![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](https://image.slidesharecdn.com/2-vectorintegration-240331135426-26cd465d/85/2-VECTOR-INTEGRATION-of-mathematics-subject-4-320.jpg)
2-VECTOR INTEGRATION of mathematics subject
- 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