⃗ (𝑥, 𝑦, 𝑧) is a vector function of position
defined and continuous over 𝑆
...
Then 𝐹⃗
...
𝒏
̂ 𝒅𝑺
...
The integral of 𝐹⃗
...

Let us associate with the differential of surface area
𝑑𝑆 a vector 𝑑𝑆⃗ (called vector area) whose magnitude
is 𝑑𝑆 and whose direction is that of 𝑛̂
...
Therefore, we can write
∬ 𝐹⃗
...
But this is possible
only when any line perpendicular to coordinate plane chosen meets the surface 𝑆 in not more than
one point
...
Then
∬ 𝐹⃗
...
Then
∬ 𝐹⃗
...
𝑘̂|

𝑅

𝑑𝑥 𝑑𝑦
|𝑛̂
...
Then
∬ 𝐹⃗
...
𝑗̂|

28 | P a g e

Example 1: Evaluate ∬𝑆 𝐴⃗
...


Solution: We have 𝐴⃗ = (𝑥 + 𝑦 2 )𝑖̂ − 2𝑥𝑗̂ + 2𝑦𝑧𝑘̂
To find 𝑛̂
A vector normal to the surface 𝑆 is given by
∇𝑆 = ∇(2𝑥 + 𝑦 + 2𝑧 − 6) = 2𝑖̂ + 𝑗̂ + 2𝑘̂
∴ 𝑛̂ = a unit vector normal to the surface 𝑆
̂
2𝑖̂ +𝑗̂ +2𝑘

= √22
𝑛̂ =

+12+22

=

̂
2𝑖̂ +𝑗̂ +2𝑘
3

̂
2𝑖̂ +𝑗̂ +2𝑘
3

∴ 𝐴⃗
...
(

̂
2𝑖̂ +𝑗̂ +2𝑘
3

)

1

= [2(𝑥 + 𝑦 2 ) − 2𝑥 + 4𝑦𝑧]
3

2
𝐴⃗
...
𝑛̂

𝑑𝑥 𝑑𝑦
̂|
|𝑛̂
...
𝑘̂ =
𝑛̂
...
𝑛̂ = ∬
𝑆

𝑅

2
𝑑𝑥 𝑑𝑦
𝑦(𝑦 + 2𝑧)
= ∬ 𝑦 (𝑦 + 2𝑧) 𝑑𝑥 𝑑𝑦
2
3
𝑅
3

= ∬𝑅 𝑦 [𝑦 + 2 (

6−2𝑥−𝑦
2

) 𝑑𝑥 𝑑𝑦]

[∵ on the surface 𝑆(2𝑥 + 𝑦 + 2𝑧 = 6): 𝑧 =

6−2𝑥−𝑦
2

]

= ∬𝑅 𝑦(6 − 2𝑥) 𝑑𝑥 𝑑𝑦
3

6−2𝑥

= ∫𝑥=0 ∫𝑦=0 𝑦(6 − 2𝑥) 𝑑𝑦 𝑑𝑥

29 | P a g e

6−2𝑥

𝑦2

3

= ∫𝑥=0(6 − 2𝑥) ( 2 )

𝑑𝑥

0

1

3

1

3

= ∫𝑥=0(6 − 2𝑥)3 𝑑𝑥
2
3

= ∫𝑥=0(6 − 2𝑥)3 𝑑𝑥 = 4 ∫0 (3 − 𝑥)3 𝑑𝑥
2
3

(3 − 𝑥)4
] = 81
= 4[
−4
0
∬ 𝐴⃗
...
𝑛̂ 𝑑𝑆, where 𝐴⃗ = 𝑧𝑖̂ + 𝑥𝑗̂ − 3𝑦 2 𝑧𝑘̂ and 𝑆 is the surface of the
cylinder 𝑥 2 + 𝑦 2 = 16 included in the first octant between 𝑧 = 0 and 𝑧 = 5
...
𝑛̂ = (𝑧𝑖̂ + 𝑥𝑗̂ − 3𝑦 2 𝑧𝑘̂ )
...
𝑛̂

𝑑𝑦 𝑑𝑧
|𝑛̂
...
𝑖̂ = (

𝑥𝑖̂ +𝑦𝑗̂
4

𝑥

)
...
𝑛̂ = ∬𝑅
4

1
4

𝑥 (𝑦 + 𝑧)

5

𝑑𝑦 𝑑𝑧
𝑥
4

= ∬𝑅 (𝑦 + 𝑧)𝑑𝑦 𝑑𝑧

4

= ∫𝑦=0 ∫𝑧=0(𝑦 + 𝑧) 𝑑𝑧 𝑑𝑦 = ∫𝑦=0 (𝑦𝑧 +
5

= (2 𝑦 2 +

25
2

𝑦)

4
𝑦=0

𝑧2

5

)

2 𝑧=0

4

𝑑𝑦 = ∫𝑦=0 (5𝑦 +

25
2

) 𝑑𝑦

= 40 + 50 = 90

∬ 𝐴⃗
...
𝑛̂ where 𝐹⃗ = 4𝑥𝑖̂ − 2𝑦 2 𝑗̂ + 𝑧 2 𝑘̂ and 𝑆 is the surface of the
bounding region 𝑥 2 + 𝑦 2 = 4, z=0, z=3
...
𝑛̂ 𝑑𝑆1 + ∬𝑆 ⃗𝐹⃗
...
𝑛̂ 𝑑𝑆3
1

2

3


...
𝑛̂ 𝑑𝑆1 = ∬𝑆 (4𝑥𝑖̂ − 2𝑦 2 𝑗̂)
...
𝑛̂ 𝑑𝑆1 = 0
∴∬ 𝐹
𝑆1

On 𝑆2 (𝑧 = 3), we have 𝑛̂ = 𝑘̂, ⃗𝐹⃗= 4𝑥𝑖̂ − 2𝑦 2 𝑗̂ + 9𝑘̂
⃗⃗
...
(𝑘̂) 𝑑𝑆2
∴ ∬𝑆 𝐹
𝑆
2

2

=∬ 9 𝑑𝑥 𝑑𝑦 = 9 ∬𝑆 𝑑𝑥 𝑑𝑦
2

= 9 × area of the surface 𝑆2 = 9(𝜋
...
𝑛̂ 𝑑𝑆2 = 36𝜋
∴∬ 𝐹
𝑆2

On 𝑆3 (𝑥 2 + 𝑦 2 = 4), we have
𝑛̂ = a unit vector normal to the surface 𝑆3 =
⃗𝐹⃗
...
(

2𝑥𝑖̂ +2𝑦𝑗̂
√4𝑥 2+4𝑦 2

=

𝑥𝑖̂ +𝑦𝑗̂
2

𝑥𝑖̂ + 𝑦𝑗̂
) = 2𝑥 2 − 𝑦 3
2

Also on 𝑆3 , 𝑖
...

⃗⃗
...
𝑛̂ 𝑑𝑆3 = 48𝜋
𝑆3

⃗⃗
...
𝑛̂ 𝑑𝑆 = 84𝜋
∬ 𝐹
𝑆

32 | P a g e

Example 4: Evaluate ∬𝑆 (𝑦𝑧𝑖̂ + 𝑧𝑥𝑗̂ + 𝑥𝑦𝑘̂ )
...


Solution: Let ∬𝑆 (𝑦𝑧𝑖̂ + 𝑧𝑥𝑗̂ + 𝑥𝑦𝑘̂ )
...
( 𝑎
𝑎

Now let 𝑅 be the projection of 𝑆 on 𝑥𝑦-plane, therefore
∬𝑆 𝐹⃗
...
𝑘

= 3 ∬𝑅 𝑥𝑦 𝑑𝑥 𝑑𝑦

𝑥𝑦 𝑑𝑦 𝑑𝑥

√𝑎 2−𝑥 2
𝑦=0

𝑥 𝑑𝑥
3 𝑎 2𝑥 2

𝑎

= 2 ∫𝑥=0(𝑎2 − 𝑥 2 )𝑥 𝑑𝑥 = 2 (

2

−

𝑥4

𝑎

) =
4
0

3𝑎 4
8

Hence, from (1) we get
∬ (𝑦𝑧𝑖̂ + 𝑧𝑥𝑗̂ + 𝑥𝑦𝑘̂ )
...


Evaluate ∬𝑆 𝐹⃗
...


2
...
𝑛̂ 𝑑𝑆 = 2 where 𝐹⃗ = 4𝑥𝑧𝑖̂ − 𝑦 2 𝑗̂ + 𝑦𝑧𝑘̂ and 𝑆 is the surface of the cube

bounded by the planes 𝑥 = 0, 𝑥 = 1, 𝑦 = 0, 𝑦 = 1, 𝑧 = 0, 𝑧 = 1
...


Evaluate ∬𝑆 𝐹⃗
...


Answers
2
...
24

3
...
3 Volume Integral: Any integral that is evaluated over a volume is called a volume integral
...
Then
The volume integral = ∭𝑉 𝐹⃗ 𝑑𝑣

Example 1
...
Firstly, we calculate
𝑖̂
𝜕
𝛻 × 𝐹⃗ = ||
𝜕𝑥
2𝑥 2 − 3𝑧
2

2−𝑥

∭𝑉 𝛻 × 𝐹⃗ = ∫ ∫
0

2

0

2−𝑥

= ∫0 ∫0
2

𝑘̂
𝜕 |
| = 𝑗̂ − 2𝑦𝑘̂
𝜕𝑧
−4𝑥

(𝑗̂ − 2𝑦𝑘̂ )𝑑𝑧𝑑𝑦𝑑𝑥

0

(𝑗̂ − 2𝑦𝑘̂)(4 − 2𝑥 − 2𝑦)𝑑𝑦𝑑𝑥

2−𝑥

= 𝑗̂ ∫0 ∫0

4−2𝑥−2𝑦

∫

𝑗̂
𝜕
𝜕𝑦
−2𝑥𝑦

2

2−𝑥

(4 − 2𝑥 − 2𝑦)𝑑𝑦𝑑𝑥 − 4𝑘̂ ∫0 ∫0

[(2 − 𝑥)𝑦 − 𝑦 2 ]𝑑𝑦𝑑𝑥

2
2 1
1
= 𝑗̂ ∫0 (2 − 𝑥)2 𝑑𝑥 − 4𝑘̂ ∫0 [2 (2 − 𝑥)3 − 3 (2 − 𝑥)3 ] 𝑑𝑥
2

2

2

= 𝑗̂ ∫0 (2 − 𝑥)2 𝑑𝑥 − 3 𝑘̂ ∫0 (2 − 𝑥)3 𝑑𝑥
1

1

= − 3 𝑗̂[(2 − 𝑥)3 ]20 + 6 𝑘̂[(2 − 𝑥)4 ]20
1
1
8
8
𝟖
̂)
= − 𝑗̂[0 − 8] + 𝑘̂[0 − 16] = 𝑗̂ − 𝑘̂ = (𝒋̂ − 𝒌
3

6

3

3

𝟑

34 | P a g e

Example 2
...

2

4

2

4

2

Solution
...
4 Gauss Divergence Theorem (Relation between surface integrals and volume
integral)
⃗⃗ is a vector point function having continuous first partial derivatives in the
Statement: If 𝐹
region 𝑉 bounded by a closed surface 𝑆, then
∬ ⃗𝐹⃗
...

Cartesian equivalent of Divergence Theorem
⃗⃗= 𝑑𝑖𝑣 ⃗𝐹⃗= 𝜕𝐹1 + 𝜕𝐹2 + 𝜕𝐹3
Let ⃗𝐹⃗= 𝐹1 𝑖̂ + 𝐹2 𝑗̂ + 𝐹3 𝑘̂
...
𝐹
𝜕𝑥

𝜕𝑦

𝜕𝑧

Therefore
∬ (𝐹1 𝑑𝑦 𝑑𝑧 + 𝐹2 𝑑𝑧 𝑑𝑥 + 𝐹3 𝑑𝑥 𝑑𝑦) = ∭ (
𝑆

𝑉

𝜕𝐹1 𝜕𝐹2 𝜕𝐹3
+
+
) 𝑑𝑥 𝑑𝑦 𝑑𝑧
𝜕𝑥
𝜕𝑦
𝜕𝑧

Example 1: Verify divergence theorem for ⃗𝐹⃗= 4𝑥𝑧𝑖̂ − 𝑦 2 𝑗̂ + 𝑦𝑧𝑘̂ taken over the cube bounded
by the planes 𝑥 = 0, 𝑥 = 1, 𝑦 = 0, 𝑦 = 1, 𝑧 = 0, 𝑧 = 1
...
𝑛̂ 𝑑𝑆 = ∭ 𝑑𝑖𝑣 ⃗𝐹⃗ 𝑑𝑉
𝑆

𝑉

where 𝑉 is the region bounded by the closed surface 𝑆
...
𝒏̂ 𝒅𝑺
To evaluate ∬𝑺 𝑭
Here 𝑆 is the surface of the cube bounded by the 6 plane surfaces
...
𝑛̂ 𝑑𝑆1 + ∬ ⃗𝐹⃗
...
𝑛̂ 𝑑𝑆3 + ∬ ⃗𝐹⃗
...
𝑛̂ 𝑑𝑆5 + ∬𝑆 ⃗𝐹⃗
...
𝑛̂ 𝑑𝑆1 = ∫1 ∫1 0 𝑑𝑥 𝑑𝑦 = 0
∴ ∬𝑆 𝐹
𝑦=0 𝑥=0
1

36 | P a g e

Over the surface 𝑆2 (𝐺𝐷𝐸𝐹): 𝑧 = 1, 𝑑𝑧 = 0,⃗⃗⃗⃗
𝐹 = 4𝑥𝑖̂ − 𝑦 2 𝑗̂ + 𝑦𝑘̂ , 𝑛̂ = 𝑘̂, 𝑑𝑆2 = 𝑑𝑥 𝑑𝑦
2

1

⃗⃗
...
𝑛̂ 𝑑𝑆3 = ∫𝑧=0 ∫𝑥=0 0 𝑑𝑥 𝑑𝑧 = 0
3

Over the surface 𝑆4 (𝐶𝐷𝐺𝐵): 𝑦 = 1, 𝑑𝑦 = 0,⃗⃗⃗⃗
𝐹 = 4𝑥𝑧𝑖̂ − 𝑗̂ + 𝑧𝑘̂ , 𝑛̂ = 𝑗̂, 𝑑𝑆4 = 𝑑𝑥 𝑑𝑧
1
1
1
1
∴ ∬𝑆 ⃗𝐹⃗
...
𝑛̂ 𝑑𝑆5 = ∫1 ∫1 4𝑧 𝑑𝑦 𝑑𝑧 = 4 ∫1 𝑧 𝑑𝑧 ∫1 𝑑𝑦 = 4 (𝑧 ) (𝑦)10 = 2
∴ ∬𝑆 𝐹
𝑧=0 𝑦=0
𝑧=0
𝑦=0
2 0

5

Over the surface 𝑆6 (𝑂𝐶𝐷𝐸): 𝑥 = 0, 𝑑𝑥 = 0,⃗⃗⃗⃗
𝐹 = −𝑦 2 𝑗̂ + 𝑦𝑧𝑘̂ , 𝑛̂ = 𝑖̂, 𝑑𝑆6 = 𝑑𝑦 𝑑𝑧
⃗⃗
...
𝑛̂ 𝑑𝑆 = 0 + 2 + 0 − 1 + 2 + 0 = 2

∴ ∬ ⃗𝐹⃗
...
(4𝑥𝑧𝑖̂ − 𝑦 2 𝑗̂ + 𝑦𝑧𝑘̂ ) = 4𝑧 − 2𝑦 + 𝑦 = 4𝑧 − 𝑦
∴ 𝑑𝑖𝑣 ⃗𝐹⃗= ∇
...
𝑛̂ 𝑑𝑆 = ∭𝑉 𝑑𝑖𝑣 ⃗𝐹⃗ 𝑑𝑉
Hence the divergence theorem is verified
...


Solution: By Gauss divergence theorem, we know that
⃗⃗
...

⃗⃗
...
𝑛̂ 𝑑𝑆 = ∬ 𝐹
⃗⃗
...
𝑛̂ 𝑑𝑆2 + ∬ 𝐹
⃗⃗
...
(1)

3

On 𝑆1 (𝑧 = 0), we have 𝑛̂ = −𝑘̂ , ⃗𝐹⃗= 4𝑥𝑖̂ − 2𝑦 2 𝑗̂
⃗⃗
...
(−𝑘̂ ) 𝑑𝑆1 = 0
∴ ∬𝑆 𝐹
𝑆
1

1

∴ ∬ ⃗𝐹⃗
...
𝑛̂ 𝑑𝑆2 = ∬ (4𝑥𝑖̂ − 2𝑦 2 𝑗̂ + 9𝑘̂ )
...
22 ) = 36𝜋

∬ ⃗𝐹⃗
...
𝑛̂ = (4𝑥𝑖̂ − 2𝑦 2 𝑗̂ + 𝑧 2 𝑘̂ )
...
𝑥 2 + 𝑦 2 = 4, 𝑥 = 2 cos 𝜃, 𝑦 = 2 sin 𝜃 and 𝑑𝑆3 = 2 𝑑𝜃 𝑑𝑧
To cover the whole surface 𝑆3 , 𝑧 varies from 0 to 3 and 𝜃 varies from 0 to 2𝜋
...
𝑛̂ 𝑑𝑆3 = ∫𝜃=0 ∫𝑧=0[2(2 cos 𝜃)2 − (2 sin 𝜃)3 ] 2 𝑑𝑧 𝑑𝜃
3

2𝜋

3

2𝜋

= 16 ∫ ∫(𝑐𝑜𝑠 2 𝜃 − 𝑠𝑖𝑛3 𝜃) 𝑑𝑧 𝑑𝜃 = 48 ∫ (𝑐𝑜𝑠 2 𝜃 − 𝑠𝑖𝑛3 𝜃) 𝑑𝜃 = 48𝜋
𝜃=0 𝑧=0

𝜃=0

39 | P a g e

⃗⃗
...
𝑛̂ 𝑑𝑆 = 0 + 36𝜋 + 48𝜋 = 84𝜋

∬ ⃗𝐹⃗
...
𝐹
⃗⃗= (𝑖̂
∴ 𝑑𝑖𝑣 𝐹

𝜕
𝜕𝑥

+ 𝑗̂

𝜕
𝜕𝑦

+ 𝑘̂

𝜕
𝜕𝑧

)
...
𝑛̂ 𝑑𝑆 = ∭ 𝑑𝑖𝑣 ⃗𝐹⃗ 𝑑𝑉
𝑆

𝑉

Hence theorem is verified
...
4
...
𝑛̂ 𝑑𝑆, where 𝑆 is the part of the sphere
𝑥 2 + 𝑦 2 + 𝑧 2 = 1 above the xy-plane and bounded by this plane
...
𝑛̂ 𝑑𝑆 = ∭ 𝑑𝑖𝑣 ⃗𝐹⃗ 𝑑𝑉
𝑆

𝑉

where 𝑉 is the region bounded by the closed surface 𝑆
...
Then by divergence theorem, we have
∬ (𝑦 2 𝑧 2 𝑖̂ + 𝑧 2 𝑥 2 𝑗̂ + 𝑧 2 𝑦 2 𝑘̂ )
...
(1)
41 | P a g e

Changing Cartesian coordinates (𝑥, 𝑦, 𝑧) to spherical polar coordinates (𝑟, 𝜃, ∅) by putting
𝑥 = 𝑟 sin 𝜃 cos ∅, 𝑦 = 𝑟 sin 𝜃 sin ∅, 𝑧 = 𝑟 cos 𝜃 , |𝐽 | = 𝑟 2 sin 𝜃,
𝑑𝑉 = 𝑑𝑟 𝑑𝜃𝑑∅
𝜋

To cover 𝑉, the limits of 𝑟 will be 0 to 1, those of 𝜃 will be 0 to 2 and those of ∅ will be 0 to 2𝜋
...
2
12
12
Therefore,
∬ (𝑦 2 𝑧 2 𝑖̂ + 𝑧 2 𝑥 2 𝑗̂ + 𝑧 2 𝑦 2 𝑘̂ )
...


Solution: By Gauss divergence theorem, we know that
∬ ⃗𝐹⃗
...

Let ∬𝑆 (𝑎2 𝑥 2 + 𝑏2 𝑦 2 + 𝑐 2 𝑧 2 )1⁄2 𝑑𝑆 = ∬𝑆 ⃗𝐹⃗
...
𝑛̂ = (𝑎2 𝑥 2 + 𝑏2 𝑦 2 + 𝑐 2 𝑧 2 )1⁄2
where 𝐹
We have
∇𝑆

𝑛̂ = a unit vector normal to surface 𝑆 = |∇𝑆| =

̂)
2(𝑎𝑥𝑖̂ +𝑏𝑦𝑗̂ +𝑐𝑧𝑘
√4𝑎 2 𝑥 2+4𝑏 2𝑦 2 +4𝑐 2𝑧 2

=

̂)
(𝑎𝑥𝑖̂ +𝑏𝑦𝑗̂ +𝑐𝑧𝑘
√𝑎 2𝑥 2+𝑏 2 𝑦 2+𝑐 2 𝑧 2

∴ ⃗𝐹⃗
...

⟹ 𝐹

√𝑎2 𝑥 2

+

𝑏2 𝑦 2

+

𝑐2𝑧2

= (𝑎2 𝑥 2 + 𝑏2 𝑦 2 + 𝑐 2 𝑧 2 )1⁄2

⟹ ⃗𝐹⃗
...
𝐹
⃗⃗= (𝑖̂
⟹ 𝑑𝑖𝑣 𝐹

=

𝜕
𝜕
𝜕
+ 𝑗̂
+ 𝑘̂ )
...


Solution: By Cartesian form of divergence theorem we know that
∬ (𝐹1 𝑑𝑦 𝑑𝑧 + 𝐹2 𝑑𝑧 𝑑𝑥 + 𝐹3 𝑑𝑥 𝑑𝑦) = ∭ (
𝑆

𝑉

𝜕𝐹1 𝜕𝐹2 𝜕𝐹3
+
+
) 𝑑𝑥 𝑑𝑦 𝑑𝑧
𝜕𝑥
𝜕𝑦
𝜕𝑧

43 | P a g e

𝜕
𝜕
𝜕
(𝑦) + (𝑧)] 𝑑𝑥 𝑑𝑦 𝑑𝑧
∴ ∬ (𝑥 𝑑𝑦 𝑑𝑧 + 𝑦 𝑑𝑧 𝑑𝑥 + 𝑧 𝑑𝑥 𝑑𝑦) = ∭ [ (𝑥) +
𝜕𝑥
𝜕𝑦
𝜕𝑧
𝑆
𝑉
6

6−𝑥 6−𝑥−2𝑦
2
3

= 3 ∭ 𝑑𝑥 𝑑𝑦 𝑑𝑧 = 3 ∫ ∫
𝑉

6

∫

𝑥=0 𝑦=0

6

𝑑𝑧 𝑑𝑦 𝑑𝑥 = 3 ∫ ∫ (

𝑧=0

6−𝑥
2

6

= ∫ ∫ (6 − 𝑥 − 2𝑦)𝑑𝑦 𝑑𝑥 = ∫ [(6 − 𝑥)𝑦 −
𝑥=0 𝑦=0

𝑥=0 𝑦=0

6−𝑥
2
𝑦 2 ]𝑦=0

𝑥=0

6

6−𝑥
2

6 − 𝑥 − 2𝑦
) 𝑑𝑦 𝑑𝑥
3

6

𝑑𝑥 = ∫ [
𝑥=0

(6 − 𝑥)2 (6 − 𝑥)2
] 𝑑𝑥
−
2
4

6

(6 − 𝑥)2
1 (6 − 𝑥)3
1
(216) = 18
] =
= ∫
𝑑𝑥 = [
4
4
−3
12
0
𝑥=0

∴ ∬ (𝑥 𝑑𝑦 𝑑𝑧 + 𝑦 𝑑𝑧 𝑑𝑥 + 𝑧 𝑑𝑥 𝑑𝑦) = 18
𝑆

Example 6: The vector field ⃗𝐹⃗= 𝑥 2 𝑖̂ + 𝑧 𝑗̂ + 𝑦𝑧𝑘̂ is defined over the volume of the cuboid
given by 0 ≤ 𝑥 ≤ 𝑎, 0 ≤ 𝑦 ≤ 𝑏, 0 ≤ 𝑧 ≤ 𝑐 enclosing the surface 𝑆, evaluate ∬𝑆 ⃗𝐹⃗
...
𝑛̂ 𝑑𝑆 = ∭ 𝑑𝑖𝑣 ⃗𝐹⃗ 𝑑𝑉
𝑆

𝑉

where 𝑉 is the region bounded by the closed surface 𝑆
...
𝑑𝑆⃗ = ∬ 𝐹
⃗⃗
...
(𝑥 2 𝑖̂ + 𝑧 𝑗̂ + 𝑦𝑧 𝑘̂ ) =
𝜕𝑥
𝜕𝑦
𝜕𝑧
𝜕𝑥
𝜕𝑦
𝜕𝑧

⃗⃗= (𝑖̂
𝑑𝑖𝑣 ⃗𝐹⃗= ∇
...
𝑑𝑆⃗ = ∭ (2𝑥 + 𝑦)𝑑𝑉 = ∫
𝑆
𝑎

=∫

𝑉

𝑐

∫ (2𝑥 + 𝑦)𝑑𝑧 𝑑𝑦 𝑑𝑥

𝑥=0 𝑦=0 𝑧=0

𝑏

∫

∫

𝑏

𝑎

(2𝑥

𝑥=0 𝑦=0

+ 𝑦)(𝑧)𝑐0

𝑎
𝑦2
𝑏
𝑑𝑦 𝑑𝑥 = 𝑐 ∫ (2𝑥𝑦 + )
𝑑𝑥 = 𝑏𝑐 ∫ (2𝑥 + ) 𝑑𝑥
2 𝑦=0
2
𝑥=0
𝑥=0

𝑏 𝑎
𝑏
= 𝑏𝑐 (𝑥 2 + 𝑥) = 𝑎𝑏𝑐 (𝑎 + )
2 0
2
𝑏
⃗⃗
...
Verify divergence theorem for 𝐹
the rectangular parallelepiped 0 ≤ 𝑥 ≤ 𝑎, 0 ≤ 𝑦 ≤ 𝑏, 0 ≤ 𝑧 ≤ 𝑐
...
Verify divergence theorem for ⃗𝐹⃗= (2𝑥 − 𝑧)𝑖̂ + 𝑥 2 𝑦𝑗̂ − 𝑥𝑧 2 𝑘̂ taken over the region
bounded by 𝑥 = 0, 𝑥 = 1, 𝑦 = 0, 𝑦 = 1, 𝑧 = 0, 𝑧 = 1
...
Use divergence theorem to evaluate∬𝑆 ⃗𝐹⃗
...

4
...

5
...
3𝜋𝑏𝑎4

4
...
0

45 | P a g e

2
...
Let 𝑀 and 𝑁 be continuous functions of 𝑥 and 𝑦 having continuous partial
derivatives

𝜕𝑀
𝜕𝑦

and

𝜕𝑁
𝜕𝑥

in 𝑅
...


Example 1: Verify Green’s theorem in the plane for ∮𝐶 [(𝑥 2 − 2𝑥𝑦)𝑑𝑥 + (𝑥 2 𝑦 + 3)𝑑𝑦] where
𝐶 is the boundary of the region defined by 𝑦 2 = 8𝑥 and 𝑥 = 2
...

Here 𝑀 = 𝑥 2 − 2𝑥𝑦 and 𝑁 = 𝑥 2 𝑦 + 3

The parabola 𝑦 2 = 8𝑥 and the straight line 𝑥 = 2 intersect at the points 𝑃(2,4) and 𝑄(2, −4)
...

46 | P a g e

We have
𝜕𝑁

𝜕𝑀

∬𝑅 ( 𝜕𝑥 −

𝜕𝑦

) 𝑑𝑥 𝑑𝑦 = ∬𝑅 [
2

𝜕
𝜕𝑥

𝜕

(𝑥 2 𝑦 + 3) − (𝑥 2 − 2𝑥𝑦)] 𝑑𝑥 𝑑𝑦
𝜕𝑦

𝑦=√8𝑥

=∬𝑅 (2𝑥𝑦 + 2𝑥)𝑑𝑥 𝑑𝑦 = ∫𝑥=0 ∫𝑦=−√8𝑥 (2𝑥𝑦 + 2𝑥)𝑑𝑦 𝑑𝑥
2

𝑦=√8𝑥

∫

2

2𝑥 (𝑦 + 1)𝑑𝑦 𝑑𝑥 = ∫

∫

𝑥=0 𝑦=−√8𝑥
2
𝑥=0

2𝑥 𝑑𝑦 𝑑𝑥

𝑥=0 𝑦=−√8𝑥

𝑦=√8𝑥
2𝑥 (𝑦)𝑦=−√8𝑥
= 2∫

=∫

𝑦=√8𝑥

∫

2

𝑥(√8𝑥 + √8𝑥) 𝑑𝑥

𝑥=0

2

2

= 4√8 ∫

𝑥 √𝑥 𝑑𝑥 = 4√8 ∫

𝑥=0

3

𝑥 2 𝑑𝑥 =

𝑥=0

5 2
16
128
√2 (𝑥 2 ) =
5
5
0

Hence,
∬ (
𝑅

𝜕𝑁 𝜕𝑀
128
−
) 𝑑𝑥 𝑑𝑦 =
𝜕𝑥
𝜕𝑦
5

--------------------------------- (1)

Verification of the theorem:
Now let us evaluate the line integral along 𝐶: 𝐶1 ∪ 𝐶2
...


𝑦4 𝑦3 1
𝑦5
− ) 𝑦 𝑑𝑦 + (
+ 3) 𝑑𝑦]
64
4 4
64
4

𝑦5
𝑦5 𝑦4
𝑦5
𝑦5 𝑦4
= ∫ [
+
−
+ 3] 𝑑𝑦 = − ∫ [
+
−
+ 3] 𝑑𝑦
256 64 16
256 64 16
𝑦=4

𝑦=−4

47 | P a g e

4

4

= −2 ∫ (−
𝑦=0

∴ 𝐼1 =

𝑦4
𝑦5
128
+ 3) 𝑑𝑦 = −2 (−
+ 3𝑦) =
− 24
16
80
5
0

128
− 24
5

Now 𝐼2 = ∫𝐶 [(𝑥 2 − 2𝑥𝑦)𝑑𝑥 + (𝑥 2 𝑦 + 3)𝑑𝑦]
2

Along 𝐶2 : 𝑥 = 2, 𝑑𝑥 = 0 and 𝑦 varies from 4 to −4
...
∮ [(𝑥 2 − 2𝑥𝑦)𝑑𝑥 + (𝑥 2 𝑦 + 3)𝑑𝑦] =
𝐶

128
5

128
5

----------------------- (3)

From (1) and (3),we see that

∮ [(𝑥 2 − 2𝑥𝑦)𝑑𝑥 + (𝑥 2 𝑦 + 3)𝑑𝑦] = ∬ (
𝐶

𝑅

𝜕𝑁 𝜕𝑀
−
) 𝑑𝑥 𝑑𝑦
𝜕𝑥
𝜕𝑦

which verifies Green’s theorem in the plane
...


Solution: By Green’s theorem in plane, we have
∮ (𝑀 𝑑𝑥 + 𝑁 𝑑𝑦) = ∬ (
𝐶

𝑅

𝜕𝑁 𝜕𝑀
−
) 𝑑𝑥 𝑑𝑦
𝜕𝑥
𝜕𝑦

where 𝑅 is the region bounded by closed curve 𝐶
...
The
positive direction in traversing 𝐶 is as shown in the figure and 𝑅 is the region bounded by the
closed curve 𝐶
...

We have
∮𝐶 [(𝑥𝑦 + 𝑦 2 )𝑑𝑥 + 𝑥 2 𝑑𝑦] = ∫𝐶1 [(𝑥𝑦 + 𝑦 2 )𝑑𝑥 + 𝑥 2 𝑑𝑦] + ∫𝐶2 [(𝑥𝑦 + 𝑦 2 )𝑑𝑥 + 𝑥 2 𝑑𝑦]
or, 𝐼 = 𝐼1 + 𝐼2

----------------------------(2)

Now 𝐼1 = ∫𝐶 [(𝑥𝑦 + 𝑦 2 )𝑑𝑥 + 𝑥 2 𝑑𝑦]
1

49 | P a g e

Along 𝐶1 : 𝑥 2 = 𝑦, 𝑑𝑦 = 2𝑥𝑑𝑥 and 𝑥 varies from 0 to 1
...

0

0

𝐼2 = ∫ [(𝑥 2 + 𝑥 2 )𝑑𝑥 + 𝑥 2 𝑑𝑥] = ∫ 3𝑥 2 𝑑𝑥 = (𝑥 3 )10 = −1
𝑥=1

𝑥=1

∴ 𝑰𝟐 = −𝟏
Therefore from (2), we get 𝐼 =

19
20

−1= −

𝑖
...


2
...
1 Application of Green’s theorem
Example 3: 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 the circle
𝒙 + 𝒚 = 𝒂𝟐
...

50 | P a g e

𝜕
𝜕
(2𝑥 2 − 𝑦 2 )] 𝑑𝑥 𝑑𝑦
∴ ∮ [(2𝑥 2 − 𝑦 2 )𝑑𝑥 + (𝑥 2 + 𝑦 2 )𝑑𝑦] = ∬ [ (𝑥 2 + 𝑦 2 ) −
𝜕𝑥
𝜕𝑦
𝐶
𝑅
𝑎 √𝑎 2−𝑥 2

= ∬ (2𝑥 + 2𝑦)𝑑𝑥 𝑑𝑦 = 2 ∫
𝑅

∫

𝑥=−𝑎 𝑦=0
𝑎

= 2 ∫ (𝑥 √𝑎2 − 𝑥 2 +

(𝑎 2 − 𝑥 2 )
2

𝑥=−𝑎

= 2 (𝑎3 −

𝑎

√𝑎 2−𝑥 2

𝑦2
(𝑥 + 𝑦)𝑑𝑥 𝑑𝑦 = 2 ∫ (𝑥𝑦 + )
2 𝑦=0
𝑥=−𝑎

𝑎

𝑎

) 𝑑𝑥 = 2 ∫ (𝑎2 − 𝑥 2 )𝑑𝑥 = 2 (𝑎2 𝑥 −
𝑥=0

𝑥3
)
3 0

𝑎3
4
) = 𝑎3
3
3

Hence,
4
∮ [(2𝑥 2 − 𝑦 2 )𝑑𝑥 + (𝑥 2 + 𝑦 2 )𝑑𝑦] = 𝑎3
3
𝐶

Example 4: By the use of Green’s theorem, show that area bounded by a simple closed
𝟏

curve 𝑪 is given by ∮𝑪 (𝒙 𝒅𝒚 − 𝒚 𝒅𝒙)
...

𝟐

Solution: By Green’s theorem in plane, we have
∮ (𝑀 𝑑𝑥 + 𝑁 𝑑𝑦) = ∬ (
𝐶

𝑅

𝜕𝑁 𝜕𝑀
−
) 𝑑𝑥 𝑑𝑦
𝜕𝑥
𝜕𝑦

------------------------ (1)

where 𝑅 is the region bounded by closed curve 𝐶
...
The parametric equation of the ellipse is given by

𝑥 = 𝑎 cos 𝜃, 𝑦 = 𝑏 sin 𝜃 , 0 ≤ 𝜃 ≤ 2𝜋

1

Hence area of the ellipse = 2 ∮𝐶 (𝑥 𝑑𝑦 − 𝑦 𝑑𝑥)
2𝜋

1
= ∫ [(𝑎 cos 𝜃)(𝑏 cos 𝜃 𝑑𝜃) − (𝑏 sin 𝜃)(𝑎 sin 𝜃 𝑑𝜃)]
2
0

2𝜋

2𝜋

0

0

1
1
= 𝑎𝑏 ∫ (𝑐𝑜𝑠 2 𝜃 + 𝑠𝑖𝑛2 𝜃)𝑑𝜃 = 𝑎𝑏 ∫ 𝑑𝜃 = 𝜋𝑎𝑏
2
2
∴ Area of the ellipse = 𝜋𝑎𝑏

52 | P a g e

Example 5: Using Green’s theorem, find the area of the region in the first
1

𝑥

𝑥

4

quadrant bounded by the curves 𝑦 = 𝑥, 𝑦 = , 𝑦 =
...
(1)

𝑥

1

4

4

Along 𝐶1 : 𝑦 = , 𝑑𝑦 = 𝑑𝑥, 𝑥 varies from 0 to 2
...

1

1

11

1

∴ 𝐼2 = ∫𝐶 (𝑥 𝑑𝑦 − 𝑦 𝑑𝑥) = ∫𝑥=2 [𝑥 (− 𝑥 2) 𝑑𝑥 − 𝑥 𝑑𝑥] = −2 ∫2 𝑥 𝑑𝑥 = −2(log 𝑥)12
2
= −2(log 1 − log 2) = 2 log 2
𝐼2 = 2 log 2

𝟏

Along 𝐶3 : 𝑦 = 𝑥, 𝑑𝑦 = 𝑑𝑥, 𝑥 varies from 1 to 0
...


Solution: By Green’s theorem in plane, we have
∮ (𝑀 𝑑𝑥 + 𝑁 𝑑𝑦) = ∬ (
𝐶

𝑅

𝜕𝑁 𝜕𝑀
−
) 𝑑𝑥 𝑑𝑦
𝜕𝑥
𝜕𝑦

where 𝑅 is the region bounded by closed curve 𝐶
...


Solution: By Green’s theorem in plane, we have
∮ (𝑀 𝑑𝑥 + 𝑁 𝑑𝑦) = ∬ (
𝐶

𝑅

𝜕𝑁 𝜕𝑀
−
) 𝑑𝑥 𝑑𝑦
𝜕𝑥
𝜕𝑦

where 𝑅 is the region bounded by closed curve 𝐶
...


Verify Green’s theorem by evaluating ∮𝐶 [(𝑥 3 − 𝑥𝑦 3 )𝑑𝑥 + (𝑦 2 − 2𝑥𝑦)𝑑𝑦] where 𝐶 is
the square having the vertices at the points (0,0), (2,0), (2,2) and (0,2)
...


Verify Green’s theorem for ∮𝐶 [(2𝑥 2 − 𝑦 2 )𝑑𝑥 + (𝑥 2 + 𝑦 2 )𝑑𝑦] where 𝐶 is the boundary
of the area enclosed by 𝑋-axis and the upper half of the circle 𝑥 2 + 𝑦 2 = 𝑎2
...


Use Green’ theorem to evaluate ∮𝐶 (2𝑦 2 𝑑𝑥 + 3𝑥 𝑑𝑦) where 𝐶 is the boundary of the
closed region bounded by 𝑦 = 𝑥 and 𝑦 = 𝑥 2
...


Evaluate by Green’s theorem ∮𝐶 (𝑒 −𝑥 sin 𝑦 𝑑𝑥 + 𝑒 −𝑥 cos 𝑦 𝑑𝑦) where 𝐶 is the rectangle
𝜋

𝜋

with vertices at (0,0), (𝜋, 0), (𝜋, 2 ), (0, 2 ) and hence verify Green’s theorem
...


7
30

4
...
6 Stoke’s Theorem (Relation between line and surface integrals)
Statement: If 𝑆 is an open surface bounded by a closed curve 𝐶 and 𝐹⃗ = ⃗⃗⃗⃗
𝐹1 𝑖̂ + ⃗⃗⃗⃗⃗
𝐹2 𝑗̂ + ⃗⃗⃗⃗⃗
𝐹3 𝑘̂ is
any vector point function having continuous first order partial derivatives, then
∮ 𝐹⃗
...
𝑛̂ 𝑑𝑆
𝐶

𝑆

where 𝑛̂ is a unit normal vector at any point of 𝑆 drawn in the sense in which a right-handed
screw would advance when rotated in the sense of description of 𝐶
...


Solution: By Stoke’s theorem, we know that
∮ 𝐹⃗
...
𝑛̂ 𝑑𝑆
𝐶

𝑆

where 𝑆 is an open surface bounded by a closed surface 𝐶
...
𝒅𝒓
⃗⃗
To evaluate: ∮𝑪 𝑭
Let 𝐶 denote the boundary of the rectangle 𝐴𝐵𝐶𝐷, then
56 | P a g e

∮ 𝐹⃗
...
(𝑖̂ 𝑑𝑥 + 𝑗̂ 𝑑𝑦) = ∮ [(𝑥 2 + 𝑦 2 )𝑑𝑥 − 2𝑥𝑦 𝑑𝑦]
𝐶

𝐶

𝐶

Here the closed curve 𝐶 consists of four lines 𝐴𝐵, 𝐵𝐶, 𝐶𝐷 and 𝐷𝐴
...

∴ ∮𝐶 𝐹⃗
...
𝑑𝑟⃗ = 𝐼1 + 𝐼2 + 𝐼3 + 𝐼4

---------------------------------- (1)

Now we have 𝐼1 = ∫𝐶 [(𝑥 2 + 𝑦 2 )𝑑𝑥 − 2𝑥𝑦 𝑑𝑦]
1

Along 𝐶1 : 𝑥 = 𝑎, 𝑑𝑥 = 0 and 𝑦 varies from 0 to 𝑏
...

−𝑎

𝑥3

∴ 𝐼2 = ∫𝑥=𝑎 (𝑥 2 + 𝑏2 )𝑑𝑥 = ( 3 + 𝑏2 𝑥)

−𝑎
𝑎

2

= − 3 𝑎3 − 2𝑎𝑏2

𝐼3 = ∫ [(𝑥 2 + 𝑦 2 )𝑑𝑥 − 2𝑥𝑦 𝑑𝑦]
𝐶3

Along 𝐶3 : 𝑥 = −𝑎, 𝑑𝑥 = 0 and 𝑦 varies from 𝑏 to 0
...

𝑥3

𝑎

∴ 𝐼4 = ∫𝑥=−𝑎 𝑥 2 𝑑𝑥 = ( 3 )

𝑎
−𝑎

2

= 3 𝑎3

∴ from (1) we get
2
2
∮ 𝐹⃗
...
𝑑𝑟⃗ = −4𝑎𝑏2
𝐶

̂ 𝒅𝑺
To evaluate: ∬𝑺 𝒄𝒖𝒓𝒍⃗⃗⃗⃗
𝑭
...
𝑛̂ = (−4𝑦𝑘̂ )
...
𝑛̂ 𝑑𝑆 = ∬𝑆 −4𝑦 𝑑𝑆 = −4 ∫𝑦=0 ∫𝑥=−𝑎 𝑦 𝑑𝑥 𝑑𝑦 = −4 ∫𝑦=0(𝑥)𝑎−𝑎 𝑦 𝑑𝑦
𝑏

𝑏

𝑦2
= −8𝑎 ∫ 𝑦 𝑑𝑦 = −8𝑎 ( ) = −4𝑎𝑏2
2 0
𝑦=0

∬ 𝑐𝑢𝑟𝑙 ⃗⃗⃗⃗
𝐹
...
𝑑𝑟⃗ = ∬ 𝑐𝑢𝑟𝑙 ⃗⃗⃗⃗
𝐹
...

58 | P a g e

Example 2: Verify Stoke’s theorem for 𝐹⃗ = 𝑧𝑖̂ + 𝑥𝑗̂ + 𝑦𝑘̂ taken over the half of the sphere
𝑥 2 + 𝑦 2 + 𝑧 2 = 𝑎2 lying above 𝑥𝑦-plane
...
𝑑𝑟⃗ = ∬ 𝑐𝑢𝑟𝑙 ⃗⃗⃗⃗
𝐹
...

Let 𝑆 be the upper half surface of the sphere 𝑥 2 + 𝑦 2 + 𝑧 2 = 𝑎2
...

⃗⃗
To evaluate: ∮𝑪 ⃗𝑭⃗
...
(𝑑𝑥 𝑖̂ + 𝑑𝑦 𝑗̂ + 𝑑𝑧 𝑘̂ ) = ∮𝐶 (𝑧 𝑑𝑥 + 𝑥 𝑑𝑦 + 𝑦 𝑑𝑧)

The equations of 𝐶 are 𝑥 2 + 𝑦 2 = 𝑎2 , 𝑧 = 0 whose parametric form is given by 𝑥 = 𝑎 cos 𝑡 , 𝑦 =
𝑎 sin 𝑡, 𝑧 = 0; 0 ≤ 𝑡 ≤ 2𝜋
...
𝑑𝑟⃗ = ∮𝐶 𝑥 𝑑𝑦

[∵ 𝑧 = 0 ⟹ 𝑑𝑧 = 0]

2𝜋

2𝜋

2𝜋 1+cos 2𝑡

= ∫0 𝑎 cos 𝑡
...
𝑑𝑟⃗ = 𝜋𝑎2
𝐶

̂ 𝒅𝑺
To evaluate: ∬𝑺 𝒄𝒖𝒓𝒍⃗⃗⃗⃗
𝑭
...
the
surface 𝜑(𝑥, 𝑦, 𝑧) ≡ 𝑥 2 + 𝑦 2 + 𝑧 2 = 𝑎2 , then
𝑛̂ =

𝑔𝑟𝑎𝑑𝜑
∇𝜑
=
|𝑔𝑟𝑎𝑑 𝜑| |∇𝜑|

∇𝜑 = (𝑖̂

∴ 𝑛̂ =

=

𝜕
𝜕
𝜕
+ 𝑗̂
+ 𝑘̂ )
...
(
𝑐𝑢𝑟𝑙 ⃗⃗⃗⃗
𝐹
...
𝑛̂ 𝑑𝑆 =
𝑆

1
𝑥+𝑦+𝑧
∬
𝑑𝑆
𝑎 𝑆
𝑎

To evaluate it we will use spherical polar coordinates (𝑟, 𝜃, 𝜑)
...

= 𝑎 𝑑𝜃
...
𝑛̂ 𝑑𝑆 = 𝑎 ∫𝜑=0 ∫𝜃=0

𝜋
2

=𝑎

2

2𝜋

∫ ∫ (𝑠𝑖𝑛2 𝜃 cos 𝜑 + 𝑠𝑖𝑛2 𝜃 sin 𝜑 + cos 𝜃 sin 𝜃) 𝑑𝜑 𝑑𝜃
𝜃=0 𝜑=0
𝜋
2

= 𝑎2 ∫ [𝑠𝑖𝑛2 𝜃 sin 𝜑 − 𝑠𝑖𝑛2 𝜃 cos φ + φ cos 𝜃 sin 𝜃]2𝜋
𝜑=0
𝜃=0
𝜋
2

= 2𝜋𝑎

2

𝜋
2

∫ sin 𝜃 cos 𝜃 𝑑𝜃 = 𝜋𝑎

2

𝜃=0

𝜋

cos 2𝜃 2
∫ sin 2𝜃 𝑑𝜃 = 𝜋𝑎 (−
) = 𝜋𝑎2
2
0
2

𝜃=0

∬ 𝑐𝑢𝑟𝑙 ⃗⃗⃗⃗
𝐹
...
𝑑𝑟⃗ = ∬ 𝑐𝑢𝑟𝑙 ⃗⃗⃗⃗
𝐹
...


Example 3: Verify Stoke’s theorem for 𝐹⃗ (𝑥, 𝑦, 𝑧) = 𝑥𝑧𝑖̂ − 𝑦𝑗̂ + 𝑥 2 𝑦𝑘̂ , where the surface 𝑆 is
the surface of the region bounded by 𝑥 = 0, 𝑦 = 0, 𝑧 = 0, 2𝑥 + 𝑦 + 2𝑧 = 8 which is not included
on 𝑥𝑧-plane
...
𝑑𝑟⃗ = ∬ 𝑐𝑢𝑟𝑙 ⃗⃗⃗⃗
𝐹
...

⃗⃗
...
𝑑𝑟⃗ = ∮ (𝑥𝑧𝑖̂ − 𝑦𝑗̂ + 𝑥 2 𝑦𝑘̂ )
...
Let these straight lines
are denotes by 𝐶1 , 𝐶2 and 𝐶3 respectively
...
𝑑𝑟⃗ = ∮𝐶 (𝑧𝑥 𝑑𝑥 − 𝑦 𝑑𝑦 + 𝑥 2 𝑦 𝑑𝑧) + ∮𝐶 (𝑧𝑥 𝑑𝑥 − 𝑦 𝑑𝑦 + 𝑥 2 𝑦 𝑑𝑧)
1

2

+ ∮𝐶 (𝑧𝑥 𝑑𝑥 − 𝑦 𝑑𝑦 + 𝑥 2 𝑦 𝑑𝑧)
3

Or, ∮𝐶 𝐹⃗
...

62 | P a g e

∴ 𝐼1 = 0
𝐼2 = ∮ (𝑧𝑥 𝑑𝑥 − 𝑦 𝑑𝑦 + 𝑥 2 𝑦 𝑑𝑧)
𝐶2

Along 𝐶2 : 𝑥 + 𝑧 = 4, 𝑦 = 0 and 𝑥 varies from 0 to 4
...

∴ 𝐼3 = 0
Therefore from (1) we get
∮ 𝐹⃗
...
𝒏
Here the surface 𝑆 consists of three plane surfaces 𝑆1 : 𝑂𝐴𝐵, 𝑆2 : 𝑂𝐵𝐶 and 𝑆3 : 𝐴𝐵𝐶 so that
∬ 𝑐𝑢𝑟𝑙 ⃗⃗⃗⃗
𝐹
...
𝑛̂ 𝑑𝑆1 + ∬ 𝑐𝑢𝑟𝑙 ⃗⃗⃗⃗
𝐹
...
𝑛̂ 𝑑𝑆3
𝑆

𝑆1

= 𝑆𝐼1 + 𝑆𝐼2 + 𝑆𝐼3

𝑆2

𝑆3

-------------------------- (3)

We have
𝑖̂
𝜕
𝑐𝑢𝑟𝑙 ⃗⃗⃗⃗
𝐹 = ||
𝜕𝑥
𝑥𝑧

𝑗̂
𝜕
𝜕𝑦
−𝑦

𝑘̂
𝜕 |
|
𝜕𝑧
𝑥2 𝑦

𝜕
𝜕
𝜕
𝜕
𝜕
𝜕
= 𝑖̂ [𝜕𝑦 (𝑥 2 𝑦) − 𝜕𝑧 (−𝑦)] − 𝑗̂ [𝜕𝑥 (𝑥 2 𝑦) − 𝜕𝑧 (𝑥𝑧)] + 𝑘̂ [𝜕𝑥 (−𝑦) − 𝜕𝑦 (𝑥𝑧)]

𝑐𝑢𝑟𝑙 ⃗⃗⃗⃗
𝐹 = 𝑥 2 𝑖̂ − (2𝑥𝑦 − 𝑥)𝑗̂
Now we have
63 | P a g e

𝑆𝐼1 = ∬ 𝑐𝑢𝑟𝑙 ⃗⃗⃗⃗
𝐹
...
𝑛̂ = [𝑥 2 𝑖̂ − (2𝑥𝑦 − 𝑥)𝑗̂]
...
𝑛̂ 𝑑𝑆2
𝑆2

On 𝑆2 : plane 𝑂𝐵𝐶: 𝑥 = 0, ̂𝑛 = −𝑖,̂ 𝑐𝑢𝑟𝑙 ⃗⃗⃗⃗
𝐹
...
(−𝑖̂) = −𝑥 2 = 0
∴ 𝑆𝐼2 = 0
𝑆𝐼3 = ∬ 𝑐𝑢𝑟𝑙 ⃗⃗⃗⃗
𝐹
...
𝑛̂ = [𝑥 2 𝑖̂ − (2𝑥𝑦 − 𝑥)𝑗̂]
...
𝑛̂ = (2𝑥 2 + 𝑥 − 2𝑥𝑦)
3
To evaluate the surface integral on the surface 𝑆3 , let us take the projection 𝑅 of 𝑆3 𝑖
...
𝑛̂ 𝑑𝑆3 = ∬ 𝑐𝑢𝑟𝑙 ⃗⃗⃗⃗
𝐹
...
𝑘 |
𝑅 3
( )
3

4 8−2𝑥

1
1
= ∬ (2𝑥 2 + 𝑥 − 2𝑥𝑦) 𝑑𝑥 𝑑𝑦 = ∫ ∫ (2𝑥 2 + 𝑥 − 2𝑥𝑦) 𝑑𝑦 𝑑𝑥
2 𝑅
2
𝑥=0 𝑦=0

4

4

𝑥=0

𝑥=0

1
3
2
= ∫ [(2𝑥 2 + 𝑥)𝑦 − 𝑥𝑦 2 ]8−2𝑥
𝑦=0 𝑑𝑥 = ∫ (−4𝑥 + 23𝑥 − 28𝑥 ) 𝑑𝑥
2
4
23 3
32
2
= (−𝑥 + 𝑥 − 14𝑥 ) =
3
3
0
4

𝑆𝐼3 =

32
3
64 | P a g e

Therefore from (3) we get

∬ 𝑐𝑢𝑟𝑙 ⃗⃗⃗⃗
𝐹
...
𝑑𝑟⃗ = ∬ 𝑐𝑢𝑟𝑙 ⃗⃗⃗⃗
𝐹
...


2
...
1 Application of Stoke’s Theorem
Example 4: Apply Stoke’s theorem to evaluate ∮𝐶 [(𝑥 + 𝑦)𝑑𝑥 + (2𝑥 − 𝑧)𝑑𝑦 + (𝑦 + 𝑧)𝑑𝑧],
where 𝐶 is the boundary of the triangle with vertices (2, 0, 0), (0, 3, 0) and (0, 0, 6)
...
𝑑𝑟⃗ = ∬ 𝑐𝑢𝑟𝑙 ⃗⃗⃗⃗
𝐹
...

We have
∮𝐶 𝐹⃗
...
(𝑖̂𝑑𝑥 + 𝑗̂ 𝑑𝑦 + 𝑘̂ 𝑑𝑧)
= ∮𝐶 [(𝑥 + 𝑦)𝑑𝑥 + (2𝑥 − 𝑧)𝑑𝑦 + (𝑦 + 𝑧)𝑑𝑧]

-------------- (1)

where 𝐹⃗ = (𝑥 + 𝑦)𝑖̂ + (2𝑥 − 𝑧)𝑗̂ + (𝑦 + 𝑧)𝑘̂ and 𝑑𝑟⃗ = 𝑖̂𝑑𝑥 + 𝑗̂ 𝑑𝑦 + 𝑘̂ 𝑑𝑧
Now let 𝑆 be the plane surface of the triangle 𝐴𝐵𝐶 bounded by 𝐶
...
the
surface 𝜑(𝑥, 𝑦, 𝑧) ≡ 3𝑥 + 2𝑦 + 𝑧 = 6, then
𝑛̂ =

𝑔𝑟𝑎𝑑𝜑
∇𝜑
=
|𝑔𝑟𝑎𝑑 𝜑| |∇𝜑|

∇𝜑 = (𝑖̂

∴ 𝑛̂ =

𝜕
𝜕
𝜕
+ 𝑗̂
+ 𝑘̂ )
...
𝑛̂ = (2𝑖̂ + 𝑘̂ )
...
𝑛̂ =

7
√14

7

√14
To evaluate the surface integral on the surface 𝑆, let us take the projection 𝑅 of 𝑆 𝑖
...
𝑛̂ 𝑑𝑆 = ∬ 𝑐𝑢𝑟𝑙 ⃗⃗⃗⃗
𝐹
...
𝑘̂|

= 7 ∬𝑅 𝑑𝑥 𝑑𝑦
1

= 7 (Area of the triangle ∆𝐴𝑂𝐵 )= 7 (2 × 2 × 3) = 21
∬ 𝑐𝑢𝑟𝑙 ⃗⃗⃗⃗
𝐹
...
𝑑𝑟⃗ by Stoke’s theorem, where 𝐹⃗ = 𝑦 2 𝑖̂ + 𝑥 2 𝑗̂ − (𝑥 + 𝑧)𝑘̂ and 𝐶 is
the boundary of the triangle with vertices at (0, 0, 0), (1, 0, 0) and (1, 1, 0)
...
𝑑𝑟⃗ = ∬ 𝑐𝑢𝑟𝑙 ⃗⃗⃗⃗
𝐹
...

We have
𝑗̂
𝑘̂
𝜕
𝜕
||
𝜕𝑦
𝜕𝑧
𝑥 2 −(𝑥 + 𝑧)

𝑖̂
𝜕
𝑐𝑢𝑟𝑙 ⃗⃗⃗⃗
𝐹 = ||
𝜕𝑥
𝑦2

𝜕
𝜕
𝜕
𝜕
𝜕
𝜕
= 𝑖̂ [𝜕𝑦 (−𝑥 − 𝑧) − 𝜕𝑧 (𝑥 2 )] − 𝑗̂ [𝜕𝑥 (−𝑥 − 𝑧) − 𝜕𝑧 (𝑦 2 )] + 𝑘̂ [𝜕𝑥 (𝑥 2 ) − 𝜕𝑦 (𝑦 2 )]

𝑐𝑢𝑟𝑙 ⃗⃗⃗⃗
𝐹 = 𝑗̂ + 2(𝑥 − 𝑦)𝑘̂

̂
To find 𝒏
Since 𝑧-coordinate of each vertex of the triangle is
zero, therefore the triangle lies in the 𝑥𝑦-plane and
𝑛̂ = 𝑘̂
...
𝑛̂ = [𝑗̂ + 2(𝑥 − 𝑦)𝑘̂]
...
𝑛̂ = 2(𝑥 − 𝑦)
Therefore, by Stoke’s theorem, we have
𝐹
...
𝑑𝑟⃗ = ∬ 𝑐𝑢𝑟𝑙 ⃗⃗⃗⃗
𝐶

𝑆

1

𝑥

= ∫𝑥=0 ∫𝑦=0 2(𝑥 − 𝑦) 𝑑𝑦 𝑑𝑥
1

= 2 ∫𝑥=0 (𝑥𝑦 −
1

= 2 ∫𝑥=0 (𝑥 2 −

𝑦2
2
𝑥2

𝑥

)

𝑦=0

𝑑𝑥
1

𝑥3

1

1

) 𝑑𝑥 = ∫0 𝑥 2 𝑑𝑥 = ( ) =
2
3
3
0

∮ 𝐹⃗
...
𝑑𝑆⃗ = −20𝜋
...
𝑑𝑟⃗ = ∬ 𝑐𝑢𝑟𝑙 ⃗⃗⃗⃗
𝐹
...


The plane 𝑧 = 2 cuts the surface 𝑆 of the paraboloid 2𝑧 = 𝑥 2 + 𝑦 2 in a circle 𝐶 whose equations
are 𝐶: 𝑥 2 + 𝑦 2 = 4, 𝑧 = 2
...


We have
𝐹⃗
...
(𝑖̂𝑑𝑥 + 𝑗̂ 𝑑𝑦 + 𝑘̂ 𝑑𝑧) = 3𝑦 𝑑𝑥 − 𝑥𝑧 𝑑𝑦 + 𝑦𝑧 2 𝑑𝑧
By Stoke’s Theorem, we have
⃗⃗ × F
⃗⃗)
...
𝑛̂ 𝑑𝑆 = ∮𝐶 (3𝑦 𝑑𝑥 − 𝑥𝑧 𝑑𝑦 + 𝑦𝑧 2 𝑑𝑧)
...

68 | P a g e

∴ from (1), we have
⃗⃗ × F
⃗⃗)
...
2
...
2 ) = −20𝜋
⃗⃗ × F
⃗⃗)
...


Verify Stoke’s theorem for 𝐹⃗ = (𝑦 − 𝑧 + 2)𝑖̂ + (𝑦𝑧 + 4)𝑗̂ − 𝑥𝑧𝑘̂ over the surface of the
cube 𝑥 = 0, 𝑦 = 0, 𝑧 = 0, 𝑥 = 2, 𝑦 = 2, 𝑧 = 2 above the XOY plane (open the bottom)
...


Verify Stoke’s theorem by evaluating the line integral ∮𝐶 (𝑥𝑒 𝑧 𝑑𝑥 + 𝑦𝑒 𝑥 𝑑𝑦 + 𝑧𝑒 𝑦 𝑑𝑧)
along the boundary of the triangle with vertices (, 0, 0, 1); (1, 0, 1); (1, 1, 1)
...


Use Stoke’s theorem to evaluate ∮𝐶 [(𝑥 + 2𝑦)𝑑𝑥 + (𝑥 − 𝑧)𝑑𝑦 + (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 anticlockwise direction
...
E-resources:
https://www
...
com/watch?v=fZ231k3zsAA&t=57s
https://www
...
com/watch?v=qOcFJKQPZfo
https://www
...
com/watch?v=3TkKm2mwR0Y
https://www
...
com/watch?v=ynzRyIL2atU
https://www
...
com/watch?v=Cxc7ihZWq5o
https://www
...
com/watch?v=vvzTEbp9lrc
https://www
...
com/watch?v=3edJPRkCV9k
https://www
...
com/watch?v=CHW6krHTtXE
https://www
...
com/watch?v=7FUNdFN6ZKI
https://www
...
com/watch?v=AqcbyjaSQ10
https://www
...
com/watch?v=y-gsqWf3Gms
https://www
...
com/watch?v=I1dfwKPV75A
https://www
...
com/watch?v=BVAbztfd2JM
https://www
...
com/watch?v=ZtQyuN7DdKE
https://www
...
com/watch?v=YXf3aKxgELY
https://www
...
com/watch?v=eKD6aDwJ2lI
https://www
...
com/watch?v=8SwKD5_VL5o

70 | P a g e



---