Green's Theorem and Circulation of Vector Field

This document explains Green's Theorem and its application in calculating the circulation of a vector field. It discusses the concept of curl and provides a step-by-step calculation of the circulation using Green's Theorem. The document also explores the area formula derived from Green's Theorem and compares the results obtained from the formula with the geometric calculation of area. References to relevant textbooks are provided for further study.

Q 3.
a)
The given vector field is:
F = h−y,xi
Green’s Theorem relates circulation of a vector field F around a closed curve
C to curl of the field over the area D enclosed by C:
Circulation of F =
I
C
F · dr =
ZZ
D
curl(F) dA
Curve C is oriented counter-clockwise.A generalclosed curve C oriented
counter-clockwise, enclosing area D is shown in figure-1 below.
Figure 1:Positively oriented curve C and the region inside it D
Curl of a vector field F = hM,N i is given by:
curl(F) = ∇ × F
= ∂
∂x , ∂
∂y × hM, Ni
= ∂N
∂x − ∂M
∂y
1

Therefore, for F = h−y,xi:
curl(F) = ∂
∂x (x) − ∂
∂y(−y)
= 1 − (−1) = 2
The Green’s Theorem applied to the field h−y,xi over a closed curve C
gives: I
C
h−y,xi · dr =
ZZ
D
curl(F) dA
r = hx,yi is the position vector of the curve and it’s differential is given by:
dr = hdx,dyi.
Therefore,
I
C
h−y,xi · hdx,dyi =
ZZ
D
curl(F) dA
Expanding the vector dot product in the line integraland substituting the
value of curl(F) in the double integral we get:
I
C
−y dx + x dy =
ZZ
D
2 dA
But, ZZ
D
dA = Area of D
It is the double integral which gives the area of region of integration.
Therefore, I
C
−y dx + x dy = 2Area(D)
This whole equation can be divided by 2 and expressed as:
Area(D) =1
2
I
C
−y dx + x dy
b)
In order to use the area formula from Green’s Theorem, let us circulate the
vector field h−y,xi around the circle C.
2

Figure 2:Circulation around positive curve C
The area inside C, that is area of D is given by:
Area(D) =1
2
I
C
−y dx + x dy
For the matter of convenience,let us switch back to vector form of the line
integral:
Area(D) =1
2
I
C
h−y,xi · dr
In order to evaluate this line integral, we require the parametric form of circle
C.
A circle oriented counter-clockwise and centered at origin with radius a has
parametric form:
r(t) = ha cos t,a sin ti [0 ≤ t ≤ 2π]
If the center is shifted to (h, k), the parametric equation is modified as:
r(t) = ha cos t + h,a sin t + ki
Therefore, the given circle, with radius = 3 and center at (6, 4), is expressed
as:
r(t) = h3 cos t + 6,3 sin t + 4i[0 ≤ t ≤ 2π]
3

Implies, dr = h−3 sin t,3 cos ti dt
Therefore, area of D is calculated as:
Area(D) =1
2
I
C
h−y,xi · dr
= 1
2
I
C
h−y,xi · h−3 sin t,3 cos ti dt
[Substitute: x = 3 cos t + 6,y = 3 sin t + 4]
= 1
2
I
C
h−3 sin t − 4,3 cos t + 6i · h−3 sin t,3 cos ti dt
= 1
2
I
C
9 sin2 t + 12 sin t + 9 cos2 t + 18 cos tdt
[Trig. Property:sin2 t + cos2 t = 1]
= 1
2
I
C
9 + 12 sin t + 18 cos tdt
[0 ≤ t ≤ 2π]
= 1
2
Z 2π
0
9 + 12 sin t + 18 cos tdt
[Periodic functions:
Z 2π
0
sin x dx =
Z 2π
0
cos x dx = 0]
∴ Area(D) =1
2
Z 2π
0
9 dt =9
2
h
t
i 2π
0
= 9π
4

c)
Area by geometry:
C is a circle with radius r = 3.
Area = π r2 = π 32 = 9π
d)
Yes, results in b) and c) agree.
References:
[1] Stewart, James.Single Variable Calculus:Early Transcendentals.Eight
edition.Boston, MA, USA: Cengage Learning, 2016.
[2] Musa, Sarhan M. and David Santos.Multivariable and Vector Calculus:
An Introduction.Dulles, Virginia:Mercury Learning and Information, 2015.
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