Begin2.DVI

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

)

.