Step by step
The question
Question: use Green's theorem to find ∮ (2y) dx + (3x) dy around the square 0 ≤ x ≤ 3, 0 ≤ y ≤ 3.
1 Find P and Q
The path goes once around the square, counterclockwise.
The part before dx is P = 2y.
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.
2y gives 2.
So ∂P/∂y = 2.
Take away: 3 minus 2.
So the inside is 1.
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 3 inside, x from 0 to 3 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 = 3, then take away the bottom y = 0.
The bottom part is 0.
So it is 3.
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 = 3, then take away the bottom x = 0.
The bottom part is 0.
So it is 9.
6 Check
Let's check by walking the four sides, one at a time.
Bottom 0, right 27, top −18, left 0.
On each side, only one of x and y moves.
Added up: 9.
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.