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Use Green's theorem on ∮ (y) dx + (x^2) dy around the rectangle 0 ≤…

Step by step, with the reason for every step · Green's theorem · Calculus 3

Step by step

The question

Question: use Green's theorem to find ∮ (y) dx + (x²) dy around the rectangle 0 ≤ x ≤ 2, 0 ≤ y ≤ 1.

∮ y dx + x2 dy
around 0 ≤ x ≤ 2, 0 ≤ y ≤ 1

1 Find P and Q

The path goes once around the rectangle, counterclockwise.

The part before dx is P = y.

P = y

The one before dy is Q = x².

Q = x2

2 Find the inside

Green's theorem: the line integral around equals the double integral inside of ∂Q/∂x minus ∂P/∂y.

∮ P dx + Q dy = ∬ (∂Q/∂x − ∂P/∂y) dA

Term by term.

x²: the power 2 comes down.

The power goes down by 1.

So 2x.

So ∂Q/∂x = 2x.

∂Q∂x = 2x

Term by term.

y gives 1.

So ∂P/∂y = 1.

∂P∂y = 1

Take away: 2x minus 1.

∂Q∂x − ∂P∂y = 2x − 1

So the inside is 2x − 1.

∂Q∂x − ∂P∂y = 2x − 1

Careful!

The order matters.

Q by x, minus P by y.

The other way round changes the sign.

Careful: ∂Q/∂x first, then minus ∂P/∂y

3 Find the limits

Over the rectangle: y from 0 to 1 inside, x from 0 to 2 outside.

∮ = ∫ from 0 to 2 of ∫ from 0 to 1 of (2x − 1) dy dx

4 The inner integral

Integrate each term: the power goes up by 1, and we divide by the new power.

∫ from 0 to 1 of (2x − 1) dy = [2xy − y] from 0 to 1

Put in the top y = 1, then take away the bottom y = 0.

∫ from 0 to 1 of (2x − 1) dy = (2x(1) − (1)) − (2x(0) − (0))

The bottom part is 0.

So it is 2x − 1.

∫ from 0 to 1 of (2x − 1) dy = 2x − 1

5 The outer integral

Integrate each term: the power goes up by 1, and we divide by the new power.

∫ from 0 to 2 of (2x − 1) dx = [x2 − x] from 0 to 2

Put in the top x = 2, then take away the bottom x = 0.

∫ from 0 to 2 of (2x − 1) dx = ((2)2 − (2)) − ((0)2 − (0))

The bottom part is 0.

So it is 2.

∫ from 0 to 2 of (2x − 1) dx = 2

6 Check

Let's check by walking the four sides, one at a time.

Bottom 0, right 4, top −2, left 0.

On each side, only one of x and y moves.

bottom 0, right 4, top −2, left 0

Added up: 2.

The same as the double integral.

Right!

0 + 4 − 2 + 0 = 2

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.

Answer
∫ from 0 to 2 of (2x − 1) dx = 2

Your turn

Try this one yourself, then check:

Use Green's theorem to find ∮ y dx + 3x2 dy, 0 ≤ x ≤ 2, 0 ≤ y ≤ 1
Show the answer
10

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