Gauss divergence theorem and stokes theorem problems

1. Trigonometry and Vector
Calculus(18UMTC42) -
Gauss divergence theorem &
Stokes theorem Problems
By
A.Sujatha M.Sc.,M.Phil.,PGDCA.,
Department of Mathematics

2. Gauss divergence theorem
If V is the volume bounded by a closed surface S and 𝑓 is a vector
valued function having continuous partial derivatives then 𝑆
𝑓 . 𝑛 𝑑𝑆 =
𝑉
𝛻. 𝑓 𝑑𝑉
Example:
Verify Gauss divergence theorem for
𝑓 = 𝑥2
− 𝑦𝑧 𝑖 + 𝑦2
− 𝑧𝑥 𝑗 + (𝑧2
− 𝑥𝑦)𝑘) taken over the rectangle
parallelepiped, 0 ≤ 𝑥 ≤ 𝑎, 0 ≤ 𝑦 ≤ 𝑏, 0 ≤ 𝑧 ≤ 𝑐.
Solution:
Gauss divergence theorem is 𝑆
𝑓 . 𝑛 𝑑𝑆 = 𝑉
𝛻. 𝑓 𝑑𝑉
Now Consider, 𝑉
𝛻. 𝑓 𝑑𝑉 = 0
𝑎
0
𝑏
0
𝑐
2 𝑥 + 𝑦 + 𝑧 𝑑𝑧 𝑑𝑦 𝑑𝑥
[∵ 𝛻. 𝑓 =
𝜕
𝜕𝑥
𝑖 +
𝜕
𝜕𝑦
𝑗 +
𝜕
𝜕𝑧
𝑘 . [ 𝑥2 − 𝑦𝑧 𝑖 + 𝑦2 − 𝑧𝑥 𝑗 + (𝑧2 − 𝑥𝑦)𝑘)]
=
𝜕
𝜕𝑥
𝑥2
− 𝑦𝑧 +
𝜕
𝜕𝑦
𝑦2
− 𝑧𝑥 +
𝜕
𝜕𝑧
𝑧2
− 𝑥𝑦 = 2(𝑥 + 𝑦 + 𝑧)]

3. = 2
0
𝑎
0
𝑏
𝑥𝑧 + 𝑦𝑧 +
𝑧2
2 0
𝑐
𝑑𝑦 𝑑𝑥
= 2
0
𝑎
0
𝑏
𝑥𝑐 + 𝑦𝑐 +
𝑐2
2
𝑑𝑦 𝑑𝑥
= 2
0
𝑎
𝑥𝑦𝑐 +
𝑦2
2
𝑐 +
𝑦𝑐2
2 0
𝑏
𝑑𝑥
= 2
0
𝑎
[𝑥𝑏𝑐 +
𝑏2
2
𝑐 +
𝑏𝑐2
2
]𝑑𝑥
= 2
𝑥2
2
𝑏𝑐 + 𝑥
𝑏2
2
𝑐 +
𝑥𝑏𝑐2
2 0
𝑎
= 2[
𝑎2
2
𝑏𝑐 + 𝑎
𝑏2
2
𝑐 +
𝑎𝑏𝑐2
2
]
𝑉
𝛻. 𝑓 𝑑𝑉 = 𝑎𝑏𝑐 𝑎 + 𝑏 + 𝑐 → 𝟏
Now Consider, 𝑆
𝑓 . 𝑛 𝑑𝑆 Where S is the surface of the
rectangle parallelepiped given by
0 ≤ 𝑥 ≤ 𝑎, 0 ≤ 𝑦 ≤ 𝑏, 0 ≤ 𝑧 ≤ 𝑐.

4. It has the following six faces OABC (xz-plane); OAFE (xy-plane); OEDC (yz-plane);
DEFG (opposite to xz-plane); AFGB (opposite to (yz-plane); BCDG (opposite to xy-
plane)
On the face OABC we have 𝑦 = 0, 𝑛 = − 𝑗, 0 ≤ 𝑥 ≤ 𝑎 and 0 ≤ 𝑧 ≤ 𝑐.
∴
𝑂𝐴𝐵𝐶
𝑓 . 𝑛 𝑑𝑆 =
0
𝑎
0
𝑐
[ 𝑥2 − 𝑦𝑧 𝑖 + 𝑦2 − 𝑧𝑥 𝑗 + (𝑧2 − 𝑥𝑦)𝑘)]. (− 𝑗) 𝑑𝑧 𝑑𝑥
=
0
𝑎
0
𝑐
𝑦2 − 𝑧𝑥 𝑑𝑧 𝑑𝑥
= 0
𝑎
0
𝑐
𝑧𝑥 𝑑𝑧 𝑑𝑥 (since y = 0)
=
0
𝑎
𝑥
𝑧2
2 0
𝑐
𝑑𝑥
=
𝑎
𝑥
𝑐2
𝑑𝑥

5. On the face DEFG we have 𝑦 = 𝑏, 𝑛 = 𝑗, 0 ≤ 𝑥 ≤ 𝑎 and 0 ≤ 𝑧 ≤ 𝑐.
∴
𝐷𝐸𝐹𝐺
𝑓 . 𝑛 𝑑𝑆 =
0
𝑎
0
𝑐
[ 𝑥2
− 𝑦𝑧 𝑖 + 𝑦2
− 𝑧𝑥 𝑗 + (𝑧2
− 𝑥𝑦)𝑘)]. ( 𝑗) 𝑑𝑧 𝑑𝑥
=
0
𝑎
0
𝑐
𝑦2
− 𝑧𝑥 𝑑𝑧 𝑑𝑥
= 0
𝑎
0
𝑐
(𝑏2
− 𝑧𝑥) 𝑑𝑧 𝑑𝑥 (since y = b)
=
0
𝑎
𝑏2
𝑧 − 𝑥
𝑧2
2 0
𝑐
𝑑𝑥
=
0
𝑎
[𝑏2 𝑐 − 𝑥
𝑐2
2
]𝑑𝑥
= 𝑥𝑏2
𝑐 −
𝑐2
2
𝑥2
2 0
𝑎
= 𝑎𝑏2
𝑐 −
𝑎2 𝑐2
4
→ 𝟑
On the face OAFE we have 𝑧 = 0, 𝑛 = −𝑘, 0 ≤ 𝑥 ≤ 𝑎 and 0 ≤ 𝑦 ≤ 𝑏.
∴ 𝑓 . 𝑛 𝑑𝑆 =
𝑎 𝑏
[ 𝑥2
− 𝑦𝑧 𝑖 + 𝑦2
− 𝑧𝑥 𝑗 + (𝑧2
− 𝑥𝑦)𝑘)]. (−𝑘) 𝑑𝑦 𝑑𝑥

6. =
𝑏2
2
𝑥2
2 0
𝑎
=
𝑎2 𝑏2
4
→ 𝟒
On the face BCDG we have 𝑧 = 𝑐, 𝑛 = 𝑘, 0 ≤ 𝑥 ≤ 𝑎 and 0 ≤ 𝑦 ≤ 𝑏.
∴
𝐵𝐶𝐷𝐺
𝑓 . 𝑛 𝑑𝑆 =
0
𝑎
0
𝑏
[ 𝑥2
− 𝑦𝑧 𝑖 + 𝑦2
− 𝑧𝑥 𝑗 + (𝑧2
− 𝑥𝑦)𝑘)]. (𝑘) 𝑑𝑦 𝑑𝑥
=
0
𝑎
0
𝑏
𝑧2
− 𝑥𝑦 𝑑𝑦 𝑑𝑥
= 0
𝑎
0
𝑏
(𝑐2
− 𝑥𝑦) 𝑑𝑦 𝑑𝑥 (since z = c)
=
0
𝑎
𝑐2 𝑦 − 𝑥
𝑦2
2 0
𝑏
𝑑𝑥
=
0
𝑎
[𝑐2 𝑏 − 𝑥
𝑏2
2
]𝑑𝑥
= 𝑥𝑐2
𝑏 −
𝑏2
2
𝑥2
2 0
𝑎
= 𝑎𝑐2
𝑏 −
𝑎2
𝑏2
4
→ 𝟓
On the face OEDC we have 𝑥 = 0, 𝑛 = − 𝑖, 0 ≤ 𝑦 ≤ 𝑏 and 0 ≤ 𝑧 ≤ 𝑐.
𝑏 𝑐
=
0
𝑎
𝑥
𝑦2
2 0
𝑏
𝑑𝑥
=
0
𝑎
𝑥
𝑏2
2
𝑑𝑥

7. = 0
𝑏
0
𝑐
𝑦𝑧 𝑑𝑧 𝑑𝑦 (since x = 0)
=
0
𝑏
𝑦
𝑧2
2 0
𝑐
𝑑𝑦
=
0
𝑏
𝑦
𝑐2
2
𝑑𝑦
=
𝑐2
2
𝑦2
2 0
𝑏
=
𝑏2 𝑐2
4
→ 𝟔
On the face ABGF we have 𝑥 = 𝑎, 𝑛 = 𝑖, 0 ≤ 𝑦 ≤ 𝑏 and 0 ≤ 𝑧 ≤ 𝑐.
∴
𝐴𝐵𝐺𝐹
𝑓 . 𝑛 𝑑𝑆 =
0
𝑏
0
𝑐
[ 𝑥2 − 𝑦𝑧 𝑖 + 𝑦2 − 𝑧𝑥 𝑗 + (𝑧2 − 𝑥𝑦)𝑘)]. ( 𝑖) 𝑑𝑧 𝑑𝑦
=
0
𝑏
0
𝑐
𝑥2
− 𝑦𝑧 𝑑𝑧 𝑑𝑦
= 0
𝑏
0
𝑐
(𝑎2
− 𝑦𝑧) 𝑑𝑧 𝑑𝑦 (since x = a)
=
0
𝑏
𝑎2
𝑧 − 𝑦
𝑧2
2 0
𝑐
𝑑𝑦

8. =
0
𝑏
[𝑎2 𝑐 − 𝑦
𝑐2
2
]𝑑𝑦
= 𝑦𝑎2
𝑐 −
𝑐2
2
𝑦2
2 0
𝑏
= 𝑎2
𝑏𝑐 −
𝑏2
𝑐2
4
→ 𝟕
Adding equations 2, 3, 4, 5, 6 & 7 we get
𝑆
𝑓 . 𝑛 𝑑𝑆 = 𝑎𝑏𝑐 𝑎 + 𝑏 + 𝑐 → 𝟖
From equations 1 & 8 we get
𝑆
𝑓 . 𝑛 𝑑𝑆 =
𝑉
𝛻. 𝑓 𝑑𝑉
Hence Gauss divergence theorem is verified.
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9. Stokes theorem:
If S is an open two sides surface bounded by a simple closed curve C
and 𝑓 is a vector valued function having continuous first order partial
derivatives then 𝐶
𝑓 . 𝑑 𝑟 = 𝑆
(𝛻 × 𝑓). 𝑛 𝑑𝑆 where C is transverse in the
anticlockwise direction.
Example:
Verify Stokes theorem for 𝑓 = 𝑥2 − 𝑦2 𝑖 + 2𝑥𝑦 𝑗 in the rectangular
region
x = 0, y = 0, x = a, y = b.
Solution:
W.K.T Stokes theorem is 𝐶
𝑓 . 𝑑 𝑟 = 𝑆
(𝛻 × 𝑓). 𝑛 𝑑𝑆
Let O = (0,0), A = (a,0), B = (a, b), C = (0,b) be the vertices of the given
rectangle
∴
𝐶
𝑓 . 𝑑 𝑟 =
𝑂𝐴
𝑓 . 𝑑 𝑟 +
𝐴𝐵
𝑓 . 𝑑 𝑟 +
𝐵𝐶
𝑓 . 𝑑 𝑟 +
𝐶𝑂
𝑓 . 𝑑 𝑟 → 𝟏

10. ∴ The parametric equation of OA can be taken as x = t, y = 0 where 0 ≤ 𝑡 ≤ 𝑎
𝑂𝐴
𝑓 . 𝑑 𝑟 =
𝑂𝐴
𝑥2
− 𝑦2
𝑖 + 2𝑥𝑦 𝑗 . [𝑑𝑥 𝑖 + 𝑑𝑦 𝑗 + 𝑑𝑧 𝑘]
=
𝑂𝐴
𝑥2
− 𝑦2
𝑑𝑥 + 2𝑥𝑦 𝑑𝑦
= 0
𝑎
𝑡2
− 0 𝑑𝑡 (since x = t, y = 0, dx = dt)
=
𝑡3
3 0
𝑎
=
𝑎3
3
→ 𝟐
∴ The parametric equation of AB can be taken as x = a, y = t where 0 ≤ 𝑡 ≤ 𝑏
𝐴𝐵
𝑓 . 𝑑 𝑟 =
𝐴𝐵
𝑥2
− 𝑦2
𝑖 + 2𝑥𝑦 𝑗 . [𝑑𝑥 𝑖 + 𝑑𝑦 𝑗 + 𝑑𝑧 𝑘]
=
𝐴𝐵
𝑥2
− 𝑦2
𝑑𝑥 + 2𝑥𝑦 𝑑𝑦
= 0
𝑏
2𝑎𝑡 𝑑𝑡 (since x = a, y = t, dy = dt)

11. The parametric equation of BC can be taken as x = t, y = b where 0 ≤ 𝑡 ≤ 𝑎
𝐵𝐶
𝑓 . 𝑑 𝑟 = −
𝐶𝐵
𝑥2 − 𝑦2 𝑖 + 2𝑥𝑦 𝑗 . [𝑑𝑥 𝑖 + 𝑑𝑦 𝑗 + 𝑑𝑧 𝑘]
= −
𝐶𝐵
𝑥2 − 𝑦2 𝑑𝑥 + 2𝑥𝑦 𝑑𝑦
= − 0
𝑎
(𝑡2
− 𝑏2
) 𝑑𝑡 (since x = t, y = b, dx = dt)
=
𝑡3
3
− 𝑏2 𝑡
0
𝑎
= −
𝑎3
3
+ 𝑎𝑏2 → 𝟒
∴ The parametric equation of BC can be taken as x = 0, y = t where 0 ≤ 𝑡 ≤ 𝑎
𝐶𝑂
𝑓 . 𝑑 𝑟 = −
𝑂𝐶
𝑓 . 𝑑 𝑟 = −
0
𝑏
0 𝑑𝑡 = 0 → 𝟓
Substitute equations 2, 3, 4&5 in equation 1 we get
𝐶
𝑓 . 𝑑 𝑟 = 2𝑎𝑏2
→ 𝟔

12. Now,
𝛻 × 𝑓 = 𝑐𝑢𝑟𝑙 𝑓 =
𝑖 𝑗 𝑘
𝜕
𝜕𝑥
𝜕
𝜕𝑦
𝜕
𝜕𝑧
𝑥2
− 𝑦2
2𝑥𝑦 0
= 𝑖 0 − 𝑗 0 + 𝑘 2𝑦 + 2𝑦 = 4𝑦𝑘
𝑆
(𝛻 × 𝑓). 𝑛 𝑑𝑆 =
𝑅
(𝛻 × 𝑓). 𝑛
𝑛. 𝑘
𝑑𝑥 𝑑𝑦
Here the surface S denotes the rectangular and the unit outward normal 𝑛 is
𝑘
𝑆
(𝛻 × 𝑓). 𝑛 𝑑𝑆 =
0
𝑏
0
𝑎
4𝑦 𝑑𝑥 𝑑𝑦
=
0
𝑏
4𝑥𝑦 0
𝑎
𝑑𝑦
=
0
𝑏
4𝑎𝑦 𝑑𝑦 = 2𝑎𝑏2
→ 𝟕
From equations 6 & 7 we get