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Green's Theorem

Relates a line integral around a closed curve to a double integral over the region it encloses.

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\oint_{C} (L d x + M d y) = \iint_{D} (\frac{\partial M}{\partial x} - \frac{\partial L}{\partial y}) d A

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Core idea

Overview

Green's Theorem establishes a fundamental connection between the line integral around a simple closed curve and the double integral over the plane region it encloses. It is essentially a two-dimensional version of the Stokes' Theorem and is used to relate local rotation or circulation in a vector field to the net curl over an area.

When to use: Apply this theorem when evaluating a line integral over a closed, piecewise-smooth curve in the xy-plane where the area integral of the curl is easier to compute. It requires the component functions L and M to have continuous first-order partial derivatives throughout the region bounded by the curve.

Why it matters: It is essential for calculating work and circulation in physics and fluid dynamics without needing to parameterize complex boundary paths individually. It also provides a mathematical basis for using line integrals to calculate the area of irregular shapes, which is the operational principle behind the planimeter.

Symbols

Variables

\text{Concept-only} = Note

Concept-only

Note

V a r iab l e

Walkthrough

Derivation

Proof of Green's Theorem for a Simple Region

We prove Green's Theorem for a simple type I and type II region by evaluating the line integral over the boundary and showing it equals the double integral of the partial derivatives.

C is a positively oriented, piecewise-smooth, simple closed curve.
P(x,y) and Q(x,y) have continuous partial derivatives on an open region containing D.

1. Decompose the Integral

We can prove the theorem in two independent parts: showing that $\oint L d x = - \iint \frac{\partial L}{\partial y} d A$ and $\oint M d y = \iint \frac{\partial M}{\partial x} d A$ .

\oint_{C} (L d x + M d y) = \oint_{C} L d x + \oint_{C} M d y

2. Setting up the Area Integral for L

Assume the region $D$ is bound by $y = g_{1} (x)$ on the bottom and $y = g_{2} (x)$ on the top, between $x = a$ and $x = b$ .

\iint_{D} \frac{\partial L}{\partial y} d A = \int_{a}^{b} \int_{g_{1} (x)}^{g_{2} (x)} \frac{\partial L}{\partial y} d y d x

3. Applying the Fundamental Theorem of Calculus

Integrating the partial derivative $\frac{\partial L}{\partial y}$ with respect to $y$ simply yields the function $L$ evaluated at the upper and lower boundaries.

\int_{a}^{b} [L (x, g_{2} (x)) - L (x, g_{1} (x))] d x

4. Relating to the Line Integral

The line integral along the bottom path $g_{1}$ goes from $a$ to $b$ , while the top path $g_{2}$ goes backwards from $b$ to $a$ (to maintain counter-clockwise orientation). Reversing the top integral's limits changes its sign.

- \oint_{C} L (x, y) d x = - (\int_{a}^{b} L (x, g_{1} (x)) d x + \int_{b}^{a} L (x, g_{2} (x)) d x)

5. Conclusion

Combining the two results derived via identical logic applied to the $x$ and $y$ axes yields the final statement of Green's Theorem.

\oint_{C} L d x = - \iint_{D} \frac{\partial L}{\partial y} d A and \oint_{C} M d y = \iint_{D} \frac{\partial M}{\partial x} d A

Result

\oint_{C} L d x = - \iint_{D} \frac{\partial L}{\partial y} d A and \oint_{C} M d y = \iint_{D} \frac{\partial M}{\partial x} d A

Source: Standard curriculum — Vector Calculus

Free formulas

Rearrangements

Solve for $\oint P d x + Q d y$

Make oint P dx + Q dy the subject

\oint P d x + Q d y = \iint (Q_{x} - P_{y}) d A

This rearrangement demonstrates common notational variations of Green's Theorem, transforming the initial form using $L$ and $M$ to a more compact form using $P$ , $Q$ , and subscript notation for partial derivatives.

Difficulty: 2/5

The static page shows the finished rearrangements. The app keeps the full worked algebra walkthrough.

Visual intuition

Graph

Graph unavailable for this formula.

The graph is a straight line passing through the origin with a slope of one, showing that the line integral and the double integral are always equal. For a student, this linear relationship means that small values of the double integral correspond to small values of the line integral, while large values scale proportionally. The most important feature is that the slope of one confirms the equivalence between the two sides of the equation, meaning that any change in the region integral results in an identical change

Graph type: linear

Why it behaves this way

Intuition

Imagine a region in the plane filled with a flowing fluid; Green's Theorem states that the total net rotation of the fluid within the entire region is exactly equal to the net flow of the fluid along its outer boundary.

\oint_{C} (L d x + M d y)

The total circulation of the 2D vector field F = <L, M> around the simple closed curve C.

Measures the net 'push' or 'flow' of the vector field along the boundary C. Imagine a tiny paddle wheel on the curve; this term quantifies its net spin as it traverses the entire boundary.

\left(\frac{\partial M}{\partial x}

The z-component of the 2D curl of the vector field F = <L, M>, representing the infinitesimal circulation density at a point.

Quantifies the 'local rotation' or 'swirliness' of the vector field at an infinitesimally small point within the region. A positive value typically indicates counter-clockwise rotation.

\iint_D \left(\frac{\partial M}{\partial x}

The total sum of all infinitesimal circulations (curl) over the entire region D enclosed by C.

Aggregates all the tiny 'swirls' or 'rotations' happening at every point inside the region D to get a total measure of the internal rotation.

Signs and relationships

(∂ M / ∂ x - ∂ L / ∂ y): This specific difference defines the scalar curl (or z-component of the 2D curl) of the vector field F = <L, M>. The order of subtraction is crucial and corresponds to the counter-clockwise orientation of the circulation

Free study cues

Insight

Canonical usage

Used to relate a line integral around a closed curve to a double integral over the enclosed region, where both sides of the equation must maintain consistent physical dimensions determined by the nature of the vector

Common confusion

Students often overlook that the physical dimensions of L and M must be identical for the line integral's integrand (L dx + M dy) to be dimensionally consistent.

Unit systems

$L, M$ Varies (e.g., N for force, m/s for velocity) · Components of the 2D vector field F = <L, M>. L and M must have the same physical dimensions, which dictate the overall dimension of the integrals.

$d x, d y$ m · Infinitesimal displacements along the coordinate axes.

$d A$ m^2 · Infinitesimal area element.

$(\partial M / \partial x - \partial L / \partial y)$ Varies (e.g., N/m for force field, s^-1 for velocity field) · The z-component of the curl of the vector field F. Its dimensions are (dimensions of L or M) / (Length).

One free problem

Practice Problem

Evaluate the line integral ∮_C (y² dx + x² dy) where C is the boundary of the rectangle defined by 0 ≤ x ≤ 2 and 0 ≤ y ≤ 3, oriented counter-clockwise.

Note-6

Solve for: $o u t$

Hint: Convert the line integral into a double integral of the expression (∂M/∂x − ∂L/∂y) over the rectangular region.

The full worked solution stays in the interactive walkthrough.

Where it shows up

Real-World Context

Calculating work done by a force field.

Study smarter

Tips

Ensure the curve is closed and oriented counter-clockwise for a positive result.
Verify that the vector field functions are continuous on the entire region enclosed by the curve.
Use the identity where the area equals the line integral of x dy or -y dx to simplify area problems.
Check that the region is simply connected before applying the standard form of the theorem.

Avoid these traps

Common Mistakes

Using for open curves.
Wrong sign (clockwise orientation).

Common questions

Frequently Asked Questions

We prove Green's Theorem for a simple type I and type II region by evaluating the line integral over the boundary and showing it equals the double integral of the partial derivatives.

Apply this theorem when evaluating a line integral over a closed, piecewise-smooth curve in the xy-plane where the area integral of the curl is easier to compute. It requires the component functions L and M to have continuous first-order partial derivatives throughout the region bounded by the curve.

It is essential for calculating work and circulation in physics and fluid dynamics without needing to parameterize complex boundary paths individually. It also provides a mathematical basis for using line integrals to calculate the area of irregular shapes, which is the operational principle behind the planimeter.

Using for open curves. Wrong sign (clockwise orientation).

Calculating work done by a force field.

Ensure the curve is closed and oriented counter-clockwise for a positive result. Verify that the vector field functions are continuous on the entire region enclosed by the curve. Use the identity where the area equals the line integral of x dy or -y dx to simplify area problems. Check that the region is simply connected before applying the standard form of the theorem.

References

Sources

Calculus: Early Transcendentals by James Stewart
Vector Calculus by Jerrold E. Marsden and Anthony J. Tromba
Wikipedia: Green's theorem
Stewart, Calculus: Early Transcendentals
Halliday, Resnick, and Walker, Fundamentals of Physics
Bird, Stewart, and Lightfoot, Transport Phenomena
Britannica, Green's theorem
Wikipedia, Green's theorem

Green's Theorem

Overview

Variables

Derivation

1. Decompose the Integral

2. Setting up the Area Integral for L

3. Applying the Fundamental Theorem of Calculus

4. Relating to the Line Integral

5. Conclusion

Rearrangements

Graph

Intuition

Insight

Practice Problem

Real-World Context

Tips

Common Mistakes

Related Formulas

Divergence Theorem

Curl (concept)

Frequently Asked Questions

Sources