Line Integral of a Vector Field

This document explains how to calculate the line integral of a vector field and its significance. It provides a step-by-step solution and examples. The subject is Vector Calculus.

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
1

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
3

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-
4

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.
5