On the curve C 1 ,where y= 0, dy = 0,the first integral reduces to
∫
C 1
Mdx +N dy =
∫a
0
x^2 dx =
a^3
3
On the curve C 2 ,where
x=acos θ, y =asin θ, 0 ≤θ≤π
4
the second integral reduces to
∫
C 2
Mdx +N dy =
∫π/ 4
0
−a^2 ·asin θ dθ +a^2 sin θcos θ·acos θ dθ
=a^3 cos θ
π/ 4
0
−
a^3
3 cos
(^3) θ π/^4
0
=a^3 (
√
2
2
−1) −a
3
3
(√
2
4
− 1
)
=a^3
(
5
√
2
12 −
2
3
)
.
On the curve C 3 ,where y=x, 0 ≤x≤
√ 2
2 a, the third integral can be expressed as
∫
C 3
Mdx +N dy =
∫ 0
√ 2
2 a
2 x^2 dx +x^2 dx =−
√
2
4 a
(^3).
Adding the three integrals give us the line integral portion of Green’s theorem
which is
∫
C
©M dx +N dy =
1
3 a
(^3) +a 3
(
5
√
2
12 −
2
3
)
−
√
2
4 a
(^3) =a^3
3
(√
2
2 −^1
)
.
The area integral representing the right-hand side of Green’s theorem is now
evaluated. One finds
∂N
∂x =y,
∂M
∂y = 2y
and ∫∫
R
(
∂N
∂x
−∂M
∂y
)
dx dy =
∫∫
−y dy dx
The geometry of the problem suggests a transformation to polar coordinates in order
to evaluate the integral. Changing to polar coordinates the above integral becomes
∫π/ 4
0
∫a
0
(rsin θ)(r drdθ ) = −r
3
3
a
0
∫π/ 4
0
sin θ dθ =a
3
3
(√
2
2
− 1
)
.