Step by step
The question
Question: use Green's theorem to find ∮ (y − x²) dx + (3x) 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 = y − x².
The one before dy is Q = 3x.
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.
3x gives 3.
So ∂Q/∂x = 3.
Term by term.
y gives 1.
x² has no y: it is a fixed number, so it gives 0.
Drop the zeros: ∂P/∂y = 1.
Take away: 3 minus 1.
So the inside is 2.
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 4.
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 8.
6 Check
Let's check by walking the four sides, one at a time.
Bottom −8/3, right 12, top −4/3, left 0.
On each side, only one of x and y moves.
Added up: 8.
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.