Line Integral of a Vector Field

Answer:
The given vector field is:
F = 〈4x − y2 sin x, z + 2y cos x, y〉
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) = 〈x(t), y(t), z(t)〉 [α ≤ t ≤ β]
For C, z is constant and equal to 2. For a clockwise oriented circle, with
radius a centered at (h, k), the parametric equation is:
r(t) = 〈a cos t + h, −a sin t + k〉 [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) = 〈3 cos t + 1, −3 sin t + 3, 2〉 [0 ≤ t ≤ 1.5π]
Implies, dr = 〈−3 sin t, −3 cos t, 0〉 dt
Then line integral is then calculated as:
∫
C
F · dr =
∫
C
〈4x − y2 sin x, z + 2y cos x, y〉 · 〈−3 sin t, −3 cos t, 0〉 dt
[Put: x = 3 cos t + 1, y = −3 sin t + 3, z = 2]
=
∫
C
〈12 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 + 3〉 · 〈−3 sin t, −3 cos t, 0〉 dt
[0 ≤ t ≤ 1.5π]
=
∫ 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