Step by step
The question
Question: use Green's theorem to find ∮ (xy) dx + (2x²) dy around the square 0 ≤ x ≤ 2, 0 ≤ y ≤ 2.
1 Find P and Q
The path goes once around the square, counterclockwise.
The part before dx is P = xy.
The one before dy is Q = 2x².
2 Find the inside
Green's theorem: the line integral around equals the double integral inside of ∂Q/∂x minus ∂P/∂y.
Term by term.
2x²: the power 2 comes down: 2 times 2 is 4.
The power goes down by 1.
So 4x.
So ∂Q/∂x = 4x.
Term by term.
In xy, x is fixed and waits in front.
y gives 1.
So x.
So ∂P/∂y = x.
Take away: 4x minus x.
So the inside is 3x.
Careful!
The order matters.
Q by x, minus P by y.
The other way round changes the sign.
3 Find the limits
Over the square: y from 0 to 2 inside, x from 0 to 2 outside.
4 The inner integral
Integrate each term: the power goes up by 1, and we divide by the new power.
Put in the top y = 2, then take away the bottom y = 0.
The bottom part is 0.
So it is 6x.
5 The outer integral
Integrate each term: the power goes up by 1, and we divide by the new power.
Put in the top x = 2, then take away the bottom x = 0.
The bottom part is 0.
So it is 12.
6 Check
Let's check by walking the four sides, one at a time.
Bottom 0, right 16, top −4, left 0.
On each side, only one of x and y moves.
Added up: 12.
The same as the double integral.
Right!
Let's look back.
First: find P and Q.
Then: find the inside.
Then: find the limits.
Then: the inner integral.
Then: the outer integral.
Last: check.