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Green's theorem

Calculus 3 · 20 problems, easy to harder

▶ Watch the first one solved

How we do it

  1. Find P and Q
  2. Find the inside
  3. Find the limits
  4. The inner integral
  5. The outer integral
  6. Check

The steps for Use Green's theorem on ∮ (x^2 - y) dx + (x + y^2) dy around the…. Every solution names its steps, so you always know what comes next.

Worked examples

Use Green's theorem on ∮ (x^2 - y) dx + (x + y^2) dy around the… Easy

The question

Question: use Green's theorem to find ∮ (x² − y) dx + (x + y²) dy around the square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1.

∮ (x2 − y) dx + (x + y2) dy
around 0 ≤ x ≤ 1, 0 ≤ y ≤ 1

1 Find P and Q

The path goes once around the square, counterclockwise.

The part before dx is P = x² − y.

P = x2 − y

The one before dy is Q = x + y².

Q = x + y2

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.

x gives 1.

y² has no x: it is a fixed number, so it gives 0.

∂Q∂x = 1 + 0

Drop the zeros: ∂Q/∂x = 1.

∂Q∂x = 1

Term by term.

x² has no y: it is a fixed number, so it gives 0.

y gives 1.

∂P∂y = 0 − 1

Drop the zeros: ∂P/∂y = −1.

∂P∂y = −1

Take away: 1 minus (−1).

∂Q∂x − ∂P∂y = 1 − (−1)

So the inside is 2.

∂Q∂x − ∂P∂y = 2

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 square: y from 0 to 1 inside, x from 0 to 1 outside.

∮ = ∫ from 0 to 1 of ∫ from 0 to 1 of 2 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 2 dy = 2y from 0 to 1

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

∫ from 0 to 1 of 2 dy = 2(1) − 2(0)

The bottom part is 0.

So it is 2.

∫ from 0 to 1 of 2 dy = 2

5 The outer integral

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

∫ from 0 to 1 of 2 dx = 2x from 0 to 1

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

∫ from 0 to 1 of 2 dx = 2(1) − 2(0)

The bottom part is 0.

So it is 2.

∫ from 0 to 1 of 2 dx = 2

6 Check

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

Bottom 1/3, right 4/3, top 2/3, left −1/3.

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

bottom 1/3, right 4/3, top 2/3, left −1/3

Added up: 2.

The same as the double integral.

Right!

1/3 + 4/3 + 2/3 − 1/3 = 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 1 of 2 dx = 2

▶ Watch it solved

Use Green's theorem on ∮ (2x - 3y) dx + (x + 4y) dy around the… Medium

The question

Question: use Green's theorem to find ∮ (2x − 3y) dx + (x + 4y) dy around the square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1.

∮ (2x − 3y) dx + (x + 4y) dy
around 0 ≤ x ≤ 1, 0 ≤ y ≤ 1

1 Find P and Q

The path goes once around the square, counterclockwise.

The part before dx is P = 2x − 3y.

P = 2x − 3y

The one before dy is Q = x + 4y.

Q = x + 4y

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.

x gives 1.

4y has no x: it is a fixed number, so it gives 0.

∂Q∂x = 1 + 0

Drop the zeros: ∂Q/∂x = 1.

∂Q∂x = 1

Term by term.

2x has no y: it is a fixed number, so it gives 0.

3y gives 3.

∂P∂y = 0 − 3

Drop the zeros: ∂P/∂y = −3.

∂P∂y = −3

Take away: 1 minus (−3).

∂Q∂x − ∂P∂y = 1 − (−3)

So the inside is 4.

∂Q∂x − ∂P∂y = 4

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 square: y from 0 to 1 inside, x from 0 to 1 outside.

∮ = ∫ from 0 to 1 of ∫ from 0 to 1 of 4 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 4 dy = 4y from 0 to 1

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

∫ from 0 to 1 of 4 dy = 4(1) − 4(0)

The bottom part is 0.

So it is 4.

∫ from 0 to 1 of 4 dy = 4

5 The outer integral

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

∫ from 0 to 1 of 4 dx = 4x from 0 to 1

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

∫ from 0 to 1 of 4 dx = 4(1) − 4(0)

The bottom part is 0.

So it is 4.

∫ from 0 to 1 of 4 dx = 4

6 Check

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

Bottom 1, right 3, top 2, left −2.

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

bottom 1, right 3, top 2, left −2

Added up: 4.

The same as the double integral.

Right!

1 + 3 + 2 − 2 = 4

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 1 of 4 dx = 4

▶ Watch it solved

Use Green's theorem on ∮ (x^2 - 2y) dx + (3x + y^2) dy around the… Harder

The question

Question: use Green's theorem to find ∮ (x² − 2y) dx + (3x + y²) dy around the rectangle 0 ≤ x ≤ 2, 0 ≤ y ≤ 1.

∮ (x2 − 2y) dx + (3x + y2) 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 = x² − 2y.

P = x2 − 2y

The one before dy is Q = 3x + y².

Q = 3x + y2

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.

3x gives 3.

y² has no x: it is a fixed number, so it gives 0.

∂Q∂x = 3 + 0

Drop the zeros: ∂Q/∂x = 3.

∂Q∂x = 3

Term by term.

x² has no y: it is a fixed number, so it gives 0.

2y gives 2.

∂P∂y = 0 − 2

Drop the zeros: ∂P/∂y = −2.

∂P∂y = −2

Take away: 3 minus (−2).

∂Q∂x − ∂P∂y = 3 − (−2)

So the inside is 5.

∂Q∂x − ∂P∂y = 5

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 5 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 5 dy = 5y from 0 to 1

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

∫ from 0 to 1 of 5 dy = 5(1) − 5(0)

The bottom part is 0.

So it is 5.

∫ from 0 to 1 of 5 dy = 5

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 5 dx = 5x from 0 to 2

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

∫ from 0 to 2 of 5 dx = 5(2) − 5(0)

The bottom part is 0.

So it is 10.

∫ from 0 to 2 of 5 dx = 10

6 Check

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

Bottom 8/3, right 19/3, top 4/3, left −1/3.

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

bottom 8/3, right 19/3, top 4/3, left −1/3

Added up: 10.

The same as the double integral.

Right!

8/3 + 19/3 + 4/3 − 1/3 = 10

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 5 dx = 10

▶ Watch it solved

Practice

Try each one, then open the answer. Stuck? Open the problem to watch every step.

Easy

  1. use Green's theorem on ∮ (x^2 - y) dx + (x + y^2) dy around the square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1
    Answer
    ∫ from 0 to 1 of 2 dx = 2
  2. use Green's theorem on ∮ (-y) dx + (x) dy around the rectangle 0 ≤ x ≤ 2, 0 ≤ y ≤ 3
    Answer
    ∫ from 0 to 2 of 6 dx = 12
  3. use Green's theorem on ∮ (y) dx + (2x) dy around the square 0 ≤ x ≤ 2, 0 ≤ y ≤ 2
    Answer
    ∫ from 0 to 2 of 2 dx = 4
  4. use Green's theorem on ∮ (3y) dx + (5x) dy around the rectangle 0 ≤ x ≤ 1, 0 ≤ y ≤ 2
    Answer
    ∫ from 0 to 1 of 4 dx = 4
  5. use Green's theorem on ∮ (x^2) dx + (xy) dy around the square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1
    Answer
    ∫ from 0 to 1 of 12 dx = 12
  6. use Green's theorem on ∮ (y^2) dx + (x^2) dy around the square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1
    Answer
    ∫ from 0 to 1 of (2x − 1) dx = 0

Medium

  1. use Green's theorem on ∮ (2x - 3y) dx + (x + 4y) dy around the square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1
    Answer
    ∫ from 0 to 1 of 4 dx = 4
  2. use Green's theorem on ∮ (y) dx + (x^2) dy around the rectangle 0 ≤ x ≤ 2, 0 ≤ y ≤ 1
    Answer
    ∫ from 0 to 2 of (2x − 1) dx = 2
  3. use Green's theorem on ∮ (-2y) dx + (x) dy around the square 0 ≤ x ≤ 2, 0 ≤ y ≤ 2
    Answer
    ∫ from 0 to 2 of 6 dx = 12
  4. use Green's theorem on ∮ (xy) dx + (2x^2) dy around the square 0 ≤ x ≤ 2, 0 ≤ y ≤ 2
    Answer
    ∫ from 0 to 2 of 6x dx = 12
  5. use Green's theorem on ∮ (xy) dx + (x^2) dy around the rectangle 0 ≤ x ≤ 2, 0 ≤ y ≤ 1
    Answer
    ∫ from 0 to 2 of x dx = 2
  6. use Green's theorem on ∮ (-y^2) dx + (x^2) dy around the square 0 ≤ x ≤ 2, 0 ≤ y ≤ 2
    Answer
    ∫ from 0 to 2 of (4x + 4) dx = 16
  7. use Green's theorem on ∮ (x - y) dx + (x + y) dy around the rectangle 0 ≤ x ≤ 3, 0 ≤ y ≤ 2
    Answer
    ∫ from 0 to 3 of 4 dx = 12
  8. use Green's theorem on ∮ (2y) dx + (3x) dy around the square 0 ≤ x ≤ 3, 0 ≤ y ≤ 3
    Answer
    ∫ from 0 to 3 of 3 dx = 9
  9. use Green's theorem on ∮ (-xy) dx + (x^2 + y) dy around the rectangle 0 ≤ x ≤ 2, 0 ≤ y ≤ 3
    Answer
    ∫ from 0 to 2 of 9x dx = 18
  10. use Green's theorem on ∮ (y - x^2) dx + (3x) dy around the square 0 ≤ x ≤ 2, 0 ≤ y ≤ 2
    Answer
    ∫ from 0 to 2 of 4 dx = 8

Harder

  1. use Green's theorem on ∮ (x^2 - 2y) dx + (3x + y^2) dy around the rectangle 0 ≤ x ≤ 2, 0 ≤ y ≤ 1
    Answer
    ∫ from 0 to 2 of 5 dx = 10
  2. use Green's theorem on ∮ (xy^2) dx + (x^2 y + x) dy around the square 0 ≤ x ≤ 2, 0 ≤ y ≤ 2
    Answer
    ∫ from 0 to 2 of 2 dx = 4
  3. use Green's theorem on ∮ (y^2 - x) dx + (2xy + x^2) dy around the rectangle 0 ≤ x ≤ 1, 0 ≤ y ≤ 2
    Answer
    ∫ from 0 to 1 of 4x dx = 2
  4. use Green's theorem on ∮ (-x^2 y) dx + (xy^2) dy around the square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1
    Answer
    ∫ from 0 to 1 of (13 + x2) dx = 23