Calculating Area with Green's Theorem: Step-by-Step Solution

This assignment provides a detailed solution for calculating the area enclosed by a curve using Green's Theorem. It begins by demonstrating the relationship between the circulation of a vector field and the area enclosed by a curve, showing that the area can be calculated using a line integral. The solution then applies this formula to a specific circle, providing a step-by-step calculation of the area using the parametric form of the circle. This result is verified by calculating the area using the standard geometric formula for the area of a circle, confirming the accuracy of Green's Theorem. This document is available on Desklib, a platform offering a range of study tools, including past papers and solved assignments to support student learning.

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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