MATH2001 Assignment 4 Solution: Vector Fields and Green's Theorem

This document provides a comprehensive solution to MATH2001 Assignment 4, focusing on vector calculus concepts. The solution begins by evaluating line integrals along different paths for a given vector field, including a parabola and a straight line, and determining if the field is conservative. It then proceeds to evaluate a line integral along a helical path for a second vector field and finds the corresponding potential function. The assignment also covers the application of Green's Theorem, calculating both line integrals and double integrals to find the flux of a vector field over a given path, including the use of parametric equations to define the path. The solution demonstrates the calculation of the flux over sub-paths and utilizes Green's Theorem to verify the results, providing a detailed analysis of vector fields, line integrals, and flux calculations.

The given vector field is:
F=¿ xy ,−x2 >¿
Path of integration C is potion of unit circle in first quadrant as shown in figure below:
Figure 1 - Path of integral C and region of integration D
a) Line integral ∮
C
F · dr
In order to evaluate the line integral, we need to parametrize curve C. The curve is divided
into 3 paths: C1, C2 and C3 as shown in the figure above.
Property of line integrals:
∮
C
F · dr =∮
C 1
F · dr+∮
C 2
F · dr+∮
C 3
F · dr

1) Path C1: line from (0,0) to (1,0)
Parametric equation for C1:
r ( t ) =¿ t , 0> ,0 ≤ t ≤ 1
⇒ dr =¿ 1 , 0>dt
therefore,
∮
C 1
F · dr=∮
C 1
¿ xy ,−x2 >·<t , 0>dt
Put: x = t and y = 0
¿∮
C 1
¿ 0 ,−t2 >·<t , 0> dt
¿∮
C 1
0 ·t−t2 ·0 dt=0
2) Path C2: circle from (1,0) to (0,1)
Parametric equation for C2:
r ( t )=¿ cos t ,sin t >, 0 ≤t ≤ π
2
⇒ dr =←sin t ,cos t>dt
therefore,
∮
C 2
F · dr=∮
C 2
¿ xy ,−x2 >←sin t ,cos t>dt
Put: x = cos t, y = sin t
¿∮
C 2
¿ cos t·sin t ,−cos2 t >·←sin t , cos t> dt
¿∫
0
π
2
−sin2 t· cos t−cos3 t dt =∫
0
π
2
−cos t ( sin2 t+cos2 t ) dt
∫
0
π
2
−cos t · dt= [ −sin t ]0
π
2 =−1
3) Path C3: line form (1,0) to (0,1)
Parametric equation for C3:
r ( t ) =¿ 0 , 1−t >, 0 ≤t ≤ 1
⇒ dr =¿ 0 ,−1> dt
therefore,

∮
C 3
F · dr=∮
C 3
¿ xy ,−x2 >·<0 ,−1> dt
Put: x = 0, y = 1-t
∮
C 3
¿ 0 , 0>·<0 ,−1>dt=0
Therefore, the line integral over path C is:
∮
C
F · dr =0+−1+0=−1
b) Double Integral:
∂ F2
∂ x − ∂ F1
∂ y = ∂
∂ x (−x2 ) − ∂
∂ y ( xy ) =−2 x −x=−3 x
therefore,
∬ ( ∂ F2
∂ x −∂ F1
∂ y )dA=∬
D
−3 x dA
Region D is the are enclosed by path of the integral (C), as shown in the figure-1.
Limits of integration in region D are:
x : 0≤ x ≤ 1 , y :0 ≤ √ 1−x2 ≤ 1
Therefore,
∬
D
−3 x dA=∫
0
1
∫
0
√1−x2
−3 x dydx=∫
0
1
[ −3 xy ]0
√1− x2
dx=∫
0
1
−3 x √1−x2 dx
Substitute:
1−x2=t ⇒−2 x dx=dt
∫
0
1
3
2 √t dt=[t
3
2 ]0
1
= [ ( 1−x2 )
3
2 ]0
1
=0−1=−1
This result is equal to the result in part a).
c) Flux ∮
C
F · dr
The flux of vector field over the entire path is again calculated by computing flux over
sub-paths (C1, C2 and C3) and adding them:

∮
C
F · n ds=∮
C 1
F · n ds+∮
C 2
F · n ds +∮
C 3
F · n ds
Parametric equation for C1, C2 and C3 is defined in part a). Normal components of
direction vector for each path is calculated as below:
Path C1:
n ds=¿ dy
dt ,− dx
dt >dt=¿ 0 ,−1>dt
Path C2:
n ds=¿ cos t , sin t>dt
Path C3:
n ds=¿−1 ,0 >dt
therefore,
∮
C 1
F · n ds=∮
C1
¿ xy ,−x2 >·<0 ,−1>dt
Put: x = t, y = 0
¿∮
C 1
¿ 0 ,−t 2 >·<0,1>dt =∫
0
1
t2 dt=[ t3
3 ]0
1
= 1
3
∮
C 2
F · n ds=∮
C 2
¿ xy ,−x2 >·<cos t , sin t> dt
Put: x=cos t, y = sin t
¿∮
C 3
¿ cos t· sin t ,−cos2 t >·<cos t ,sin t >dt
¿∮
C 3
cos2 t· sin t−cos2 t· sint dt=0
∮
C 3
F · n ds=∮
C 3
¿ xy ,−x2>·<−1,0>dt
Put: x = 0, y = 1-t
¿∮
C 3
¿ 0,0>·<−1,0>dt=0
Therefore, the flux of the field over path C is:
∮
C
F · n ds= 1
3 + 0+0= 1
3

d) Flux by Green’s Theorem:
Divergence of the field is given as:
¿ ( F )=∇ · F= ∂ F1
∂ x + ∂ F2
∂ y = ∂
∂ x ( xy )+ ∂
∂ y ( −x2 ) = y
By Green’s Theorem:
∮
C
F · n ds=∬
D
¿ ( F ) dA
¿∬
D
y dA=∫
0
1
∫
0
√ 1−x2
y dydx=∫
0
1
0.5 [ y2 ] 0
√ 1−x2
dx=∫
0
1
0.5 ( 1−x2 ) dx= [ 0.5 x− x3
6 ] 0
1
= 1
3
A curve defined by general equation r ( t ) =¿ x ( t ) , y ( t ) >¿, is shown in figure below. The direction
along curve is t which is positive anti-clockwise while the outward direction of normal (n) is positive.
Given vector filed is:
F=a=¿ a1 , a2 >¿
Flux of the vector field is evaluated using integral:
Flux = ∮
C
F · n ds
The component of curve along its direction is calculated as:
t ds=dr=¿ x ' ( t ) , y ' (t ) >dt
The normal component of curve is obtained by rotating the above vector by 90 degree. This
transformation is given as:
n ds=¿ y ' ( t ) , x ( t ) >dt
Therefore,
Flux =∮
C
¿ a1 ,a2> ·< y ' ( t ) ,−x ' ( t ) >dt=∫
a
b
a1 y ' ( t ) −a2 x ' ( t ) dt
Divergence of Constant vector field is given as:

¿ ( F ) = ∂
∂ x a1+ ∂
∂ y a2=0
When the curve C is closed, we can apply Green’s Theorem to obtain flux of the field over the curve:
Flux =∮
C
F ·n ds=∬
D
¿ ( F ) dA
where, D is the region enclosed by the curve C.
As divergence of the field is 0, the flux will also be 0 by Green’s Theorem.
Figures form:
https://ocw.mit.edu/courses/mathematics/18-02sc-multivariable-calculus-fall-2010/3.-double-
integrals-and-line-integrals-in-the-plane/part-c-greens-theorem/session-69-flux-in-2d/
MIT18_02SC_MNotes_v3.pdf