Proving Green's Theorem

Answer
The given vector field is:
F = 〈P, Q〉 = 〈−y2, x2〉
Circulation of a vector field around a closed curve C, by Green’s Theorem,
is given as: ∮
C
P dx + Q dy =
∫ ∫
D
Qx − Py dA
where, C is oriented counter-clockwise and D is the region enclosed by C.
In the given problem, C is composed of a semi-circle C1 from (2, 0) to (−2, 0)
and a segment C2 from (−2, 0) to (2, 0), as shown in figure-1 below.

Figure 1: Path of integral (C)
a) Left side of Green’s Theorem:
Using the property of line integrals, circulation is calculated as:
∮
C
P dx + Q dy =
∫
C1
P dx + Q dy +
∫
C2
P dx + Q dy
In order to evaluate these line integrals, we need to first define the parametric
equations for curve C1 and C2.
Path C1: from (2, 0) to (−2, 0)
It is a semi-circle centered at origin with radius=2. Therefore,
r(t) = 〈2 cos t, 2 sin t [0 ≤ t ≤ π]
Implies, dr = 〈−2 sin t, 2 cos t〉 dt
Therefore,
∫
C1
P dx + Q dy =
∫
C1
〈P, Q〉 · dr
=
∫
C1
〈−y2, x2〉 · 〈−2 sin t, 2 cos t〉 dt
[Put: y = 2 sin t, x = 2 cos t]
=
∫
C1
〈−4 sin2 t, 4 cos2 t〉 · 〈−2 sin t, 2 cos t〉 dt
=
∫
C1
8 sin3 t + 8 cos3 t dt

[0 ≤ t ≤ π]
=
∫ π
0
8 sin3 t + 8 cos3 t dt
[Trig. Identity: sin3 x = 3
4 sin x − 1
4 sin 3x]
[Trig. Identity: cos3 x = 3
4 cos x + 1
4 cos 3x]
= 8
∫ π
0
3
4 sin t − 1
4 sin 3t + 3
4 cos t + 1
4 cos 3t dt
= 8
[
−3
4 cos t + cos 3t
12 + 3
4 sin t + sin 3t
12
]π
0
= 8
[3
2 − 1
6 + 0 + 0
]
= 32
3
Path C2: from (−2, 0) to (2, 0)
It is straight line connecting two points along the x- axis. Therefore,
r(t) = 〈t, 0〉 [−2 ≤ t ≤ 2]
Implies, dr = 〈1, 0〉 dt Therefore,
∫
C2
P dx + Q dy =
∫
C2
〈P, Q〉 · dr
=
∫
C2
〈−y2, x2〉 · 〈1, 0〉 dt