1 stokes theorem
DESCRIPTION
stokeTRANSCRIPT
Stokes’ Theorem
At the end of the lesson, you should be able to
Apply Stokes’ theorem and determine the direction of unit
normal vectors to a surface.
Introduction
If F is a vector field existing over an open surface S and
around its boundary, closed curve c, then
s
curl F dS = 𝑐
F ∙ dr
This mean that we can express a surface integral in terms
of a line integral round the boundary curve.
Example 1
A hemisphere S is defined by 𝒙𝟐 + y2 + z2 = 4 z ≥ 0 . A vector field 𝑭 = 𝟐𝒚𝒊 − 𝒙𝒋 + 𝒙𝒛 𝒌 exists over the surface and around its boundary c.
verify the Stokes’ theorem, s
curl F dS = 𝑐
F ∙ dr
S: 𝑥2 + 𝑦2 + 𝑧2 − 4 = 0𝐹 = 2𝑦𝑖 − 𝑥𝑗 + 𝑥𝑧𝑘c is the circle 𝑥2 + 𝑦2 = 4
Answer
We need to show that s curl F dS = c F ∙ dr
a) c F ∙ dr
= 𝑐 (2𝑦𝑖 − 𝑥𝑗 + 𝑥𝑧𝑘) ∙ (𝑖 𝑑𝑥 + 𝑗 𝑑𝑦 + 𝑘𝑑𝑧)
= 𝑐 (2𝑦𝑑𝑥 − 𝑥 𝑑𝑦 + 𝑥𝑧 𝑑𝑧)
Converting to polar coordinates
𝑥 = 2 cos 𝜃 ; 𝑦 = 2 sin 𝜃 ; 𝑧 = 0𝑑𝑥 = −2sin 𝜃 𝑑𝜃 ; 𝑑𝑦 = 2 cos 𝜃 𝑑𝜃 ; limit 𝜃 = 0 𝑡𝑜 2𝜋
c
F ∙ d𝑟 =
0
2𝜋
2(2 sin 𝜃)( − 2 sin 𝜃 𝑑𝜃) − 2 cos 𝜃 (2 cos 𝜃 𝑑𝜃) +(2 cos 𝜃)𝑧 𝑑𝑧
= 02𝜋
(4 sin 𝜃[− 2 sin 𝜃 𝑑𝜃] − 2 cos 𝜃 2 cos 𝜃 𝑑𝜃)
𝜃
= −4 02𝜋
2 sin2 𝜃 + cos2 𝜃 𝑑𝜃
= −4 0
2𝜋(1 + sin2 𝜃) 𝑑𝜃
= −2 0
2𝜋(3 − 𝑐𝑜𝑠 2𝜃) 𝑑𝜃
= −2 3𝜃 −(sin 2𝜃)
2
2𝜋0
= −12𝜋
Now we determine for s curl F dS
b) s curl F dS = s curl F 𝑛 ds
curl F =
𝑖 𝑗 𝑘𝜕
𝜕𝑥
𝜕
𝜕𝑦
𝜕
𝜕𝑧
2𝑦 −𝑥 𝑥𝑧= −zj − 3k
𝑛 =∇𝑆
∇S=
2𝑥𝑖 + 2𝑦𝑗 + 2𝑧𝑘
2𝑥 2 + 2𝑦 2 + 2𝑧 2=
𝑥𝑖 + 𝑦𝑗 + 𝑧𝑘
2
s
curl F 𝑛 ds =
s
(−zj − 3k) ∙𝑥𝑖 + 𝑦𝑗 + 𝑧𝑘
2𝑑𝑠
=1
2 s −𝑦𝑧 − 3𝑧 𝑑𝑠
Converting to spherical polar coordinates
𝑥 = 2 sin 𝜃 cos∅ ; 𝑦 = 2 sin 𝜃 sin ∅ ; 𝑧 = 2 cos 𝜃 ;𝑑𝑠 = 4 sin 𝜃 𝑑𝜃 𝑑∅
=1
2
0
2𝜋
0
𝜋2
− 2 sin 𝜃 sin∅ 2 cos 𝜃 − 3 2 cos 𝜃 4 sin 𝜃 𝑑𝜃 𝑑∅
= −4
0
2𝜋
0
𝜋2
2 sin2 𝜃 sin ∅ cos 𝜃 + 3 sin 𝜃 cos 𝜃 𝑑𝜃 𝑑∅
= −4
0
2𝜋
0
𝜋2
2 sin2 𝜃 sin ∅ cos 𝜃 + 3 sin 𝜃 cos 𝜃 𝑑𝜃 𝑑∅
= −12π
So we can verify that s curl F dS = c F ∙ dr
Direction Of Unit Normal Vectors to A Surface S
When we dealing with divergence theorem, the normal vectors
were drawn in a direction outward from the enclosed region.
With open surface, there is in fact no inward or outward
direction. With any general surface, a normal vector can be
drawn in either of 2 opposite direction.
A unit normal 𝑛 is drawn perpendicular to the surface S at any
point in the direction indicated by applying a right handed
screw sense to the direction of integration round the boundary
c.
Let see example
Example 2
A surface consists of five section formed by the planes 𝑥 = 0;𝑥 = 1; 𝑦 = 0; 𝑦 = 3; 𝑧 = 2 in first octant. If the vector field F = 𝑦 𝑖 + 𝑧2𝑗 + 𝑥𝑦 𝑘 exist over the surface and its boundary, verify the Stokes’ theorem
So we have to verify that s
curl F dS = c
F ∙ dr
a) we start with c F ∙ dr = 𝑦 𝑑𝑥 + 𝑧2𝑑𝑦 + 𝑥𝑦 𝑑𝑧
i) along c1: 𝑦 = 0; 𝑧 = 0; 𝑑𝑦 = 0; 𝑑𝑧 = 0
∴ c1
F ∙ dr = 0 + 0 + 0 = 0
ii) along c2: 𝑥 = 1; 𝑧 = 0; 𝑑𝑥 = 0; 𝑑𝑧 = 0
∴ c2
F ∙ dr = 0 + 0 + 0 = 0
iii) along c3: 𝑦 = 3; 𝑧 = 0; 𝑑𝑦 = 0; 𝑑𝑧 = 0
∴ c3
F ∙ dr = 1
03𝑑𝑥 + 0 + 0 = 3𝑥
01
= −3
iv) along c4: 𝑥 = 1; 𝑧 = 0; 𝑑𝑥 = 0; 𝑑𝑧 = 0
∴ c4
F ∙ dr = 0 + 0 + 0 = 0
So c
F ∙ dr = 0 + 0 − 3 + 0 = −3
c
F ∙ dra
b) we continue with s curl F dS
curl F =
𝑖 𝑗 𝑘𝜕
𝜕𝑥
𝜕
𝜕𝑦
𝜕
𝜕𝑧
𝑦 𝑧2 𝑥𝑦
= 𝑥 − 2𝑧 𝑖 − 𝑦𝑗 − 𝑘
s
curl F dS =
s
curl F ∙ 𝑛 ds
i) 𝑆1 𝑡𝑜𝑝 : 𝑛 = 𝑘
s1curl F ∙ 𝑛 ds = s1
𝑥 − 2𝑧, −𝑦 − 1 ∙ 0,0,1
𝑠1−1𝑑𝑠 = −3
ii) 𝑆2 𝑟𝑖𝑔ℎ𝑡 ℎ𝑎𝑛𝑑 𝑒𝑛𝑑 : 𝑛 = 𝑗
s2curl F ∙ 𝑛 ds = s2
𝑥 − 2𝑧, −𝑦 − 1 ∙ 0,1,0
𝑠2
−𝑦𝑑𝑠
but 𝑦 = 3 for this section
𝑠2
−3𝑑𝑠 = −6
iii) 𝑆3 𝑙𝑒𝑓𝑡 ℎ𝑎𝑛𝑑 𝑒𝑛𝑑 : 𝑛 = −𝑗
s3curl F ∙ 𝑛 ds = s3
𝑥 − 2𝑧, −𝑦 − 1 ∙ 0,−1,0
𝑠3
𝑦 𝑑𝑠
but 𝑦 = 0 for this section
𝑠3
0𝑑𝑠 = 0
iv) 𝑆4 𝑓𝑟𝑜𝑛𝑡 : 𝑛 = 𝑖
s4curl F ∙ 𝑛 ds = s4
𝑥 − 2𝑧, −𝑦 − 1 ∙ 1,0,0
= 𝑠4𝑥 − 2𝑧 𝑑𝑠
but x = 1 for this section
𝑠41 − 2𝑧 𝑑𝑠 = 0
3 021 − 2𝑧 𝑑𝑧 𝑑𝑦 = −6
iii) 𝑆5 𝑏𝑎𝑐𝑘 : 𝑛 = −𝑖
𝑆5
curl F ∙ 𝑛 ds = 𝑆5
𝑥 − 2𝑧, −𝑦 − 1 ∙ −1,0,0
= 𝑆5
−𝑥 + 2𝑧 𝑑𝑠
but x = 0 for this section
𝑆5
2𝑧 𝑑𝑠 = 0
3 0
22𝑧 𝑑𝑧 𝑑𝑦 = 12
𝑆
curl F ∙ 𝑛 ds = −3 − 6 + 0 − 6 + 12 = −3
Exercise
A surface S consist of that part of the cylinder x2 + y2 = 9between z = 0 and z = 4 for y > 0 and 2 semicircles of radius 3
in the planes z = 0 and z = 4. If F = zi + xy j + xz k, evaluate
S curl F ∙ dS over the surface.
Answer = −24