Green's theorem! Will rate best answer!

Evaluate the line integral xy^2dx+2x^2ydy where AD is the ellipse 1/9x^2+1/4y^2 =1

I have tried to use Greens theorem... so i get double integral of (4xy-2xy)dxdy

Just have no where what to do from there. please help!

You can compute this with or without Green's Theorem. The answer is zero either way.

To use Green's Theorem, you can use a sort of modified polar coordinates. If you let

x/3 = r cosΘ and y/2 = r sinΘ,

then the boundary of the ellipse occurs when r = 1. The interior of the ellipse is mapped out when 0 ≤ r < 1 if you let Θ vary from 0 to 2π.

This changes the differential area element a little bit. With regular polar coordinates, dx dy = r dr dΘ. But the extra factors of 3 and 2 give the Jacobian

dx dy = (∂x/∂r ∂y/∂Θ - ∂x/∂Θ ∂y/∂r) dr dΘ = 6 r dr dΘ.

The area integral is

2π 1
∫ . . ∫ 2xy dA =
0 . 0

2π . 1
∫ . . . ∫ 2(3r cosΘ)(2 r sinΘ) 6 r dr dΘ = 0.
0 . . 0

The actual integration is pretty straight forward.

To set up the integral without Green's Theorem, the same approach applies. Use

x = 3 cosΘ, y = 2sinΘ ==> dx = -3sinΘ dΘ and dy = 2cosΘ dΘ.

The line integral ends up being

2π
∫ (-36sin^3Θ cosΘ + 72cos^3ΘsinΘ) dΘ = 0.
0