VECTOR ANALYSIS
Dr. Mohammed Yousuf Kamil
Examples:
Ex. (1): Verify Green's theorem in the plane ∮𝐶 (𝑥𝑦 + 𝑦 2 ) 𝑑𝑥 + 𝑥 2 𝑑𝑦
where C is the closed curve of the region bounded by 𝑦 = 𝑥 and 𝑦 = 𝑥 2 ?
Solution: Along 𝑦 = 𝑥 2 , from (0,0) to (1,1) the line integral equals
1
∫
(𝑥𝑥 2
+𝑥
4)
𝑑𝑥 + 𝑥
2 (2𝑥
1
𝑑𝑥) = ∫ (3𝑥 3 + 𝑥 4 ) 𝑑𝑥 =
0
0
19
20
Along 𝑦 = 𝑥 from (1,1) to (0,0) the line integral equals
0
∫ (𝑥𝑥 + 𝑥
2)
𝑑𝑥 + 𝑥
2 (𝑥
0
𝑑𝑥) = ∫ 3𝑥 3 𝑑𝑥 = −1
1
1
Then the required line integral =
19
1
−1 =−
20
20
𝜕𝛹 𝜕(𝑥𝑦 + 𝑦 2 )
∮(𝛷𝑑𝑥 + 𝛹𝑑𝑦) = ∬ (
−
) 𝑑𝑥𝑑𝑦
𝜕𝑥
𝜕𝑦
𝐶
𝑅
𝜕𝛹 𝜕𝛷
𝜕𝑥 2 𝜕𝛷
∬(
−
−
) 𝑑𝑥𝑑𝑦 = ∬ (
) 𝑑𝑥𝑑𝑦 = ∬(𝑥 − 2𝑦)𝑑𝑥𝑑𝑦
𝜕𝑥 𝜕𝑦
𝜕𝑥
𝜕𝑦
𝑅
𝑅
1
𝑥
𝑅
1
1
= ∫ 𝑑𝑥 ∫ (𝑥 − 2𝑦)𝑑𝑦 = ∫[𝑥𝑦 − 𝑦 2 ]𝑥𝑥2 𝑑𝑥 = ∫(𝑥 4 − 𝑥 3 )𝑑𝑥
𝑥=0
𝑦=𝑥 2
𝑥=0
𝑥=0
1
𝑥5 𝑥4
1
=[ − ] =−
5
4 0
20
Ex. (2): (a) Show that the area of a region R enclosed by a simple closed
curve C is given by 𝐴 = 1⁄2 ∮𝐶 𝑥 𝑑𝑦 − 𝑦 𝑑𝑥? (b) Calculate the area of the
ellipse 𝑥 = 𝑎 cos 𝜃, 𝑦 = 𝑏 sin 𝜃 ?
Page 75
Second Class in Department of Physics
VECTOR ANALYSIS
Dr. Mohammed Yousuf Kamil
𝜕𝛹
𝜕𝛷
Solution: (a) in Green's theorem ∮𝐶 (𝛷𝑑𝑥 + 𝛹𝑑𝑦) = ∬𝑅 ( − ) 𝑑𝑥𝑑𝑦
𝜕𝑥
𝜕𝑦
𝜕𝑥 𝜕(−𝑦)
∮𝑥 𝑑𝑦 − 𝑦 𝑑𝑥 = ∬ ( −
) 𝑑𝑥𝑑𝑦 = 2 ∬ 𝑑𝑥𝑑𝑦 = 2𝐴
𝜕𝑥
𝜕𝑦
𝐶
𝑅
𝑅
1
∴ 𝐴 = ∮𝑥 𝑑𝑦 − 𝑦 𝑑𝑥
2 𝐶
1
(𝐛) 𝐴 = ∮𝑥 𝑑𝑦 − 𝑦 𝑑𝑥
2 𝐶
2𝜋
1
𝐴 = ∫ (𝑎 cos 𝜃) (𝑏 cos 𝜃)𝑑𝜃 − (𝑏 sin 𝜃)(−𝑎 sin 𝜃)𝑑𝜃
2
0
2𝜋
2𝜋
0
0
1
1
1
𝐴 = ∫ 𝑎𝑏(𝑐𝑜𝑠 2 𝜃 + 𝑠𝑖𝑛2 𝜃) 𝑑𝜃 = ∫ 𝑎𝑏 𝑑𝜃 = 𝑎𝑏(2𝜋) = 𝑎𝑏𝜋
2
2
2
Ex. (3): A vector field F is given 𝐅 = sin 𝑦 𝑖 + 𝑥(1 + cos 𝑦)𝑗. Evaluate the
line integral ∫𝐶 𝐅 . 𝑑𝐫 where C is the circular path given 𝑥 2 + 𝑦 2 = 𝑎2 ?
Solution:
∫ 𝐅 . 𝑑𝐫 = ∫ (sin 𝑦 𝑖 + 𝑥(1 + cos 𝑦)𝑗) . (𝑑𝑥 𝑖 + 𝑑𝑦 𝑗)
𝐶
𝐶
∫ 𝐅 . 𝑑𝐫 = ∫ sin 𝑦 𝑑𝑥 + 𝑥(1 + cos 𝑦)𝑑𝑦
𝐶
𝐶
Using Green′s theorem ∮(𝛷𝑑𝑥 + 𝛹𝑑𝑦) = ∬ (
𝐶
𝑅
𝜕𝛹 𝜕𝛷
−
) 𝑑𝑥𝑑𝑦
𝜕𝑥 𝜕𝑦
𝜕(𝑥(1 + cos 𝑦)) 𝜕 sin 𝑦
∫ sin 𝑦 𝑑𝑥 + 𝑥(1 + cos 𝑦)𝑑𝑦 = ∬ (
−
) 𝑑𝑥𝑑𝑦
𝜕𝑥
𝜕𝑦
𝐶
𝑅
Page 76
Second Class in Department of Physics
VECTOR ANALYSIS
Dr. Mohammed Yousuf Kamil
= ∬((1 + cos 𝑦) − cos 𝑦)𝑑𝑥𝑑𝑦 = ∬ 𝑑𝑥𝑑𝑦 = ∬ 𝜌 𝑑𝜌𝑑𝜑
𝑅
𝑅
2𝜋
𝑎
𝑅′
𝑎
𝜌2
𝑎2
2𝜋
∴ ∫ 𝐅 . 𝑑𝐫 = ∫ 𝑑𝜑 ∫ 𝜌𝑑𝜌 = [𝜑]0 [ ] = (2𝜋) ( ) = 𝜋𝑎2
2 0
2
𝐶
0
0
Ex. (4): using Green's theorem, evaluate ∮𝐶 𝑥 2 𝑦 𝑑𝑥 + 𝑥 2 𝑑𝑦, where C is
boundary described counter clockwise of the triangle with vertices (0, 0),
(1, 0), (1, 1) ?
Solution:
Using Green′s theorem ∮(𝛷𝑑𝑥 + 𝛹𝑑𝑦) = ∬ (
𝐶
𝑅
𝜕𝛹 𝜕𝛷
−
) 𝑑𝑥𝑑𝑦
𝜕𝑥 𝜕𝑦
𝜕𝑥 2 𝜕𝑥 2 𝑦
∮𝑥 𝑦 𝑑𝑥 + 𝑥 𝑑𝑦 = ∬ (
−
) 𝑑𝑥𝑑𝑦 = ∬(2𝑥 − 𝑥 2 )𝑑𝑥𝑑𝑦
𝜕𝑥
𝜕𝑦
𝐶
2
2
𝑅
1
𝑅
𝑥
1
1
= ∫ 𝑑𝑥 ∫ (2𝑥 − 𝑥 2 )𝑑𝑦 = ∫[2𝑥𝑦 − 𝑥 2 𝑦]0𝑥 𝑑𝑥 = ∫(2𝑥 2 − 𝑥 3 )𝑑𝑥
𝑥=0
𝑦=0
𝑥=0
𝑥=0
1
2 3 𝑥4
2 1
5
=[ 𝑥 − ] = − =
3
4 0 3 4 12
Ex. (5): Using Green's theorem to evaluate ∮𝐶 (3𝑥 2 − 8𝑦 2 )𝑑𝑥 + (4𝑦 −
6𝑥𝑦)𝑑𝑦 where C is boundary of the region defined by 𝑦 = √𝑥 and𝑦 = 𝑥 2 ?
Solution:
Page 77
Second Class in Department of Physics
VECTOR ANALYSIS
Dr. Mohammed Yousuf Kamil
Using Green′s theorem ∮(𝛷𝑑𝑥 + 𝛹𝑑𝑦) = ∬ (
𝐶
𝑅
𝜕𝛹 𝜕𝛷
−
) 𝑑𝑥𝑑𝑦
𝜕𝑥 𝜕𝑦
𝜕(4𝑦 − 6𝑥𝑦) 𝜕(3𝑥 2 − 8𝑦 2 )
∮ (3𝑥 − 8𝑦 𝑑𝑥 + (4𝑦 − 6𝑥𝑦)𝑑𝑦 = ∬ (
) 𝑑𝑥𝑑𝑦
−
𝜕𝑥
𝜕𝑦
𝐶
2
2)
𝑅
1
1
√𝑥
𝑦2
= ∬ 10𝑦 𝑑𝑥𝑑𝑦 = 10 ∫ [ ] 𝑑𝑥 = 5 ∫(𝑥 − 𝑥 4 )𝑑𝑥
2 𝑥2
𝑅
𝑥=0
𝑥=0
1
𝑥2 𝑥5
1 1
3
= 5[ − ] = 5( − ) =
2
5 0
2 5
2
Ex. (6): Apply Green's theorem to evaluate ∮𝐶 (2𝑥 2 − 𝑦 2 )𝑑𝑥 + (𝑥 2 −
𝑦 2 )𝑑𝑦 where C is the boundary of the area enclosed by the x-axis and the
upper half of circle 𝑥 2 + 𝑦 2 = 𝑎2 ?
Solution:
Using Green′s theorem ∮(𝛷𝑑𝑥 + 𝛹𝑑𝑦) = ∬ (
𝐶
𝑅
∮ (2𝑥 2 − 𝑦 2 )𝑑𝑥 + (𝑥 2 − 𝑦 2 )𝑑𝑦 = ∬ (
𝐶
𝑅
𝑎
𝜕(𝑥 2 − 𝑦 2 ) 𝜕(2𝑥 2 − 𝑦 2 )
) 𝑑𝑥𝑑𝑦
−
𝜕𝑥
𝜕𝑦
√𝑎2 −𝑥 2
= ∬ 2(𝑥 + 𝑦) 𝑑𝑥𝑑𝑦 = 2 ∫ 𝑑𝑥 ∫
𝑅
𝑥=−𝑎
𝑎
√𝑎2 −𝑥 2
𝑦2
= 2 ∫ [𝑥𝑦 + ]
2 0
𝑥=−𝑎
Page 78
𝜕𝛹 𝜕𝛷
−
) 𝑑𝑥𝑑𝑦
𝜕𝑥 𝜕𝑦
(𝑥 + 𝑦)𝑑𝑦
𝑦=0
𝑎
𝑑𝑥 = 2 ∫ (𝑥 √𝑎2 − 𝑥 2 +
𝑥=−𝑎
𝑎2 − 𝑥 2
) 𝑑𝑥
2
Second Class in Department of Physics
VECTOR ANALYSIS
Dr. Mohammed Yousuf Kamil
𝑎
𝑎
𝑎
= 2 ∫ (𝑥 √𝑎2 − 𝑥 2 ) 𝑑𝑥 + ∫(𝑎2 − 𝑥 2 )𝑑𝑥 = 0 + 2 ∫(𝑎2 − 𝑥 2 )𝑑𝑥
−𝑎
−𝑎
−𝑎
𝑎
𝑥3
𝑎3
4
2
3
= 2 [𝑎 𝑥 − ] = 2 (𝑎 − ) = 𝑎3
3 0
3
3
Ex. (7): Let C be the curve 𝑥 = 1 − 𝑦 2 from (0, -1) to (0, 1). Evaluate
∮𝐶 𝑦 3 𝑑𝑥 + 𝑥 2 𝑑𝑦, by using Green's theorem?
Solution:
Using Green′s theorem ∮(𝛷𝑑𝑥 + 𝛹𝑑𝑦) = ∬ (
𝐶
𝑅
𝜕𝛹 𝜕𝛷
−
) 𝑑𝑥𝑑𝑦
𝜕𝑥 𝜕𝑦
𝜕𝑥 2 𝜕𝑦 3
∮𝑦 𝑑𝑥 + 𝑥 𝑑𝑦 = ∬ (
−
) 𝑑𝑥𝑑𝑦 = ∬(2𝑥 − 2𝑦 2 ) 𝑑𝑥𝑑𝑦
𝜕𝑥
𝜕𝑦
𝐶
3
2
𝑅
𝑅
√1−𝑥
1
= ∫ 𝑑𝑥
∫
(2𝑥 − 2𝑦 2 )𝑑𝑦 = ∫ 𝑑𝑥[2𝑥𝑦 − 𝑦 3 ]√−1−𝑥
√1−𝑥
𝑥=0
𝑦=−√1−𝑥
1
𝑥=0
1
= ∫ (2𝑥√1 − 𝑥 − (1 − 𝑥)
3⁄
2
+ 2𝑥√1 − 𝑥 − (1 − 𝑥)
3⁄
2 ) 𝑑𝑥
𝑥=0
1
= ∫ (4𝑥√1 − 𝑥 − 2(1 − 𝑥)
3⁄
2 ) 𝑑𝑥
𝑥=0
Let 𝑥 = 1 − 𝑡 2 ⟹ 𝑑𝑥 = −2𝑡 𝑑𝑡
0
∴ ∮𝑦 3 𝑑𝑥 + 𝑥 2 𝑑𝑦 = ∫(12 𝑡 4 − 8 𝑡 2 )𝑑𝑡
𝐶
Page 79
𝑡=1
Second Class in Department of Physics
VECTOR ANALYSIS
Dr. Mohammed Yousuf Kamil
0
𝑡5 8 3
12 8
4
= [12 − 𝑡 ] = − + =
5 3 1
5 3 15
Ex. (8): Let C be the circle 𝑥 2 + 𝑦 2 = 4, oriented counterclockwise. Use
Green's Theorem to evaluate ∮𝐶 (cos 𝑥 2 − 𝑦 3 )𝑑𝑥 + 𝑥 3 𝑑𝑦 ?
Solution:
Using Green′s theorem ∮(𝛷𝑑𝑥 + 𝛹𝑑𝑦) = ∬ (
𝐶
2
∮(cos 𝑥 − 𝑦
3 )𝑑𝑥
𝐶
𝑅
𝜕𝛹 𝜕𝛷
−
) 𝑑𝑥𝑑𝑦
𝜕𝑥 𝜕𝑦
𝜕𝑥 3 𝜕(cos 𝑥 2 − 𝑦 3 )
+ 𝑥 𝑑𝑦 = ∬ (
−
) 𝑑𝑥𝑑𝑦
𝜕𝑥
𝜕𝑦
3
𝑅
= ∬(3𝑥 2 + 3𝑦 2 ) 𝑑𝑥𝑑𝑦 = 3 ∬(𝑥 2 + 𝑦 2 ) 𝑑𝑥𝑑𝑦
𝑅
𝑅
Put 𝑥 = 𝜌 cos 𝜑 , 𝑦 = 𝜌 sin 𝜑,
𝑑𝑥 𝑑𝑦 = 𝜌 𝑑𝜌 𝑑𝜑
2𝜋
2
∴ ∮(cos 𝑥 2 − 𝑦 3 )𝑑𝑥 + 𝑥 3 𝑑𝑦 = 3 ∫ 𝑑𝜑 ∫ 𝜌3 𝑑𝜌 = 3(2𝜋)(4) = 24𝜋
𝐶
0
0
Ex. (9): Evaluate ∬𝑆 𝐅 ∙ 𝑛̂ 𝑑𝑠, where 𝐅 = 4𝑥𝑧 𝑖 − 𝑦 2 𝑗 + 𝑦𝑧 𝑘 and S is
the surface of the cube bounded by x =0, x=1, y=0, y=1, z=0, z=1 ?
Solution:
𝑈𝑠𝑖𝑛𝑔 𝑑𝑖𝑣𝑒𝑟𝑔𝑒𝑛𝑐𝑒 𝑡ℎ𝑒𝑜𝑟𝑒𝑚 ∬ 𝐅 ∙ 𝑛̂ 𝑑𝑠 = ∭ (𝛁 ⋅ 𝐅) 𝑑𝑉
𝑆
∭ (𝛁 ⋅ 𝐅) 𝑑𝑉 = ∭ ((𝑖
𝑉
Page 80
𝑉
𝑉
𝜕
𝜕
𝜕
+𝑗
+ 𝑘 ) ⋅ (4𝑥𝑧 𝑖 − 𝑦 2 𝑗 + 𝑦𝑧 𝑘)) 𝑑𝑉
𝜕𝑥
𝜕𝑦
𝜕𝑧
Second Class in Department of Physics
VECTOR ANALYSIS
Dr. Mohammed Yousuf Kamil
1
1
𝑥=0
𝑦=0
∬ 𝐅 ∙ 𝑛̂ 𝑑𝑠 = ∭ (4𝑧 − 𝑦) 𝑑𝑉 = ∫ 𝑑𝑥 ∫
𝑆
𝑉
1
1
= ∫ 𝑑𝑥 ∫
𝑥=0
[2𝑧 2
−
𝑦=0
𝑦𝑧]10 𝑑𝑦
1
𝑑𝑦 ∫ (4𝑧 − 𝑦)𝑑𝑧
𝑧=0
1
1
=∫
𝑑𝑥 ∫
𝑥=0
(2 − 𝑦)𝑑𝑦
𝑦=0
1
1
𝑦2
3 1
3
3
= ∫ [2𝑦 − ] 𝑑𝑥 = ∫ 𝑑𝑥 = [𝑥]10 =
2 0
2 𝑥=0
2
2
𝑥=0
Ex. (10): Evaluate ∬𝑆 𝐅 ∙ 𝑛̂ 𝑑𝑠 , where 𝐅 = 4𝑥 𝑖 − 2𝑦 2 𝑗 + 𝑧 2 𝑘 taken
over the region bounded by 𝑥 2 + 𝑦 2 = 4, 𝑧 = 0 and 𝑧 = 3 ?
Solution:
𝑈𝑠𝑖𝑛𝑔 𝑑𝑖𝑣𝑒𝑟𝑔𝑒𝑛𝑐𝑒 𝑡ℎ𝑒𝑜𝑟𝑒𝑚 ∬ 𝐅 ∙ 𝑛̂ 𝑑𝑠 = ∭ (𝛁 ⋅ 𝐅) 𝑑𝑉
𝑆
∭ (𝛁 ⋅ 𝐅) 𝑑𝑉 = ∭ ((𝑖
𝑉
𝑉
𝑉
𝜕
𝜕
𝜕
+𝑗
+ 𝑘 ) ⋅ (4𝑥 𝑖 − 2𝑦2 𝑗 + 𝑧2 𝑘 )) 𝑑𝑉
𝜕𝑥
𝜕𝑦
𝜕𝑧
√4−𝑥 2
2
= ∭ (4 − 4𝑦 + 2𝑧) 𝑑𝑉 = ∫
𝑉
=∫
𝑑𝑥 ∫
𝑥=−2
√4−𝑥 2
2
𝑑𝑥 ∫
𝑦=−√4−𝑥 2
𝑥=−2
[4𝑧 − 4𝑦𝑧 +
𝑦=−√4−𝑥 2
𝑧 2 ]30 𝑑𝑦
=∫
𝑑𝑥 ∫
=∫
2
= 42 ∫
𝑧=0
√4−𝑥 2
𝑑𝑥 ∫
𝑥=−2
2
𝑦=−√4−𝑥 2
𝑥=−2
𝑑𝑦 ∫ (4 − 4𝑦 + 2𝑧)𝑑𝑧
2
√4−𝑥 2
2
3
(21 − 12𝑦)𝑑𝑦 = ∫
𝑥=−2
𝑦=−√4−𝑥 2
(21 − 12𝑦)𝑑𝑦
2
𝑑𝑥 [21𝑦 − 6𝑦 2 ]√4−𝑥
−√4−𝑥 2
√4 − 𝑥 2 𝑑𝑥
𝑥=−2
Let 𝑥 = 2 sin 𝜃
Page 81
⟹ 𝑑𝑥 = 2 cos 𝜃 𝑑𝜃
Second Class in Department of Physics
VECTOR ANALYSIS
Dr. Mohammed Yousuf Kamil
𝜋⁄
2
∴ ∬ 𝐅 ∙ 𝑛̂ 𝑑𝑠 = 42 ∫
−𝜋⁄2
𝑆
√4 − 4𝑠𝑖𝑛2 𝜃 2 cos 𝜃 𝑑𝜃 = 168 ∫
𝜋⁄
2
𝑐𝑜𝑠 2 𝜃𝑑𝜃
−𝜋⁄2
𝜋
𝜃 ⁄2
𝜋 𝜋
= 168 [ ]
= 168 [ + ] = 84𝜋
2 −𝜋⁄2
4 4
Ex. (11): Evaluate∬𝑆 𝐫 ∙ 𝑛̂ 𝑑𝑠, where S is a closed surface?
Solution:
𝑈𝑠𝑖𝑛𝑔 𝑑𝑖𝑣𝑒𝑟𝑔𝑒𝑛𝑐𝑒 𝑡ℎ𝑒𝑜𝑟𝑒𝑚 ∬ 𝐫 ∙ 𝑛̂ 𝑑𝑠 = ∭ (𝛁 ⋅ 𝐫) 𝑑𝑉
𝑆
∭ (𝛁 ⋅ 𝐫) 𝑑𝑉 = ∭ ((𝑖
𝑉
𝑉
𝑉
𝜕
𝜕
𝜕
+𝑗
+ 𝑘 ) ⋅ (𝑥 𝑖 + 𝑦 𝑗 + 𝑥 𝑘 )) 𝑑𝑉
𝜕𝑥
𝜕𝑦
𝜕𝑧
= ∭ (1 + 1 + 1) 𝑑𝑉 = 3 ∭ 𝑑𝑉 = 3𝑉
𝑉
𝑉
∴ ∬ 𝐫 ∙ 𝑛̂ 𝑑𝑠 = 3𝑉
𝑆
where V is the volume enclosed by S.
Ex. (12): Use Divergence theorem to evaluate ∬𝑆 𝐅 ∙ 𝑑𝐒 , where 𝐅 =
𝑥 3 𝑖 + 𝑦 3 𝑗 + 𝑧 3 𝑘 and S is the surface of the sphere 𝑥 2 + 𝑦 2 + 𝑧 2 = 𝑎2 ?
Solution:
𝑈𝑠𝑖𝑛𝑔 𝑑𝑖𝑣𝑒𝑟𝑔𝑒𝑛𝑐𝑒 𝑡ℎ𝑒𝑜𝑟𝑒𝑚 ∬ 𝐅 ∙ 𝑑𝐒 = ∭ (𝛁 ⋅ 𝐅) 𝑑𝑉
𝑆
∭ (𝛁 ⋅ 𝐅) 𝑑𝑉 = ∭ ((𝑖
𝑉
Page 82
𝑉
𝑉
𝜕
𝜕
𝜕
+𝑗
+ 𝑘 ) ⋅ (𝑥3 𝑖 + 𝑦3 𝑗 + 𝑧3 𝑘 )) 𝑑𝑉
𝜕𝑥
𝜕𝑦
𝜕𝑧
Second Class in Department of Physics
VECTOR ANALYSIS
Dr. Mohammed Yousuf Kamil
= ∭ (3𝑥 2 + 3𝑦 2 + 3𝑧 2 ) 𝑑𝑉 = 3 ∭ (𝑥 2 + 𝑦 2 + 𝑧 2 ) 𝑑𝑉
𝑉
𝑉
Put 𝑥 = 𝑟 sin 𝜃 cos ∅ , 𝑦 = 𝑟 sin 𝜃 sin ∅ , 𝑧 = 𝑟 cos 𝜃
𝑑𝑥 𝑑𝑦 𝑑𝑧 = 𝑟 2 sin 𝜃 𝑑𝑟 𝑑𝜃 𝑑∅
∬ 𝐅 ∙ 𝑑𝐒 = 3 ∭ (𝑟 2 ) 𝑟 2 sin 𝜃 𝑑𝑟 𝑑𝜃 𝑑∅
𝑆
𝑉
2𝜋
𝜋
𝑎
2𝜋
𝑎
𝜋
𝑟5
4
= 3 ∫ 𝑑∅ ∫ sin 𝜃 𝑑𝜃 ∫ 𝑟 𝑑𝑟 = 3 ∫ 𝑑∅ ∫ sin 𝜃 𝑑𝜃 [ ]
5 0
0
0
0
0
0
=
3𝑎5 2𝜋
6𝑎5 2𝜋
6𝑎5
12 5
[𝜑]2𝜋
∫ [− cos 𝜃]𝜋0 𝑑∅ =
∫ 𝑑∅ =
𝜋𝑎
0 =
5 0
5 0
5
5
Ex. (13): Apply the Divergence theorem to compute ∬𝑆 𝐅 ∙ 𝑑𝐒, where S
is the surface of cylinder 𝑥 2 + 𝑦 2 = 𝑎2 bounded by the planes 𝑧 = 0,
𝑧 = 𝑏 and 𝐅 = 𝑥 𝑖 − 𝑦 𝑗 + 𝑧 𝑘 ?
Solution:
𝑈𝑠𝑖𝑛𝑔 𝑑𝑖𝑣𝑒𝑟𝑔𝑒𝑛𝑐𝑒 𝑡ℎ𝑒𝑜𝑟𝑒𝑚 ∬ 𝐅 ∙ 𝑑𝐒 = ∭ (𝛁 ⋅ 𝐅) 𝑑𝑉
𝑆
∭ (𝛁 ⋅ 𝐅) 𝑑𝑉 = ∭ ((𝑖
𝑉
𝑉
𝑉
𝜕
𝜕
𝜕
+𝑗
+ 𝑘 ) ⋅ (𝑥 𝑖 − 𝑦 𝑗 + 𝑧 𝑘 )) 𝑑𝑉
𝜕𝑥
𝜕𝑦
𝜕𝑧
= ∭ (1 − 1 + 1) 𝑑𝑉 = ∭ 𝑑𝑉 = ∭ 𝑑𝑥𝑑𝑦𝑑𝑧
𝑉
𝑉
𝑉
𝑑𝑥 𝑑𝑦 𝑑𝑧 = 𝜌 𝑑𝜌 𝑑𝜑 𝑑𝑧
𝑏
2𝜋
𝑎
𝑏
2𝜋
𝑎
𝜌2
∬ 𝐅 ∙ 𝑑𝐒 = ∫ 𝑑𝑧 ∫ 𝑑𝜑 ∫ 𝜌𝑑𝜌 = ∫ 𝑑𝑧 ∫ 𝑑𝜑 [ ]
2 0
𝑆
0
0
0
0
0
Page 83
Second Class in Department of Physics
VECTOR ANALYSIS
Dr. Mohammed Yousuf Kamil
2𝜋
𝑏
𝑎2 𝑏
𝑎2 𝑏
2𝜋
2
= ∫ 𝑑𝑧 ∫ 𝑑𝜑 = ∫ 𝑑𝑧[𝜑]0 = 𝜋𝑎 ∫ 𝑑𝑧 = 𝜋𝑎2 [𝜑]𝑏0 = 𝜋𝑎2 𝑏
2 0
2 0
0
0
Ex. (14): ): Evaluate∬𝑆 𝐅 ∙ 𝑛̂ 𝑑𝑠, where 𝐅 = (2𝑥 + 3𝑧) 𝑖 − (𝑥𝑧 + 𝑦) 𝑗 +
(𝑦 2 + 2𝑧) 𝑘 and S is the surface of the sphere having centre (3, -1, 2) and
radius 3 ?
Solution:
𝑈𝑠𝑖𝑛𝑔 𝑑𝑖𝑣𝑒𝑟𝑔𝑒𝑛𝑐𝑒 𝑡ℎ𝑒𝑜𝑟𝑒𝑚 ∬ 𝐅 ∙ 𝑑𝐒 = ∭ (𝛁 ⋅ 𝐅) 𝑑𝑉
𝑆
𝑉
∭ (𝛁 ⋅ 𝐅) 𝑑𝑉 = ∭ (𝛁 ⋅ (2𝑥 + 3𝑧) 𝑖 − (𝑥𝑧 + 𝑦) 𝑗 + (𝑦 2 + 2𝑧)) 𝑑𝑉
𝑉
𝑉
= ∭ (2 − 1 + 2) 𝑑𝑉 = 3 ∭ 𝑑𝑉 = ∭ 𝑑𝑥𝑑𝑦𝑑𝑧
𝑉
𝑉
𝑉
Put 𝑥 = 𝑟 sin 𝜃 cos ∅ , 𝑦 = 𝑟 sin 𝜃 sin ∅ , 𝑧 = 𝑟 cos 𝜃
𝑑𝑥 𝑑𝑦 𝑑𝑧 = 𝑟 2 sin 𝜃 𝑑𝑟 𝑑𝜃 𝑑∅
∬ 𝐅 ∙ 𝑑𝐒 = 3 ∭ 𝑟 2 sin 𝜃 𝑑𝑟 𝑑𝜃 𝑑∅
𝑆
𝑉
2𝜋
𝜋
3
2𝜋
𝜋
3
𝑟3
= 3 ∫ 𝑑∅ ∫ sin 𝜃 𝑑𝜃 ∫ 𝑟 𝑑𝑟 = 3 ∫ 𝑑∅ ∫ sin 𝜃 𝑑𝜃 [ ]
3 0
0
0
0
0
0
2𝜋
= 27 ∫
0
[− cos 𝜃]𝜋0
2
2𝜋
𝑑∅ = 54 ∫ 𝑑∅ = 54 [𝜑]2𝜋
0 = 108 𝜋
0
Ex. (15): Show that∬𝑆 𝛁(𝑥 2 + 𝑦 2 + 𝑧 2 ) ∙ 𝑑𝐒 = 6𝑉, where S is any closed
surface enclosing volume V. Using Divergence theorem?
Solution:
Page 84
Second Class in Department of Physics
VECTOR ANALYSIS
Dr. Mohammed Yousuf Kamil
𝑈𝑠𝑖𝑛𝑔 𝑑𝑖𝑣𝑒𝑟𝑔𝑒𝑛𝑐𝑒 𝑡ℎ𝑒𝑜𝑟𝑒𝑚 ∬ 𝐅 ∙ 𝑑𝐒 = ∭ (𝛁 ⋅ 𝐅) 𝑑𝑉
𝑆
𝑉
∭ (𝛁 ⋅ 𝐅) 𝑑𝑉 = ∭ (𝛁 ⋅ 𝛁(𝑥 2 + 𝑦 2 + 𝑧 2 )) 𝑑𝑉
𝑉
𝑉
= ∭ (𝛁 𝟐 (𝑥 2 + 𝑦 2 + 𝑧 2 )) 𝑑𝑉
𝑉
𝜕2
𝜕2
𝜕2
= ∭ (( 2 + 2 + 2 ) (𝑥 2 + 𝑦 2 + 𝑧 2 )) 𝑑𝑉
𝜕𝑥
𝜕𝑦
𝜕𝑧
𝑉
= ∭ (2 + 2 + 2) 𝑑𝑉 = 6 ∭ 𝑑𝑉 = 6 𝑉
𝑉
𝑉
Ex. (16): Evaluate∬𝑆 (𝑦 2 𝑧 2 𝑖 + 𝑧 2 𝑥 2 𝑗 + 𝑧 2 𝑦 2 𝑘) ∙ 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?
Solution:
𝑈𝑠𝑖𝑛𝑔 𝑑𝑖𝑣𝑒𝑟𝑔𝑒𝑛𝑐𝑒 𝑡ℎ𝑒𝑜𝑟𝑒𝑚 ∬ 𝐅 ∙ 𝑑𝐒 = ∭ (𝛁 ⋅ 𝐅) 𝑑𝑉
𝑆
𝑉
∭ (𝛁 ⋅ 𝐅) 𝑑𝑉 = ∭ (𝛁 ⋅ (𝑦 2 𝑧 2 𝑖 + 𝑧 2 𝑥 2 𝑗 + 𝑧 2 𝑦 2 𝑘)) 𝑑𝑉 = ∭ (2z𝑦 2 ) 𝑑𝑉
𝑉
𝑉
𝑉
Put 𝑥 = 𝑟 sin 𝜃 cos ∅ , 𝑦 = 𝑟 sin 𝜃 sin ∅ , 𝑧 = 𝑟 cos 𝜃
𝑑𝑥 𝑑𝑦 𝑑𝑧 = 𝑟 2 sin 𝜃 𝑑𝑟 𝑑𝜃 𝑑∅
∭ (2z𝑦 2 ) 𝑑𝑉 = 2 ∭(𝑟 cos 𝜃)(𝑟 sin 𝜃 sin ∅)2 𝑟2 sin 𝜃 𝑑𝑟 𝑑𝜃 𝑑∅
𝑉
𝑉
2𝜋
𝜋
Page 85
0
3
1
= 2 ∫ 𝑠𝑖𝑛 ∅ 𝑑∅ ∫ 𝑠𝑖𝑛 𝜃 cos 𝜃 𝑑𝜃 ∫ 𝑟 5 𝑑𝑟
0
2
0
Second Class in Department of Physics
VECTOR ANALYSIS
2𝜋
Dr. Mohammed Yousuf Kamil
𝜋
1
𝜋
𝑟6
1 2𝜋 2
𝑠𝑖𝑛4 𝜃
= 2 ∫ 𝑠𝑖𝑛 ∅ 𝑑∅ ∫ 𝑠𝑖𝑛 𝜃 cos 𝜃 𝑑𝜃 [ ] = ∫ 𝑠𝑖𝑛 ∅ 𝑑∅ [
]
6
3
4
0
0
0
0
0
2
3
1 2𝜋 2
1 2𝜋
𝜋
=
∫ 𝑠𝑖𝑛 ∅ 𝑑∅ =
=
12 0
12 2
12
Ex. (17): the vector field 𝐅 = 𝑥 2 𝑖 − 𝑧 𝑗 + 𝑦𝑧 𝑘 is defined over the volume
of the cuboid given by 0 ≤ 𝑥 ≤ 𝑎, 0 ≤ 𝑦 ≤ 𝑏, 0 ≤ 𝑧 ≤ 𝑐, enclosing the
surface S. Evaluate the surface integral ∬𝑆 𝐅 ∙ 𝑛̂ 𝑑𝑠 ?
Solution:
𝑈𝑠𝑖𝑛𝑔 𝑑𝑖𝑣𝑒𝑟𝑔𝑒𝑛𝑐𝑒 𝑡ℎ𝑒𝑜𝑟𝑒𝑚 ∬ 𝐅 ∙ 𝑑𝐒 = ∭ (𝛁 ⋅ 𝐅) 𝑑𝑉
𝑆
𝑉
∭ (𝛁 ⋅ 𝐅) 𝑑𝑉 = ∭ (𝛁 ⋅ (𝑥2 𝑖 − 𝑧 𝑗 + 𝑦𝑧 𝑘)) 𝑑𝑉 = ∭ (2x + y) 𝑑𝑉
𝑉
𝑉
𝑎
𝑉
𝑏
𝑐
𝑎
𝑏
∭ (2x + y) 𝑑𝑉 = ∫ 𝑑𝑥 ∫ 𝑑𝑦 ∫ (2x + y)𝑑𝑧 = ∫ 𝑑𝑥 ∫ 𝑑𝑦[2𝑥𝑧 + 𝑦𝑧]𝑐0
0
𝑉
0
0
0
0
𝑎
𝑦2 𝑏
𝑏2
= 𝑐 ∫ 𝑑𝑥 ∫ (2𝑥 + 𝑦)𝑑𝑦 = 𝑐 ∫ 𝑑𝑥 [2𝑥𝑦 + ] = 𝑐 ∫ (2𝑏𝑥 + ) 𝑑𝑥
2 0
2
0
0
0
0
𝑎
𝑏
𝑎
𝑎
𝑏2
𝑏2
𝑏
2
= 𝑐 ∫ (2𝑏𝑥 + ) 𝑑𝑥 = 𝑐 [𝑏𝑥 + 𝑥] = 𝑎𝑏𝑐 (𝑎 + )
2
2
2
0
0
𝑎
Ex. (18): Let S be the boundary of the region𝑥 2 + 𝑦 2 ≤ 4, 0 ≤ 𝑧 ≤ 3,,
oriented with unit normal pointing outwards. Consider the vector 𝐅 =
(𝑥 3 + cos(𝑦 2 )) 𝑖 + 𝑦𝑧 𝑗 + (3𝑦 2 𝑧 + cos 𝑥𝑦) 𝑘
Use
the
divergence
theorem to evaluate the following integral ∬𝑆 𝐅 ∙ 𝑑𝐀 ?
Solution:
Page 86
Second Class in Department of Physics
VECTOR ANALYSIS
Dr. Mohammed Yousuf Kamil
𝑈𝑠𝑖𝑛𝑔 𝑑𝑖𝑣𝑒𝑟𝑔𝑒𝑛𝑐𝑒 𝑡ℎ𝑒𝑜𝑟𝑒𝑚 ∬ 𝐅 ∙ 𝑑𝐀 = ∭ (𝛁 ⋅ 𝐅) 𝑑𝑉
𝑆
𝑉
∭ (𝛁 ⋅ 𝐅) 𝑑𝑉 = ∭ (𝛁 ⋅ (𝑥 3 + cos(𝑦 2 )) 𝑖 + 𝑦𝑧 𝑗 + (3𝑦 2 𝑧 + cos 𝑥𝑦) 𝑘 ) 𝑑𝑉
𝑉
𝑉
= ∭ (3x 2 + z + 3y 2 ) 𝑑𝑉
𝑉
Put 𝑥 = 𝜌 cos 𝜑 , 𝑦 = 𝜌 sin 𝜑 , 𝑧 = 𝑧,
2𝜋
3
𝑑𝑥 𝑑𝑦𝑑𝑧 = 𝜌 𝑑𝜌 𝑑𝜑 𝑑𝑧
2
∴ ∬ 𝐅 ∙ 𝑑𝐀 = ∫ 𝑑𝜑 ∫ 𝑑𝑧 ∫(3𝜌2 + z)𝜌 𝑑𝜌
𝑆
0
0
2𝜋
3
0
2
2𝜋
3
3𝜌 4
𝜌2
= ∫ 𝑑𝜑 ∫ 𝑑𝑧 [
+ 𝑧 ] = ∫ 𝑑𝜑 ∫(12 + 2𝑧)𝑑𝑧
4
2 0
0
0
0
2𝜋
0
2𝜋
= ∫ 𝑑𝜑 [12z + 𝑧2 ]30 = 45 ∫ 𝑑𝜑 = 45(2𝜋) = 90 𝜋
0
0
Ex. (19): Given the vector field 𝐅 = 𝑦 𝑖 − 𝑥 𝑗 + 𝑧 𝑘 , verify Stokes’
theorem for the hemispherical surface 𝑥 2 + 𝑦 2 + 𝑧 2 = 𝑎2 , 𝑧 ≥ 0 ?
Solution:
∮𝑭 ⋅ 𝑑𝒓 = ∬ 𝜵 × 𝑭 ⋅ 𝑛̂ 𝑑𝑠
𝐶
𝑆
𝜵 × 𝑭 = 𝜵 × (𝑦 𝑖 − 𝑥 𝑗 + 𝑧 𝑘) = −2 𝑘
𝛁(𝑥 2 + 𝑦 2 + 𝑧 2 ) = (𝑖
𝑛̂ =
𝜕∅
𝜕∅
𝜕∅
+𝑗
+ 𝑘 ) (𝑥 2 + 𝑦 2 + 𝑧 2 ) = 2𝑥𝑖 + 2𝑦𝑗 + 2𝑧𝑘
𝜕𝑥
𝜕𝑦
𝜕𝑧
𝛁𝑓
2𝑥𝑖 + 2𝑦𝑗 + 2𝑧𝑘
𝑥𝑖 + 𝑦𝑗 + 𝑧𝑘
𝑥𝑖 + 𝑦𝑗 + 𝑧𝑘
=
=
=
|𝛁𝑓| |2𝑥𝑖 + 2𝑦𝑗 + 2𝑧𝑘| √𝑥 2 + 𝑦 2 + 𝑧 2
𝑎
Page 87
Second Class in Department of Physics
VECTOR ANALYSIS
𝑛̂ . 𝑘̂ = (
𝑑𝑠 =
Dr. Mohammed Yousuf Kamil
𝑥𝑖 + 𝑦𝑗 + 𝑧𝑘
𝑧
) ∙ 𝑘̂ =
𝑎
𝑎
𝑑𝑥 𝑑𝑦 𝑑𝑥 𝑑𝑦 𝑎
= 𝑧
= 𝑑𝑥 𝑑𝑦
𝑧
(𝑛̂ . 𝑘̂ )
𝑎
∴ ∬ 𝜵 × 𝑭 ⋅ 𝑛̂ 𝑑𝑠 = ∬ (−2 𝑘 ∙ (
𝑆
𝑆
𝑥𝑖 + 𝑦𝑗 + 𝑧𝑘 𝑎
)) 𝑑𝑥 𝑑𝑦
𝑎
𝑧
= −2 ∬ 𝑑𝑥 𝑑𝑦
𝑆
Put 𝑥 = 𝜌 cos 𝜑 , 𝑦 = 𝜌 sin 𝜑,
2𝜋
𝑑𝑥 𝑑𝑦 = 𝜌 𝑑𝜌 𝑑𝜑
𝑎
𝑎2
∴ ∬ 𝜵 × 𝑭 ⋅ 𝑛̂ 𝑑𝑠 = −2 ∫ 𝑑𝜑 ∫ 𝜌 𝑑𝜌 = −2 (2𝜋) ( ) = −2𝜋𝑎2
2
𝑆
0
0
On the circle C, 𝑥 2 + 𝑦 2 = 𝑎2 , 𝑧 = 0 in the xy-plane.
∮𝐶 𝑭 ⋅ 𝑑𝒓 = ∮𝐶 (𝑦 𝑖 − 𝑥 𝑗 + 𝑧 𝑘) ⋅ (𝑑𝑥 𝑖 + 𝑑𝑦 𝑗 + 𝑑𝑧 𝑘) = ∮𝐶 (𝑦𝑑𝑥 − 𝑥𝑑𝑦)
Let 𝑥 = 𝑎 cos 𝜑 , 𝑑𝑥 = −𝑎 sin 𝜑 𝑑𝜑 , 𝑦 = 𝑎 sin 𝜑 , 𝑑𝑦 = 𝑎 cos 𝜑 𝑑𝜑
2𝜋
2𝜋
∮𝑭 ⋅ 𝑑𝒓 = −𝑎2 ∫ (𝑠𝑖𝑛2 𝜑 + 𝑐𝑜𝑠 2 𝜑) 𝑑𝜑 = −𝑎2 ∫ 𝑑𝜑 = −2𝜋𝑎2
𝐶
0
0
Ex. (20): Verify Stokes' theorem for 𝐅 = (2𝑥 − 𝑦) 𝑖 − 𝑦𝑧 2 𝑗 − 𝑦 2 𝑧 𝑘 ,
where S is the upper half surface of the sphere 𝑥 2 + 𝑦 2 + 𝑧 2 = 1 and C is
its boundary?
Solution:
∮𝑭 ⋅ 𝑑𝒓 = ∬ 𝜵 × 𝑭 ⋅ 𝑛̂ 𝑑𝑠
𝐶
𝑆
𝜵 × 𝑭 = 𝜵 × ((2𝑥 − 𝑦) 𝑖 − 𝑦𝑧 2 𝑗 − 𝑦 2 𝑧 𝑘) = 𝑘
Page 88
Second Class in Department of Physics
VECTOR ANALYSIS
𝛁(𝑥 2 + 𝑦 2 + 𝑧 2 ) = (𝑖
𝑛̂ =
Dr. Mohammed Yousuf Kamil
𝜕∅
𝜕∅
𝜕∅
+𝑗
+ 𝑘 ) (𝑥 2 + 𝑦 2 + 𝑧 2 ) = 2𝑥𝑖 + 2𝑦𝑗 + 2𝑧𝑘
𝜕𝑥
𝜕𝑦
𝜕𝑧
𝛁𝑓
2𝑥𝑖 + 2𝑦𝑗 + 2𝑧𝑘
𝑥𝑖 + 𝑦𝑗 + 𝑧𝑘
=
=
= 𝑥𝑖 + 𝑦𝑗 + 𝑧𝑘
|𝛁𝑓| |2𝑥𝑖 + 2𝑦𝑗 + 2𝑧𝑘| √𝑥 2 + 𝑦 2 + 𝑧 2
𝑛̂ . 𝑘̂ = (𝑥𝑖 + 𝑦𝑗 + 𝑧𝑘) ∙ 𝑘̂ = 𝑧
𝑑𝑥 𝑑𝑦 𝑑𝑥 𝑑𝑦 𝑑𝑥 𝑑𝑦
=
=
𝑧
𝑧
(𝑛̂ . 𝑘̂ )
𝑑𝑠 =
∴ ∬ 𝜵 × 𝑭 ⋅ 𝑛̂ 𝑑𝑠 = ∬ (𝑘 ∙ (𝑥𝑖 + 𝑦𝑗 + 𝑧𝑘))
𝑆
𝑆
𝑑𝑥 𝑑𝑦
𝑧
= ∬ 𝑑𝑥 𝑑𝑦
𝑆
Put 𝑥 = 𝜌 cos 𝜑 , 𝑦 = 𝜌 sin 𝜑,
2𝜋
𝑑𝑥 𝑑𝑦 = 𝜌 𝑑𝜌 𝑑𝜑
1
1
∴ ∬ 𝜵 × 𝑭 ⋅ 𝑛̂ 𝑑𝑠 = ∫ 𝑑𝜑 ∫ 𝜌 𝑑𝜌 = (2𝜋) ( ) = 𝜋
2
𝑆
0
0
On the circle C, 𝑥 2 + 𝑦 2 = 1, 𝑧 = 0 in the xy-plane.
∮𝑭 ⋅ 𝑑𝒓 = ∮((2𝑥 − 𝑦) 𝑖 − 𝑦𝑧 2 𝑗 − 𝑦 2 𝑧 𝑘) ⋅ (𝑑𝑥 𝑖 + 𝑑𝑦 𝑗 + 𝑑𝑧 𝑘)
𝐶
𝐶
= ∮𝐶 (2𝑥 − 𝑦)𝑑𝑥
Let 𝑥 = cos 𝑡 , 𝑑𝑥 = − sin 𝑡 𝑑𝑡 , 𝑦 = sin 𝑡
2𝜋
2𝜋
∮𝑭 ⋅ 𝑑𝒓 = ∫ (2cos 𝑡 − sin 𝑡) (− sin 𝑡 𝑑𝑡) = ∫ (−2sin 𝑡 cos 𝑡 + 𝑠𝑖𝑛2 𝑡) 𝑑𝑡
𝐶
0
0
2𝜋
2𝜋
∴ ∮𝑭 ⋅ 𝑑𝒓 = ∫ −2sin 𝑡 cos 𝑡 𝑑𝑡 + ∫ 𝑠𝑖𝑛2 𝑡 𝑑𝑡 = 0 +
𝐶
Page 89
0
0
2𝜋
=𝜋
2
Second Class in Department of Physics
VECTOR ANALYSIS
Dr. Mohammed Yousuf Kamil
Ex. (21): Evaluate by Stokes' theorem ∮𝐶(𝑦𝑧 𝑑𝑥 + 𝑧𝑥 𝑑𝑦 + 𝑥𝑦 𝑑𝑧), where
C is the curve 𝑥 2 + 𝑦 2 = 1, 𝑧 = 𝑦 2 ?
Solution:
∮𝑭 ⋅ 𝑑𝒓 = ∮(𝑦𝑧 𝑑𝑥 + 𝑧𝑥 𝑑𝑦 + 𝑥𝑦 𝑑𝑧)
𝐶
𝐶
∴ 𝑭 = 𝑦𝑧 𝑖 + 𝑧𝑥 𝑗 + 𝑥𝑦 𝑘
Stokes′ theorem ∮ 𝑭 ⋅ 𝑑𝒓 = ∬ 𝜵 × 𝑭 ⋅ 𝑛̂ 𝑑𝑠
𝐶
𝑆
𝜵 × 𝑭 = 𝜵 × (𝑦𝑧 𝑖 + 𝑧𝑥 𝑗 + 𝑥𝑦 𝑘) = 0
∬ 𝜵 × 𝑭 ⋅ 𝑛̂ 𝑑𝑠 = 0
𝑆
∴ ∮(𝑦𝑧 𝑑𝑥 + 𝑧𝑥 𝑑𝑦 + 𝑥𝑦 𝑑𝑧) = 0
𝐶
Ex. (22): Evaluate ∮𝐶 𝑭 ⋅ 𝑑𝒓, where 𝑭 = −𝑦2 𝑖 + 𝑥 𝑗 + 𝑧2 𝑘 and C is the
curve of intersection of the plane 𝑦 + 𝑧 = 2 and the cylinder𝑥 2 + 𝑦 2 = 1?
Solution:
Stokes′ theorem ∮ 𝑭 ⋅ 𝑑𝒓 = ∬ 𝜵 × 𝑭 ⋅ 𝑛̂ 𝑑𝑠
𝐶
𝑆
𝜵 × 𝑭 = 𝜵 × (−𝑦2 𝑖 + 𝑥 𝑗 + 𝑧2 𝑘) = (1 + 2𝑦) 𝑘
𝛁(𝑦 + 𝑧) = (𝑖
𝑛̂ =
𝜕∅
𝜕∅
𝜕∅
+𝑗
+ 𝑘 ) (𝑦 + 𝑧 ) = 𝑗 + 𝑘
𝜕𝑥
𝜕𝑦
𝜕𝑧
𝛁𝑓
𝑗+𝑘
𝑗+𝑘
=
=
|𝛁𝑓| |𝑗 + 𝑘|
√2
𝑛̂ . 𝑘̂ = (
Page 90
𝑗+𝑘
√2
) ∙ 𝑘̂ =
1
√2
Second Class in Department of Physics
VECTOR ANALYSIS
𝑑𝑠 =
Dr. Mohammed Yousuf Kamil
𝑑𝑥 𝑑𝑦 𝑑𝑥 𝑑𝑦
= 1 = √2 𝑑𝑥 𝑑𝑦
(𝑛̂ . 𝑘̂ )
√2
∴ ∬ 𝜵 × 𝑭 ⋅ 𝑛̂ 𝑑𝑠 = ∬ ((1 + 2𝑦) 𝑘 ∙ (
𝑆
𝑗+𝑘
𝑆
√2
)) √2 𝑑𝑥 𝑑𝑦
= ∬ (1 + 2𝑦) 𝑑𝑥 𝑑𝑦
𝑆
Put 𝑥 = 𝜌 cos 𝜑 , 𝑦 = 𝜌 sin 𝜑,
2𝜋
𝑑𝑥 𝑑𝑦 = 𝜌 𝑑𝜌 𝑑𝜑
1
∴ ∬ 𝜵 × 𝑭 ⋅ 𝑛̂ 𝑑𝑠 = ∫ 𝑑𝜑 ∫(𝜌 + 2𝜌2 sin 𝜑) 𝑑𝜌 =
𝑆
0
2𝜋
0
2𝜋
1
2𝜋
𝜌2
2𝜌3
1 2
𝜑 2
= ∫ 𝑑𝜑 [ + sin 𝜑
] = ∫ ( + sin 𝜑) 𝑑𝜑 = [ − cos 𝜑]
2
3 0
2 3
2 3
0
0
0
2
2
∴ ∮𝑭 ⋅ 𝑑𝒓 = ∬ 𝜵 × 𝑭 ⋅ 𝑛̂ 𝑑𝑠 = (𝜋 − − 0 + ) = 𝜋
3
3
𝐶
𝑆
Ex. (23): Evaluate ∮𝐶 𝑭 ⋅ 𝑑𝒓 by Stokes' theorem, where 𝑭 = 𝑦2 𝑖 + 𝑥2 𝑗 −
(𝑥 + 𝑧) 𝑘 and C is the boundary of triangle with vertices at (0, 0, 0), (1, 0,
0) and (1, 1,1 0)?
Solution:
Stokes′ theorem ∮ 𝑭 ⋅ 𝑑𝒓 = ∬ 𝜵 × 𝑭 ⋅ 𝑛̂ 𝑑𝑠
𝐶
𝑆
𝜵 × 𝑭 = 𝜵 × (𝑦2 𝑖 + 𝑥2 𝑗 − (𝑥 + 𝑧) 𝑘 ) = 𝑗 + 2(𝑥 − 𝑦) 𝑘
We observe that z coordinate of each vertex of the triangle is zero, (xy-plane).
𝑛̂ = 𝑘
Page 91
Second Class in Department of Physics
VECTOR ANALYSIS
𝑑𝑠 =
Dr. Mohammed Yousuf Kamil
𝑑𝑥 𝑑𝑦 𝑑𝑥 𝑑𝑦
=
= 𝑑𝑥 𝑑𝑦
(𝑛̂ . 𝑘̂ ) (𝑘 . 𝑘̂ )
∴ ∬ 𝜵 × 𝑭 ⋅ 𝑛̂ 𝑑𝑠 = ∬ (𝑗 + 2(𝑥 − 𝑦) 𝑘 ∙ (𝑘)) 𝑑𝑥 𝑑𝑦
𝑆
𝑆
1
𝑥
= ∬ 2(𝑥 − 𝑦) 𝑑𝑥 𝑑𝑦 = ∫ 𝑑𝑥 ∫ 2(𝑥 − 𝑦)𝑑𝑦
𝑆
0
0
1
𝑦2 𝑥
𝑥3 1 1
2
= 2 ∫ 𝑑𝑥 [𝑥𝑦 − ] = ∫ 𝑥 𝑑𝑥 = [ ] =
2 0
3 0 3
0
0
1
Ex. (24): Evaluate ∮𝐶 𝑭 ⋅ 𝑑𝒓 by Stokes' theorem, where 𝑭 = (𝑥2 + 𝑦2 ) 𝑖 −
2𝑥𝑦 𝑗 and C is the boundary of the rectangle = ∓𝑎 , 𝑦 = 0 and 𝑦 = 𝑏 ?
Solution:
Stokes′ theorem ∮ 𝑭 ⋅ 𝑑𝒓 = ∬ 𝜵 × 𝑭 ⋅ 𝑛̂ 𝑑𝑠
𝐶
𝑆
𝜵 × 𝑭 = 𝜵 × (𝑥2 + 𝑦2 ) 𝑖 − 2𝑥𝑦 𝑗 = −4𝑦 𝑘
Since the z coordinate of each vertex of the rectangle is zero, (xy-plane).
𝑛̂ = 𝑘
𝑑𝑠 =
𝑑𝑥 𝑑𝑦 𝑑𝑥 𝑑𝑦
=
= 𝑑𝑥 𝑑𝑦
(𝑛̂ . 𝑘̂ ) (𝑘 . 𝑘̂ )
∴ ∬ 𝜵 × 𝑭 ⋅ 𝑛̂ 𝑑𝑠 = ∬ (−4𝑦 𝑘 ∙ (𝑘)) 𝑑𝑥 𝑑𝑦
𝑆
𝑆
𝑎
𝑏
= ∬ −4𝑦 𝑑𝑥 𝑑𝑦 = −4 ∫ 𝑑𝑥 ∫ 𝑦𝑑𝑦
𝑆
−𝑎
0
𝑎
𝑦2 𝑏
2
= −4 ∫ 𝑑𝑥 [ ] = −2𝑏 ∫ 𝑑𝑥 = −4𝑎𝑏2
2 0
−𝑎
−𝑎
𝑎
Page 92
Second Class in Department of Physics
VECTOR ANALYSIS
Dr. Mohammed Yousuf Kamil
Ex. (25): Verify Stokes' theorem for 𝐅 = (𝑥 2 + 𝑦 − 4) 𝑖 + 3𝑥𝑦 𝑗 +
(2𝑥𝑧 + 𝑧 2 ) 𝑘 , over the surface of the hemisphere 𝑥 2 + 𝑦 2 + 𝑧 2 = 16
above the xy-plane?
Solution:
∮𝑭 ⋅ 𝑑𝒓 = ∬ 𝜵 × 𝑭 ⋅ 𝑛̂ 𝑑𝑠
𝐶
𝑆
𝜵 × 𝑭 = 𝜵 × ((𝑥 2 + 𝑦 − 4) 𝑖 + 3𝑥𝑦 𝑗 + (2𝑥𝑧 + 𝑧 2 ) 𝑘)
= −2𝑧 𝑗 + (3𝑦 − 1)𝑘
𝛁(𝑥 2 + 𝑦 2 + 𝑧 2 ) = (𝑖
𝑛̂ =
𝜕∅
𝜕∅
𝜕∅
+𝑗
+ 𝑘 ) (𝑥 2 + 𝑦 2 + 𝑧 2 ) = 2𝑥𝑖 + 2𝑦𝑗 + 2𝑧𝑘
𝜕𝑥
𝜕𝑦
𝜕𝑧
𝛁𝑓
2𝑥𝑖 + 2𝑦𝑗 + 2𝑧𝑘
𝑥𝑖 + 𝑦𝑗 + 𝑧𝑘
𝑥𝑖 + 𝑦𝑗 + 𝑧𝑘
=
=
=
|𝛁𝑓| |2𝑥𝑖 + 2𝑦𝑗 + 2𝑧𝑘| √𝑥 2 + 𝑦 2 + 𝑧 2
4
𝑛̂ . 𝑘̂ = (
𝑑𝑠 =
𝑥𝑖 + 𝑦𝑗 + 𝑧𝑘
𝑧
) ∙ 𝑘̂ =
4
4
𝑑𝑥 𝑑𝑦 𝑑𝑥 𝑑𝑦 4
= 𝑧
= 𝑑𝑥 𝑑𝑦
𝑧
(𝑛̂ . 𝑘̂ )
4
∬ 𝜵 × 𝑭 ⋅ 𝑛̂ 𝑑𝑠 = ∬ ((−2𝑧 𝑗 + (3𝑦 − 1)𝑘 ) ∙ (
𝑆
𝑆
𝑥𝑖 + 𝑦𝑗 + 𝑧𝑘 4
)) 𝑑𝑥 𝑑𝑦
4
𝑧
= ∬ [−2𝑦 + (3𝑦 − 1)] 𝑑𝑥 𝑑𝑦 = ∬ (𝑦 − 1) 𝑑𝑥 𝑑𝑦
𝑆
𝑆
Put 𝑥 = 𝜌 cos 𝜑 , 𝑦 = 𝜌 sin 𝜑,
2𝜋
𝑑𝑥 𝑑𝑦 = 𝜌 𝑑𝜌 𝑑𝜑
4
∴ ∬ 𝜵 × 𝑭 ⋅ 𝑛̂ 𝑑𝑠 = ∫ 𝑑𝜑 ∫(𝜌2 sin 𝜑 − 𝜌) 𝑑𝜌
𝑆
0
2𝜋
0
4
2𝜋
𝜌3
𝜌2
64
= ∫ 𝑑𝜑 [ sin 𝜑 − ] = ∫ 𝑑𝜑 ( sin 𝜑 − 8)
3
2 0
3
0
Page 93
0
Second Class in Department of Physics
VECTOR ANALYSIS
Dr. Mohammed Yousuf Kamil
2𝜋
64
64
64
= [− cos 𝜑 − 8𝜑] = − − 16𝜋 +
= −16𝜋
3
3
3
0
On the circle C, 𝑥 2 + 𝑦 2 = 16, 𝑧 = 0 in the xy-plane.
∮𝑭 ⋅ 𝑑𝒓 = ∮((𝑥 2 + 𝑦 − 4) 𝑖 + 3𝑥𝑦 𝑗 + (2𝑥𝑧 + 𝑧 2 ) 𝑘) ⋅ (𝑑𝑥 𝑖 + 𝑑𝑦 𝑗)
𝐶
𝐶
= ∮𝐶 (𝑥 2 + 𝑦 − 4) 𝑑𝑥 + 3𝑥𝑦 𝑑𝑦
Let 𝑥 = 4cos 𝑡 , 𝑑𝑥 = − 4sin 𝑡 𝑑𝑡 , 𝑦 = 4sin 𝑡 , 𝑑𝑦 = 4cos 𝑡 𝑑𝑡
2𝜋
∫ (16 𝑐𝑜𝑠 2 𝑡 + 4sin 𝑡 − 4) (− 4sin 𝑡 𝑑𝑡) + 3(4cos 𝑡)(4sin 𝑡) (4cos 𝑡 𝑑𝑡)
0
2𝜋
∴ ∮𝑭 ⋅ 𝑑𝒓 = 16 ∫ (−4 𝑐𝑜𝑠 2 𝑡 sin 𝑡 − 𝑠𝑖𝑛2 𝑡 + sin 𝑡 + 12 sin 𝑡 𝑐𝑜𝑠 2 𝑡)𝑑𝑡
𝐶
0
2𝜋
2𝜋
= 16 ∫ (8 sin 𝑡 𝑐𝑜𝑠2 𝑡 − 𝑠𝑖𝑛2 𝑡 + sin 𝑡)𝑑𝑡 = −16 ∫ 𝑠𝑖𝑛2 𝑡𝑑𝑡 = −16 𝜋
0
0
∫ sin2 𝑡 𝑑𝑡 =
where:
𝑡 sin 2𝑡
−
2
4
2𝜋
∫ sin 𝑡 𝑐𝑜𝑠 𝑛 𝑡 𝑑𝑡 = 0
0
Page 94
2𝜋
𝑎𝑛𝑑
∫ sin 𝑡 𝑑𝑡 = 0
0
Second Class in Department of Physics