Vectors

1. VECTOR INTEGRATION (2019)
By
Dr Thamma Koteswara Rao
Line Integral – work done – area – Surface
and Volume integrals – vector integral
theorems: Greens, Stokes and Gauss
Divergence theorems (without proof).
1

2. Gauss Divergence Theorem:
( Transformation between surface integral and volume integral )
Let S be a closed surface enclosing a volume V. If 𝐹 is a
continuously differential vector function, then
𝑺
𝑭 ∙ 𝒏 𝒅𝑺 = 𝑽
𝒅𝒊𝒗 𝑭 𝒅𝒗 𝒐𝒓
𝑺
𝒇𝟏𝒅𝒚 𝒅𝒛 + 𝒇𝟐𝒅𝒛 𝒅𝒙 + 𝒇𝟑𝒅𝒙 𝒅𝒚 = 𝑽
𝝏𝒇𝟏
𝝏𝒙
+
𝝏𝒇𝟐
𝝏𝒚
+
𝝏𝒇𝟑
𝝏𝒛
𝒅𝒙 𝒅𝒚 𝒅𝒛 ,
where 𝑛 is the outward drawn unit normal vector at any point of S.
2

3. Example 1: Using Gauss Divergence theorem (or) By transforming into triple
integral evaluate 𝑺
𝑭 ∙ 𝒏 𝒅𝑺 , if 𝐹 = 𝑥3𝑖 + 𝑥2𝑦𝑗 + 𝑥2𝑧𝑘 (or)
𝑆
𝑥3𝑑𝑦 𝑑𝑧 + 𝑥2𝑦𝑑𝑧 𝑑𝑥 + 𝑥2𝑧𝑑𝑥 𝑑𝑦 , where S is the closed surface
consisting of the cylinder 𝑥2 + 𝑦2 = 𝑎2 and the circular discs z=0, z=b .
Solution: Gauss Divergence theorem 𝑺
𝑭 ∙ 𝒏 𝒅𝑺 = 𝑽
𝒅𝒊𝒗 𝑭 𝒅𝒗 𝒐𝒓
𝑺
𝒇𝟏𝒅𝒚 𝒅𝒛 + 𝒇𝟐𝒅𝒛 𝒅𝒙 + 𝒇𝟑𝒅𝒙 𝒅𝒚 =
𝑽
𝝏𝒇𝟏
𝝏𝒙
+
𝝏𝒇𝟐
𝝏𝒚
+
𝝏𝒇𝟑
𝝏𝒛
𝒅𝒙 𝒅𝒚 𝒅𝒛
Given 𝐹 = 𝑥3𝑖 + 𝑥2𝑦𝑗 + 𝑥2𝑧𝑘
 𝑓1 = 𝑥3
, 𝑓2 = 𝑥2
𝑦 , 𝑓3 = 𝑥2
𝑧

𝜕𝑓1
𝜕𝑥
= 3𝑥2 ,
𝜕𝑓2
𝜕𝑦
= 3𝑥2 ,
𝜕𝑓3
𝜕𝑥
= 𝑥2
Now 𝑑𝑖𝑣𝐹 =
𝜕𝑓1
𝜕𝑥
+
𝜕𝑓2
𝜕𝑦
+
𝜕𝑓3
𝜕𝑧
= 5𝑥2
Given S is the closed surface consisting of the
cylinder 𝑥2
+ 𝑦2
= 𝑎2
and the circular discs z=0, z=b
 y : − 𝑎2 − 𝑥2 → 𝑎2 − 𝑥2 , x : -a → a and z : 0 → b
3

4. 𝑆
𝑥3𝑑𝑦 𝑑𝑧 + 𝑥2𝑦𝑑𝑧 𝑑𝑥 + 𝑥2𝑧𝑑𝑥 𝑑𝑦 = 5𝑥2𝑑𝑥 𝑑𝑦 𝑑𝑧
= 5
𝑥=−𝑎
𝑎
𝑦=− 𝑎2−𝑥2
𝑎2−𝑥2
𝑧=0
𝑏
𝑥2𝑑𝑥 𝑑𝑦 𝑑𝑧 = 5
𝑥=−𝑎
𝑎
𝑦=− 𝑎2−𝑥2
𝑎2−𝑥2
𝑥2 𝑧 0
𝑏
𝑑𝑥𝑑𝑦
= 5𝑏 𝑥=−𝑎
𝑎
𝑦=− 𝑎2−𝑥2
𝑎2−𝑥2
𝑥2 𝑑𝑥𝑑𝑦 = 20𝑏 𝑥=0
𝑎
𝑦=0
𝑎2−𝑥2
𝑥2 𝑑𝑥𝑑𝑦
= 20𝑏 𝑥=0
𝑎
𝑥2 𝑦 0
𝑎2−𝑥2
𝑑𝑥 = 20𝑏 𝑥=0
𝑎
𝑥2 𝑎2 − 𝑥2 𝑑𝑥
= 20𝑎4
𝑏 𝜃=0
𝜋
2
𝑠𝑖𝑛2
𝜃 𝑐𝑜𝑠2
𝜃 𝑑𝜃 [ 𝑥 = 𝑎 𝑐𝑜𝑠𝜃, 𝑑𝑥 = 𝑎 𝑐𝑜𝑠𝜃 𝑑𝜃, 𝜃: 0 →
𝜋
2
]
= 5𝑎4𝑏 𝜃=0
𝜋
2
𝑠𝑖𝑛22𝜃 𝑑𝜃 = 5𝑎4𝑏 𝜃=0
𝜋
2 1−𝑐𝑜𝑠4𝜃
2
𝑑𝜃
=
5𝑎4𝑏
2
𝜃 −
𝑠𝑖𝑛4𝜃
4 0
𝜋
2
=
5𝑎4𝑏
4
𝜋
Therefore 𝑆
𝐹 ∙ 𝑛 𝑑𝑆 𝑜𝑟 𝑆
𝑥3𝑑𝑦 𝑑𝑧 + 𝑥2𝑦𝑑𝑧 𝑑𝑥 + 𝑥2𝑧𝑑𝑥 𝑑𝑦 =
5
4
𝜋𝑎4𝑏
4

5. Problems:
1) Using Gauss Divergence theorem (or) By transforming into triple integral
evaluate 𝑺
𝑭 ∙ 𝒏 𝒅𝑺 , if 𝐹 = 𝑥3𝑖 + 𝑥2𝑦𝑗 + 𝑥2𝑧𝑘 (or)
𝑆
𝑥3𝑑𝑦 𝑑𝑧 + 𝑥2𝑦𝑑𝑧 𝑑𝑥 + 𝑥2𝑧𝑑𝑥 𝑑𝑦 , where S is the closed surface
consisting of the cylinder 𝑥2
+ 𝑦2
= 𝑎2
and the circular discs z=0, z=b
in first octant.
2) Show that 𝑆
𝑎𝑥𝑖 + 𝑏𝑦𝑗 + 𝑐𝑧𝑘 ∙ 𝑛 𝑑𝑠 =
4𝜋
3
𝑎 + 𝑏 + 𝑐 , where S is the
surface of the sphere 𝑥2 + 𝑦2 + 𝑧2 = 1 (or)
Compute 𝑆
𝑎𝑥2 + 𝑏𝑦2 + 𝑐𝑧2 𝑑𝑠 over the surface of the
sphere 𝑥2 + 𝑦2 + 𝑧2 = 1
3) Evaluate 𝑆
𝑥 𝑑𝑦 𝑑𝑧 + 𝑦 𝑑𝑧 𝑑𝑥 + 𝑧 𝑑𝑥𝑑𝑦 , 𝑤ℎ𝑒𝑟𝑒 𝑆: 𝑥2 + 𝑦2 + 𝑧2 = 𝑎2
4) Evaluate 𝑆
𝑥 + 𝑧 𝑑𝑦 𝑑𝑧 + 𝑦 + 𝑧 𝑑𝑧 𝑑𝑥 + 𝑥 +
5

6. Example 2: Verify Gauss Divergence theorem for 𝐹 = 4𝑥𝑧 𝑖 − 𝑦2𝑗 + 𝑦𝑧 𝑘
taken over the surface of the cube bounded by the planes x = y = z = a and
coordinate planes. (or)
If 𝐹 = 4𝑥𝑧 𝑖 − 𝑦2𝑗 + 𝑦𝑧 𝑘, 𝑡ℎ𝑒𝑛 𝑒𝑣𝑎𝑙𝑢𝑎𝑡𝑒 𝑠
𝐹 ∙ 𝑛 𝑑𝑠 using Gauss
Divergence theorem . Where S is the surface of the cube bounded by x=0, x=a,
y=0, y=a, z=0, z=a.
Solution: Gauss Divergence theorem 𝑺
𝑭 ∙ 𝒏 𝒅𝑺 = 𝑽
𝒅𝒊𝒗 𝑭 𝒅𝒗
Given 𝐹 = 4𝑥𝑧 𝑖 − 𝑦2𝑗 + 𝑦𝑧 𝑘
 𝑓1 = 4𝑥𝑧 , 𝑓2 = −𝑦2 , 𝑓3 = 𝑦𝑧

𝜕𝑓1
𝜕𝑥
= 4𝑧 ,
𝜕𝑓2
𝜕𝑦
= −2𝑦 ,
𝜕𝑓3
𝜕𝑥
= 𝑦
Now 𝑑𝑖𝑣𝐹 =
𝜕𝑓1
𝜕𝑥
+
𝜕𝑓2
𝜕𝑦
+
𝜕𝑓3
𝜕𝑧
= 4𝑧 − 𝑦
Given S is the Cube consisting six planes
x = 0, y = 0, z = 0, x = a, y = a, z = a.
 x : 0 → a , x : 0 → a and z : 0 → a
6

7. Now from Gauss Divergence theorem
𝑺
𝑭 ∙ 𝒏 𝒅𝑺 = 𝑽
𝒅𝒊𝒗 𝑭 𝒅𝒗 = 𝒙=𝟎
𝒂
𝒚=𝟎
𝒂
𝒛=𝟎
𝒂
𝟒𝒛 − 𝒚 𝒅𝒙 𝒅𝒚 𝒅𝒛
= 𝑥=0
𝑎
𝑦=0
𝑎
2𝑧2
− 𝑦𝑧 𝑧=0
𝑎
𝑑𝑥 𝑑𝑦 = 𝑥=0
𝑎
𝑦=0
𝑎
2𝑎2
− 𝑎𝑦 𝑑𝑥 𝑑𝑦
= 𝑥=0
𝑎
2𝑎2𝑦 − 𝑎
𝑦2
2 𝑦=0
𝑎
𝑑𝑥
=
𝑥=0
𝑎
3 𝑎3
2
𝑑𝑥 =
3 𝑎3
2
𝑥 𝑥=0
𝑎
=
3 𝑎4
2
Therefore 𝑺
𝑭 ∙ 𝒏 𝒅𝑺 =
𝟑 𝒂𝟒
𝟐
7

8. Verification: Consider the cube OABCDEFP
bounded by the planes
x=0, x=a, y=0, y=a, z=0, z=a
Now 𝑠
𝐹 ∙ 𝑛 𝑑𝑠 = 𝐴𝐹𝑃𝐷
𝐹 ∙ 𝑛 𝑑𝑠
+ 𝑂𝐵𝐸𝐶
𝐹 ∙ 𝑛 𝑑𝑠 + 𝐵𝐷𝑃𝐸
𝐹 ∙ 𝑛 𝑑𝑠
+ 𝑂𝐴𝐹𝐶
𝐹 ∙ 𝑛 𝑑𝑠 + 𝐶𝐸𝑃𝐹
𝐹 ∙ 𝑛 𝑑𝑠
+ 𝑂𝐵𝐷𝐴
𝐹 ∙ 𝑛 𝑑𝑠
Given 𝐹 = 4𝑥𝑧 𝑖 − 𝑦2𝑗 + 𝑦𝑧 𝑘
1)To evaluate 𝐴𝐹𝑃𝐷
𝐹 ∙ 𝑛 𝑑𝑠:
For the surface AFPD  x = a, y : 0 → a, z : 0 → a
and unit outward drawn normal 𝒏 = 𝒊
Now 𝐹 ∙ 𝑛 𝑑𝑠 = 4xz dy dz = 4az dy dz
8

9. Now 𝐴𝐹𝑃𝐷
𝐹 ∙ 𝑛 𝑑𝑠 = 𝑦=0
𝑎
𝑧=0
𝑎
4𝑎𝑧 𝑑𝑦 𝑑𝑧
= 4𝑎 𝑦 𝑦=0
𝑎 𝑧2
2 𝑧=0
𝑎
= 2 𝑎4
.
2) To evaluate 𝑂𝐵𝐸𝐶
𝐹 ∙ 𝑛 𝑑𝑠:
For the surface OBEC  x = 0, y : 0 → a, z : 0 → a
and unit outward drawn normal 𝒏 = −𝒊
Now 𝐹 ∙ 𝑛 𝑑𝑠 = −4xz dy dz = 0 dy dz
Now 𝐴𝐹𝑃𝐷
𝐹 ∙ 𝑛 𝑑𝑠 = 0
3) To evaluate 𝐵𝐷𝑃𝐸
𝐹 ∙ 𝑛 𝑑𝑠:
For the surface BDPE  y = a, z : 0 → a, x : 0 → a
and unit outward drawn normal 𝒏 = 𝒋
Now 𝐹 ∙ 𝑛 𝑑𝑠 = −𝑦2
dz dx = −𝑎2
dz dx
Now 𝐵𝐷𝑃𝐸
𝐹 ∙ 𝑛 𝑑𝑠 = 𝑥=0
𝑎
𝑧=0
𝑎
−𝑎2
dz dx
= −𝑎2
𝑥 𝑥=0
𝑎
𝑧 𝑧=0
𝑎
= −𝑎4
.
4) To evaluate 𝑂𝐴𝐹𝐶
𝐹 ∙ 𝑛 𝑑𝑠:
For the surface OAFC  y = 0, y : 0 → a, z : 0 → a
and unit outward drawn normal 𝒏 = −j
Now 𝐹 ∙ 𝑛 𝑑𝑠 = 𝑦2
dz dx = 0 dz dx
Now 𝑂𝐴𝐹𝐶
𝐹 ∙ 𝑛 𝑑𝑠 = 0
9

10. 5) To evaluate 𝐶𝐸𝑃𝐹
𝐹 ∙ 𝑛 𝑑𝑠:
For the surface CEPF  z = a, x : 0 → a, y : 0 → a
and unit outward drawn normal 𝒏 = 𝒌
Now 𝐹 ∙ 𝑛 𝑑𝑠 = yz dx dy = ay dx dy
Now 𝐶𝐸𝑃𝐹
𝐹 ∙ 𝑛 𝑑𝑠 = 𝑥=0
𝑎
𝑦=0
𝑎
ay dx dy
= 𝑎 𝑥 𝑥=0
𝑎 𝑦2
2 𝑧=0
𝑎
=
𝑎4
2
.
6) To evaluate 𝑂𝐵𝐷𝐴
𝐹 ∙ 𝑛 𝑑𝑠:
For the surface OBDA  z = 0, x : 0 → a, y : 0 → a
and unit outward drawn normal 𝒏 = −𝒌
Now 𝐹 ∙ 𝑛 𝑑𝑠 = yz dx dy = 0 dx dy
Now 𝑂𝐵𝐷𝐴
𝐹 ∙ 𝑛 𝑑𝑠 = 0
Now 𝑠
𝐹 ∙ 𝑛 𝑑𝑠 = 2 𝑎4
+ 0 − 𝑎4
+ 0 +
𝑎4
2
+ 0 =
3𝑎4
2
Hence G D T verified.
10

11. Problems:
1) 𝐹 = 𝑥3 − 𝑦𝑧 𝑖 − 2𝑥2𝑦 𝑗 + 𝑧 𝑘, cube bounded by the planes x = y = z = a
and coordinate planes.
2) 𝐹 = 𝑥3 𝑖 + 𝑦3 𝑗 + 𝑧3 𝑘 , cube bounded by the planes x = 0, x = a, y = 0,
y = a, z = 0, z = a.
3) 𝐹 = 𝑥2 𝑖 + 𝑦2 𝑗 + 𝑧2 𝑘 , paralallow piped bounded by the planes x = 0,
x = a, y = 0, y = b, z = 0, z = c .
4) 𝐹 = 2𝑥2𝑦 𝑖 − 𝑦2 𝑗 + 4𝑥𝑧2 𝑘 , the region of first octant of the cylinder
𝑦2 + 𝑥2 = 9 𝑎𝑛𝑑 𝑥 = 2.
5) 𝐹 = 𝑥2 𝑖 + 𝑦2 𝑗 + 𝑧2 𝑘 , cut off by the plane x + y + z = a in the first
octant.
11

12. Green’s Theorem in the XY - Plane:
( Transformation between Line integral and double integral )
Let R be a closed region in XY – Plane bounded by a simple closed
curve C and M and N are continuous functions of x and y having continuous
partial derivatives in R, then
𝐶
𝑀 𝑑𝑥 + 𝑁 𝑑𝑦 = 𝑅
𝜕𝑁
𝜕𝑥
−
𝜕𝑀
𝜕𝑦
𝑑𝑥 𝑑𝑦
Where C is traversed in the positive direction.
12

13. Example 3: Using Green’s theorem evaluate 𝐶
2𝑥𝑦 − 𝑥2 𝑑𝑥 + 𝑥2 +
13
Y=x2
X=y2
O
P

14. Verification: Since c is piece-wise smooth curve, 𝐶
𝐹 ∙ 𝑑𝑟 = 𝑂𝑃
𝐹 ∙ 𝑑𝑟 + 𝑃𝑂
𝐹 ∙ 𝑑𝑟
Given vector function 𝐹 = 2𝑥𝑦 − 𝑥2 𝑖 + 𝑥2 + 𝑦2 𝑗 , we know 𝑑𝑟 = 𝑑𝑥 𝑖 +
𝑑𝑦 𝑗 + 𝑑𝑧 𝑘  𝐹 ∙ 𝑑𝑟 = 2𝑥𝑦 − 𝑥2
𝑑𝑥 + 𝑥2
+ 𝑦2
𝑑𝑦
Given curve 𝑦 = 𝑥2
, 𝑦2
= 𝑥 =⇒ 𝑥4
= 𝑥 =⇒ 𝑥 = 0, 𝑥 = 1 =⇒ 𝑦 = 0, 𝑦 =
1
The points of intersection are O (0, 0) and P (1, 1)
Along the OP, the curve is 𝑦 = 𝑥2
𝑑𝑦 = 2𝑥 𝑑𝑥 and x varies from 0 to 1  𝐹 ∙ 𝑑𝑟 = − 𝑥2
+ 4𝑥3
+ 2 𝑥5
𝑑𝑥
𝑂𝑃
𝐹 ∙ 𝑑𝑟 = 𝑂𝑃
2𝑥𝑦 − 𝑥2 𝑑𝑥 + 𝑥2 + 𝑦2 𝑑𝑦 = 𝑥=0
1
− 𝑥2 + 4𝑥3 + 2 𝑥5 𝑑𝑥
= −
𝑥3
3
+ 𝑥4
+
𝑥6
3 0
1
= 1
Along the PO, the curve is 𝑥 = 𝑦2
𝑑𝑥 = 2𝑦 𝑑𝑦 and y varies from 1 to 0  𝐹 ∙ 𝑑𝑟 = 𝑦2
+ 5 𝑦4
− 2𝑦5
𝑑𝑦
Now 𝑃𝑂
𝐹 ∙ 𝑑𝑟 = 𝑃𝑂
2𝑥𝑦 − 𝑥2
𝑑𝑥 + 𝑥2
+ 𝑦2
𝑑𝑦 = 𝑦=1
0
𝑦2
+ 5 𝑦4
−
14

15. Example 4: Verify Green’s theorem in the plane for
𝐶
𝑥2 − 𝑥𝑦3 𝑑𝑥 + 𝑦2 − 2𝑥𝑦 𝑑𝑦 , where C is a square with vertices
(0, 0), (2, 0), (2, 2), (0, 2) .
Solution: Green’s theorem 𝐶
𝑀 𝑑𝑥 + 𝑁 𝑑𝑦 = 𝑅
𝜕𝑁
𝜕𝑥
−
𝜕𝑀
𝜕𝑦
𝑑𝑥 𝑑𝑦
Given Integral 𝐶
𝑥2 − 𝑥𝑦3 𝑑𝑥 + 𝑦2 − 2𝑥𝑦 𝑑𝑦
Here 𝑀 = 𝑥2
− 𝑥𝑦3
and 𝑁 = 𝑦2
− 2𝑥𝑦
𝜕𝑁
𝜕𝑥
−
𝜕𝑀
𝜕𝑦
= −2𝑦 + 3𝑥𝑦2
Given C is a square with vertices (0, 0), (2, 0), (2, 2), (0, 2)
clearly x : 0 → 2, y : 0 → 2
Now 𝐶
𝑥2 − 𝑥𝑦3 𝑑𝑥 + 𝑦2 − 2𝑥𝑦 𝑑𝑦 = 𝑥=0
2
𝑦=0
2
−2𝑦 + 3𝑥𝑦2 𝑑𝑥 𝑑𝑦
=
𝑥=0
2
−𝑦2
+ 𝑥𝑦3
𝑦=0
2
𝑑𝑥 =
𝑥=0
2
8𝑥 − 4 𝑑𝑥 = 4𝑥2
− 4𝑥 𝑥=0
2
= 8
15
O
(0, 0) A
(2, 0)
B(2, 2)
C (0, 2)

16. Verification: since the square is a piece-wise smooth curve
𝐶
𝐹 ∙ 𝑑𝑟 = 𝑂𝐴
𝐹 ∙ 𝑑𝑟 + 𝐴𝐵
𝐹 ∙ 𝑑𝑟 + 𝐵𝐶
𝐹 ∙ 𝑑𝑟 + 𝐶𝑂
𝐹 ∙ 𝑑𝑟
Along OA: y = 0  dy = 0 , and x : 0 → 2
𝑂𝐴
𝑥2
− 𝑥𝑦3
𝑑𝑥 + 𝑦2
− 2𝑥𝑦 𝑑𝑦 = 𝑥=0
2
𝑥2
𝑑𝑥 =
8
3
Along AB: x = 2  dx = 0, and y : 0 → 2
𝐴𝐵
𝑥2 − 𝑥𝑦3 𝑑𝑥 + 𝑦2 − 2𝑥𝑦 𝑑𝑦 = 𝑦=0
2
𝑦2 − 4𝑦 𝑑𝑦 = −
16
3
Along BC: y = 2  dy = 0 , and x : 2 → 0
𝐵𝐶
𝑥2 − 𝑥𝑦3 𝑑𝑥 + 𝑦2 − 2𝑥𝑦 𝑑𝑦 = 𝑥=2
0
𝑥2 − 8𝑥 𝑑𝑥 =
40
3
Along CO: x = 0  dx = 0, and y : 2 → 0
𝐶𝑂
𝑥2
− 𝑥𝑦3
𝑑𝑥 + 𝑦2
− 2𝑥𝑦 𝑑𝑦 = 𝑦=2
0
𝑦2
𝑑𝑦 = −
8
3
Now 𝐶
𝑥2 − 𝑥𝑦3 𝑑𝑥 + 𝑦2 − 2𝑥𝑦 𝑑𝑦 =
8
3
−
16
3
+
40
3
−
8
3
= 8
Therefore Green’s theorem verified.
16

17. Example 5: Show that area bounded by a simple closed curve C is given by
1
2 𝐶
𝑥 𝑑𝑦 − 𝑦 𝑑𝑥 .
Solution: Green’s theorem 𝐶
𝑀 𝑑𝑥 + 𝑁 𝑑𝑦 = 𝑅
𝜕𝑁
𝜕𝑥
−
𝜕𝑀
𝜕𝑦
𝑑𝑥 𝑑𝑦
Given Integral 𝐶
𝑥 𝑑𝑦 − 𝑦 𝑑𝑥
Here M = −𝑦 and N = 𝑥

𝜕𝑁
𝜕𝑥
−
𝜕𝑀
𝜕𝑦
= 2
Now 𝐶
𝑥 𝑑𝑦 − 𝑦 𝑑𝑥 = 𝑅
2 𝑑𝑥 𝑑𝑦 = 2 𝑅
𝑑𝑥 𝑑𝑦 = 2 𝐴
Therefore A =
1
2 𝐶
𝑥 𝑑𝑦 − 𝑦 𝑑𝑥
17

18. Problems:
1) 𝐶
3𝑥2 − 8𝑦2 𝑑𝑥 + 4𝑦 − 6𝑥𝑦 𝑑𝑦 , C is 𝑦 = 𝑥 𝑎𝑛𝑑 𝑦 = 𝑥2.
2) 𝐶
𝑦 − sin 𝑥 𝑑𝑥 + cos 𝑥 𝑑𝑦 , C is the triangle enclosed by the lines y =
0, 2x = п, пy = 2x .
3) 𝐶
𝑥2 − cosh 𝑦 𝑑𝑥 + 𝑦 + sin 𝑥 𝑑𝑦 , C is the rectangle with vertices
(0, 0), (п, 0), (п, 1), (0, 1) .
4) 𝐶
𝑥𝑦 + 𝑦2 𝑑𝑥 + 𝑥2𝑑𝑦 , C is y = x and 𝑦 = 𝑥2.
5) 𝐶
3𝑥2 − 8𝑦2 𝑑𝑥 + 4𝑦 − 6𝑥𝑦 𝑑𝑦 , C is x = 0, y = 0 and x + y =1.
6) 𝐶
2𝑥2
− 𝑦2
𝑑𝑥 + 𝑥2
+ 𝑦2
𝑑𝑦 , C is the boundary of the area
enclosed by the X-axis and upper half of the circle 𝑥2 + 𝑦2 = 𝑎2 .
18

19. Stokes’s Theorem:
( Transformation between Line integral and surface integral )
Let S be a open surface bounded by a simple closed, non-intersecting
curve C. If 𝐹 is any differentiable vector point function, then
𝐶
𝐹 ∙ 𝑑𝑟 = 𝑆
𝑐𝑢𝑟𝑙𝐹 ∙ 𝑛 𝑑𝑠
Where C is traversed in the positive direction and 𝑛 is unit outward drawn
normal at any point of the surface.
19

20. Example 6: Verify Stoke’s theorem for 𝐹 = 𝑥2 − 𝑦2 𝑖 + 2𝑥𝑦 𝑗 over the
surface bounded by the planes x = 0, x = a, y = 0, y = b, z = c .
Solution: Stoke’s theorem 𝐶
𝐹 ∙ 𝑑𝑟 = 𝑆
𝑐𝑢𝑟𝑙𝐹 ∙ 𝑛 𝑑𝑠
Given 𝐹 = 𝑥2 − 𝑦2 𝑖 + 2𝑥𝑦 𝑗
𝑪𝒖𝒓𝒍 𝑭 =
𝒊 𝒋 𝒌
𝝏
𝝏𝒙
𝝏
𝝏𝒚
𝝏
𝝏𝒛
𝒙𝟐 − 𝒚𝟐 𝟐𝒙𝒚 𝟎
= 𝟒𝒚 𝒌
Given surface x = 0, x = a, y = 0, y = b, z = c
clearly its normal vector 𝑛 = 𝑘
Now 𝑐𝑢𝑟𝑙𝐹 ∙ 𝑛 = 4𝑦
x : 0 → a, y : 0 → b
Now 𝑆
𝑐𝑢𝑟𝑙𝐹 ∙ 𝑛 𝑑𝑠 = 𝑥=0
𝑎
𝑦=0
𝑏
4𝑦 𝑑𝑥 𝑑𝑦 = 4 𝑥=0
𝑎 𝑦2
2 𝑦=0
𝑏
𝑑𝑥
= 2𝑏2
𝑥=0
𝑎
𝑑𝑥 = 2𝑏2 𝑥 𝑥=0
𝑎
= 2𝑎𝑏2
20

21. Verification:
Given 𝐹 = 𝑥2 − 𝑦2 𝑖 + 2𝑥𝑦 𝑗
𝐹 ∙ 𝑑𝑟 = 𝑥2 − 𝑦2 𝑑𝑥 + 2𝑥𝑦 𝑑𝑦
Given surface x = 0, x = a, y = 0, y = b, z = c
clearly its boundary C is a rectangle QSPR
Now 𝐶
𝐹 ∙ 𝑑𝑟 = 𝑄𝑆
𝐹 ∙ 𝑑𝑟 + 𝑆𝑃
𝐹 ∙ 𝑑𝑟 + 𝑃𝑅
𝐹 ∙ 𝑑𝑟 + 𝑅𝑄
𝐹 ∙ 𝑑𝑟
Along QS: x : 0 → a, y = 0, z = c
dy = 0, dz = 0
𝑄𝑆
𝐹 ∙ 𝑑𝑟 = 𝑥=0
𝑎
𝑥2 𝑑𝑥 =
𝑥3
3 0
𝑎
=
𝑎3
3
Along SP: x = a, y = 0 → b, z = c
dx = 0, dz = 0
𝑆𝑃
𝐹 ∙ 𝑑𝑟 = 𝑦=0
𝑏
2𝑎𝑦 𝑑𝑦 = 2𝑎
𝑦2
2 0
𝑏
= 𝑎𝑏2
Along PR: x : a → 0, y = b, z = c
dy = 0, dz = 0
𝑃𝑅
𝐹 ∙ 𝑑𝑟 = 𝑥=𝑎
0
𝑥2 − 𝑏2 𝑑𝑥 =
𝑥3
3
− 𝑏2𝑥
𝑎
0
= −
𝑎3
3
+ 𝑎𝑏2
21

22. Along RQ: x = a, y = b → 0, z = c
dx = 0, dz = 0
𝑅𝑄
𝐹 ∙ 𝑑𝑟 = 𝑦=𝑏
0
0 𝑑𝑦 = 0
Now 𝐶
𝐹 ∙ 𝑑𝑟 =
𝑎3
3
+ 𝑎𝑏2
−
𝑎3
3
+ 𝑎𝑏2
+ 0 = 2𝑎𝑏2
Therefore Stoke’s theorem verified .
22

23. Example 8: Verify Stoke’s theorem for 𝐹 = −𝑦3𝑖 + 𝑥3𝑗 , where S is the
circular disc 𝑥2 + 𝑦2 ≤ 1, 𝑧 = 0 .
Solution: Stoke’s theorem 𝐶
𝐹 ∙ 𝑑𝑟 = 𝑆
𝑐𝑢𝑟𝑙𝐹 ∙ 𝑛 𝑑𝑠
Given 𝐹 = −𝑦3𝑖 + 𝑥3𝑗
𝑪𝒖𝒓𝒍 𝑭 =
𝒊 𝒋 𝒌
𝝏
𝝏𝒙
𝝏
𝝏𝒚
𝝏
𝝏𝒛
−𝒚𝟑 𝒙𝟑 𝟎
= 𝟑 𝒙𝟐 + 𝒚𝟐 𝒌
Given surface is 𝑥2 + 𝑦2 = 1, 𝑧 = 0, it is a circle having centre at O and radius1
clearly its normal vector 𝑛 = 𝑘
Now 𝑐𝑢𝑟𝑙𝐹 ∙ 𝑛 = 3 𝑥2 + 𝑦2
Polar coordinates of the circle 𝑥 = 𝑟 cos 𝜃 , 𝑦 = 𝑟 sin 𝜃 𝑎𝑛𝑑 𝑑𝑥 𝑑𝑦 = 𝑟 𝑑𝑟 𝑑𝜃
r : 0 → 1, θ : 0 → 2П
Now 𝑆
𝑐𝑢𝑟𝑙𝐹 ∙ 𝑛 𝑑𝑠 = 𝑅
3 𝑥2
+ 𝑦2
𝑑𝑥 𝑑𝑦 = 𝑟=0
1
𝜃=0
2𝜋
3𝑟2
𝑟 𝑑𝑟 𝑑𝜃
= 3
𝑟4
4 𝑟=0
1
𝜃 𝜃=0
2𝜋
=
3𝜋
2
23

24. Verification: Given 𝐹 = −𝑦3𝑖 + 𝑥3𝑗
𝐹 ∙ 𝑑𝑟 = −𝑦3𝑑𝑥 + 𝑥3𝑑𝑦
Given surface is 𝑥2 + 𝑦2 = 1, 𝑧 = 0, it is a circle having centre at O
and radius1.
Its parametric equations 𝑥 = cos 𝑡 , 𝑦 = sin 𝑡
𝑑𝑥 = − sin 𝑡 𝑑𝑡, 𝑑𝑦 = cos 𝑡 𝑑𝑡 and t : 0 → 2П
Now 𝐹 ∙ 𝑑𝑟 = 𝑐𝑜𝑠4
𝑡 + 𝑠𝑖𝑛4
𝑡 𝑑𝑡
Now 𝐶
𝐹 ∙ 𝑑𝑟 = 𝑡=0
2𝜋
𝑐𝑜𝑠4𝑡 + 𝑠𝑖𝑛4𝑡 𝑑𝑡 = 𝑡=0
2𝜋
1 − 2 𝑠𝑖𝑛2𝑡 𝑐𝑜𝑠2𝑡 𝑑𝑡
= 𝑡=0
2𝜋
𝑑𝑡 −
1
2 𝑡=0
2𝜋
𝑠𝑖𝑛2
2𝑡 𝑑𝑡 = 𝑡=0
2𝜋
𝑑𝑡 −
1
4 𝑡=0
2𝜋
1 − 𝑐𝑜𝑠4𝑡 𝑑𝑡
=
3
4 𝑡=0
2𝜋
𝑑𝑡 + 𝑡=0
2𝜋
𝑐𝑜𝑠4𝑡 𝑑𝑡 =
3
4
𝑡 𝑡=0
𝑡=2𝜋
+
𝑠𝑖𝑛4𝑡
4 𝑡=0
𝑡=2𝜋
=
3𝜋
2
Hence Stoke’s theorem verified.
24

25. Problems:
1) 𝐹 = 𝑥2𝑖 + 𝑥𝑦𝑗 , C is the square in the plane z = 0 whose sides are along
the lines x = 0, y = 0, x = a, y = a .
2) 𝐶
𝑥 + 𝑦 𝑑𝑥 + 2𝑥 − 𝑧 𝑑𝑦 + 𝑦 + 𝑧 𝑑𝑧 , C is the boundary of the
triangle with vertices (0, 0, 0), (1, 0, 0) and (1, 1, 0) .
3) 𝐹 = 𝑥2 + 𝑦2 𝑖 − 2𝑥𝑦𝑗 , C is the rectangle in the plane z = 0 whose sides
are along the lines x = ± a, y = 0, y = b .
4) 𝐹 = 2𝑥 − 𝑦 𝑖 − 𝑦𝑧2𝑗 − 𝑦2𝑧𝑘 over the upper half surface of the sphere
𝑥2 + 𝑦2 + 𝑧2 = 1 bounded by the projection of the XY – plane.
5) 𝐹 = 𝑦𝑖 + 𝑧𝑗 + 𝑥𝑘 over the part of the sphere 𝑥2 + 𝑦2 + 𝑧2 = 1 above
the XY – plane.
25

26. THANK YOU
ALL THE BEST
26