Detailed Analysis of Line Integrals and Vector Fields in Calculus

This document presents a detailed solution to a calculus homework assignment involving line integrals and vector fields. The solution begins by finding a parametric equation for a given curve C, a circle with radius 3. It then calculates the line integral of a given vector field F along C. The solution proceeds to compute the curl of the vector field, determining if it's conservative. Because the curl is zero, the solution finds the potential function f, demonstrating that F is a conservative field. Finally, the line integral is evaluated using the Fundamental Theorem of Calculus (FTC), comparing the result with the direct line integral calculation from the first part. The document includes references to calculus textbooks and online resources, providing a comprehensive analysis of the problem and its solution.

Answer:
The given vector field is:
F = h4x − y2 sin x,z + 2y cos x,yi
Curve C is a circle with radius 3 centered at (1, 3, 2),traversed clockwise,
from (4, 3, 2) to (1, 6, 2).It can be imagined as a portion of circle with radius
3, centered at point (1, 3), lying in a plane z = 2, parallel to xy- plane.It is
plotted and shown in figure-1 below.
Figure 1:Path of integration C
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a)
In order to evaluate the line integral,we first need to find a parametric
equation for curve C.
Parametric form of a curve, in terms of a parameter t, is given as:
r(t) = hx(t),y(t), z(t)i [α ≤ t ≤ β]
For C, z is constant and equalto 2. For a clockwise oriented circle,with
radius a centered at (h, k), the parametric equation is:
r(t) = ha cos t + h,−a sin t + ki[α ≤ t ≤ β]
For C: h = 1, k = 3, & a = 3. Also, t goes from 0 to 1.5π,as C moves
from (4, 3) to (1, 6), clockwise.Therefore, parametric equation of C, in plane
z = 2 is:
r(t) = h3 cos t + 1, −3 sin t + 3,2i [0 ≤ t ≤ 1.5π]
Implies, dr = h−3 sin t, −3 cos t,0i dt
Then line integral is then calculated as:
Z
C
F · dr =
Z
C
h4x − y2 sin x,z + 2y cos x,yi · h−3 sin t, −3 cos t,0i dt
[Put: x = 3 cos t + 1,y = −3 sin t + 3,z = 2]
=
Z
C
h12 cos t + 4 − (−3 sin t + 3)2 sin (3 cos t + 1),
2 + 6 cos (3 cos t + 1) − 6 sin t cos (3 cos t + 1),
− 3 sin t + 3i · h−3 sin t, −3 cos t,0i dt
[0 ≤ t ≤ 1.5π]
=
Z 2π
0
−36 sin t cos t − 12 sin t + 3 sin t(−3 sin t + 3)2 sin (3 cos t + 1)
− 6 cos t − 18 cos t cos (3 cos t + 1) + 18 cos t sin t cos (3 cos t + 1)dt
2

[Using Integral Calculator, we get]
Z 1.5π
0
F · dr ≈ 1.3337
b
Curl is calculated as:
curl(F) = ∂
∂x , ∂
∂y, ∂
∂z × h4x − y2 sin x,z + 2y cos x,yi
=
ˆi ˆj ˆk
∂
∂x
∂
∂y
∂
∂z
4x − y2 sin x z + 2y cos xy
= ∂
∂y(y) − ∂
∂z (z + 2y cos x)ˆi − ∂
∂x (y) − ∂
∂z (4x − y2 sin x) ˆj
+ ∂
∂x (z + 2y cos x) −
∂
∂y(4x − y2 sin x) ˆk
= (1 − 1)ˆi − (0 − 0)ˆj + (−2y sin x + 2y sin x)ˆk
= h0,0, 0i = 0
c)
Zero curlindicates that the field is a conservative field (irrotational) and
there exists a potential f such that F = ∇f .
d)
Potential function f :
F = ∇f
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Implies,
hfx, f y, f zi = h4x − y2 sin x,z + 2y cos x,yi
We have,
f z = y =⇒ ∂f
∂z = y
Integrating we get:
f = yz + h(x, y)
Now,
f y = z + 2y cos x =⇒ f = yz + y2 cos x + g(x)
Taking derivative of f with respect to x:
f x = −y2 sin(x) + g0
(x)
But, fx = 4x − y2 sin x.Comparing we get:
g0
(x) = 4x =⇒ g(x) = 2x2 + C1
C1 is a constant.
Therefore, finally we get:
f = 2x2 + y2 cos x + yz + C1
which is the potential function for the given vector field.
e)
Integralin a conservative field (with potentialf ) along a curve C,by FTC
is evaluated as: Z
C
F · dr =
Z
C
∇f dr = f b − fa
where, a and b are starting and ending points of C.
In the given problem,C starts from (4, 3, 2) and ends at (1, 6, 2).There-
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fore:
Z
C
F · dr = f(1,6,2) − f(4,3,2)
= (2x2 + y2 cos x + yz)(1,6,2)− (2x2 + y2 cos x + yz)(4,3,2)
= (2 + 36 cos 1 + 12) − (32 + 9 cos 4 + 6)
= −9 cos 4 + 36 cos 1 − 24 ≈ 1.3337
f )
Results from a) and e) agree completely.Part e), using FTC, was much
easier.
References:
1. Stewart, J. (2016).Single variable calculus.Boston, MA: Cengage Learn-
ing.
2. Jacks, K. (2012).Vector Calculus.New Delhi, World Technologies.
http://public.eblib.com/choice/publicfullrecord.aspx?p=841282.
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