Green's theorem

In

• Green's Theorem on MathWorld

External links

• Calculus (5th edition), F. Ayres, E. Mendelson, Schaum's Outline Series, 2009, ISBN 978-0-07-150861-2.
• Advanced Calculus (3rd edition), R. Wrede, M.R. Spiegel, Schaum's Outline Series, 2010, ISBN 978-0-07-162366-7.

Further reading

References

• Planimeter
• Method of image charges – A method used in electrostatics that takes advantage of the uniqueness theorem (derived from Green's theorem)
• Shoelace formula - A special case of Green's theorem for simple polygons

See also

A=\oint_{C} x\, dy = -\oint_{C} y\, dx = \tfrac 12 \oint_{C} (-y\, dx + x\, dy).

Possible formulas for the area of D include:^[4]

A = \oint_{C} (L\, dx + M\, dy).

Then the area is given by:

\frac{\partial M}{\partial x} - \frac{\partial L}{\partial y} = 1.

Provided we choose L and M such that:

A = \iint_{D}dA.

Green's theorem can be used to compute area by line integral.^[4] The area of D is given by:

Area Calculation

\oint_{C} (M, -L) \cdot \mathbf{\hat n}\,ds = \iint_D\left(\nabla \cdot (M, -L)\right)dA = \iint_D \left(\frac{\partial M}{\partial x} - \frac{\partial L}{\partial y}\right) \, dA.

Applying the two-dimensional divergence theorem with \mathbf{F} = (M, -L), we get the right side of Green's theorem:

\oint_{C} (L\, dx + M\, dy) = \oint_{C} (M, -L) \cdot (dy, -dx) = \oint_{C} (M, -L) \cdot \mathbf{\hat n}\,ds.

Start with the left side of Green's theorem:

To see this, consider the unit normal \mathbf{\hat n} in the right side of the equation. Since in Green's theorem d\mathbf{r} = (dx, dy) is a vector pointing tangential along the curve, and the curve C is the positively-oriented (i.e. counterclockwise) curve along the boundary, an outward normal would be a vector which points 90° to the right of this; one choice would be (dy, -dx). The length of this vector is \sqrt{dx^2 + dy^2} = ds. So (dy, -dx) = \mathbf{\hat n}\,ds.

where \nabla\cdot\mathbf{F} is the divergence on the two-dimensional vector field \mathbf{F}, and \mathbf{\hat n} is the outward-pointing unit normal vector on the boundary.

\iint_D\left(\nabla\cdot\mathbf{F}\right)dA=\oint_C \mathbf{F} \cdot \mathbf{\hat n} \, ds,

Considering only two-dimensional vector fields, Green's theorem is equivalent to the two-dimensional version of the divergence theorem:

Relationship to the divergence theorem

\oint_C L \,dx + M \,dy = \oint_{\partial D} \omega = \int_D \,d\omega = \int_D \frac{\partial L}{\partial y} \,dy \wedge \,dx + \frac{\partial M}{\partial x} \,dx \wedge \,dy = \iint_D \left(\frac{\partial M}{\partial x} - \frac{\partial L}{\partial y}\right) \,dx \,dy.

Green's theorem is also a straightforward result of the general Stokes' theorem using differential forms and exterior derivatives:

\iint_S \nabla \times \mathbf{F} \cdot \mathbf{\hat n} \, dS = \iint_D \left(\frac{\partial M}{\partial x} - \frac{\partial L}{\partial y}\right) \, dA.

Thus we get the right side of Green's theorem

\nabla \times \mathbf{F} \cdot \mathbf{\hat n} = \left[ \left(\frac{\partial 0}{\partial y} - \frac{\partial M}{\partial z}\right) \mathbf{i} + \left(\frac{\partial L}{\partial z} - \frac{\partial 0}{\partial x}\right) \mathbf{j} + \left(\frac{\partial M}{\partial x} - \frac{\partial L}{\partial y}\right) \mathbf{k} \right] \cdot \mathbf{k} = \left(\frac{\partial M}{\partial x} - \frac{\partial L}{\partial y}\right).

The expression inside the integral becomes

The surface S is just the region in the plane D, with the unit normals \mathbf{\hat n} pointing up (in the positive z direction) to match the "positive orientation" definitions for both theorems.

\oint_{C} \mathbf{F} \cdot d\mathbf{r} = \iint_S \nabla \times \mathbf{F} \cdot \mathbf{\hat n} \, dS.

Kelvin–Stokes Theorem:

\oint_{C} (L\, dx + M\, dy) = \oint_{C} (L, M, 0) \cdot (dx, dy, dz) = \oint_{C} \mathbf{F} \cdot d\mathbf{r}.

We can augment the two-dimensional field into a three-dimensional field with a z component that is always 0. Write F for the vector-valued function \mathbf{F}=(L,M,0). Start with the left side of Green's theorem:

Green's theorem is a special case of the Kelvin–Stokes theorem, when applied to a region in the xy-plane:

Relationship to the Stokes theorem

Combining (3) with (4), we get (1) for regions of type I. A similar treatment yields (2) for regions of type II. Putting the two together, we get the result for regions of type III.

\begin{align} \int_{C} L\, dx & = \int_{C_1} L(x,y)\, dx + \int_{C_2} L(x,y)\, dx + \int_{C_3} L(x,y)\, dx + \int_{C_4} L(x,y)\, dx \\ & = -\int_a^b L(x,g_2(x))\, dx + \int_a^b L(x,g_1(x))\, dx.\qquad\mathrm{(4)} \end{align}

Therefore,

\int_{C_4} L(x,y)\, dx = \int_{C_2} L(x,y)\, dx = 0.

The integral over C₃ is negated because it goes in the negative direction from b to a, as C is oriented positively (counterclockwise). On C₂ and C₄, x remains constant, meaning

\int_{C_3} L(x,y)\, dx = -\int_{-C_3} L(x,y)\, dx = - \int_a^b L(x,g_2(x))\, dx.

With C₃, use the parametric equations: x = x, y = g₂(x), a ≤ x ≤ b. Then

\int_{C_1} L(x,y)\, dx = \int_a^b L(x,g_1(x))\, dx.

With C₁, use the parametric equations: x = x, y = g₁(x), a ≤ x ≤ b. Then

Now compute the line integral in (1). C can be rewritten as the union of four curves: C₁, C₂, C₃, C₄.

\begin{align} \iint_D \frac{\partial L}{\partial y}\, dA & =\int_a^b\,\int_{g_1(x)}^{g_2(x)} \frac{\partial L}{\partial y} (x,y)\,dy\,dx \\ & = \int_a^b \Big\{L(x,g_2(x)) - L(x,g_1(x)) \Big\} \, dx.\qquad\mathrm{(3)} \end{align}

where g₁ and g₂ are continuous functions on [a, b]. Compute the double integral in (1):

D = \{(x,y)|a\le x\le b, g_1(x) \le y \le g_2(x)\}

Assume region D is a type I region and can thus be characterized, as pictured on the right, by

are true, then Green's theorem follows immediately for the region D. We can prove (1) easily for regions of type I, and (2) for regions of type II. Green's theorem then follows for regions of type III.

\oint_{C} M\, dy = \iint_{D} \left(\frac{\partial M}{\partial x}\right)\, dA\qquad\mathrm{(2)}

and

\oint_{C} L\, dx = \iint_{D} \left(- \frac{\partial L}{\partial y}\right)\, dA\qquad\mathrm{(1)}

If it can be shown that

The following is a proof of half of the theorem for the simplified area D, a type I region where C₂ and C₄ are vertical lines (possibly of zero length). A similar proof exists for the other half of the theorem when D is a type II region where C₁ and C₃ are horizontal lines (again, possibly of zero length). Putting these two parts together, the theorem is thus proven for regions of type III (defined as regions which are both type I and type II). The general case can then be deduced from this special case by decomposing D into a set of type III regions.

If D is a simple region with its boundary consisting of the curves C₁, C₂, C₃, C₄, half of Green's theorem can be demonstrated.

Proof when D is a simple region

In physics, Green's theorem is mostly used to solve two-dimensional flow integrals, stating that the sum of fluid outflows at any point inside a volume is equal to the total outflow summed about an enclosing area. In plane geometry, and in particular, area surveying, Green's theorem can be used to determine the area and centroid of plane figures solely by integrating over the perimeter.

where the path of integration along C is counterclockwise.

\oint_{C} (L\, dx + M\, dy) = \iint_{D} \left(\frac{\partial M}{\partial x} - \frac{\partial L}{\partial y}\right)\, dx\, dy

Let C be a positively oriented, piecewise smooth, simple closed curve in a plane, and let D be the region bounded by C. If L and M are functions of (x, y) defined on an open region containing D and have continuous partial derivatives there, then^[2][3]

Theorem

. Kelvin–Stokes theorem and is the two-dimensional special case of the more general [1]