Green's Theorem

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Green's Theorem, or "Green's Theorem in a plane," has two formulations: one formulation to find the circulation of a two-dimensional function around a closed contour (a loop), and another formulation to find the flux of a two-dimensional function around a closed contour. Applications of Green's Theorem include finding the area enclosed by a two-dimensional curve, as well as many engineering applications. Green's Theorem originated in 1825.

The "circulation" formulation of Green's Theorem expresses the line integral of a vector function Pi + Qj over a closed curve in terms of the double integral of the partial derivatives of those same functions:

\oint_{C} (P\, \mathrm{d}x + Q\, \mathrm{d}y) = \iint_{R} \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right)\, \mathrm{d}x{d}y.

Simply stated, Green's Theorem converts a line integral over a closed curve (a loop) into a double integral that is often easier to solve. This theorem is an extension of calculus to the context of integrals in planar regions.

Another way of stating Green's Theorem is this: the macroscopic circulation of a vector function around a closed contour C is equal to the sum of all the microscopic circulations of the same function within the region enclosed by C.

The "flux" formulation of Green's Theorem equates the outward flux of a vector function over a closed contour to the double integral of its divergence over the enclosed region:

\oint_C \vec F \cdot \vec n{\mathrm{d}s}\ = \iint_R (\nabla \cdot \vec F) \vec{\mathrm{d}A},

Green's Theorem is a popular topic in advanced calculus courses. But in physics and engineering its three-dimensional counterpart, the Divergence Theorem, is more useful.

Stokes' Theorem is the generalization of Green's Theorem to non-planar surfaces. Stated another way, Green's Theorem is a special case of Stokes' Theorem where the capping surface is limited to the xy plane, and the normal vector for the surface integral points solely in the z direction.

Example

Problem: Calculate the following along the contour from the origin to (2,0) to (2,6):

\oint_{C} (x^2y\, \mathrm{d}x + x^2y^4\, \mathrm{d}y)

Solution: Using Green's Theorem, P=x2y and Q=x2y4, and the above line integral equals:

=\iint_{R} \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right)\, \mathrm{d}x{d}y.
=\iint_{R} (2xy^4 - x^2)\, \mathrm{d}y{d}x.

Next we have to determine the limits in terms of x and y for the surface R. The surface is a triangle for which y=3x, so the inner integral should be from y=0 to y=3x. The outer integral can then be from x=0 to x=2. This yields:

=\int_{(x=0)}^{(x=2)} \int_{(y=0)}^{(y=3x)}(2xy^4 - x^2)\, \mathrm{d}y{d}x.

Solving the integrals equals:

= \ \left [\frac{2}{5}*\frac{3^5}{7}x^7 - \frac{3}{4}*x^4 \right]^2_0
= \frac{2}{5}*\frac{3^5}{7}*2^7 - \frac{3}{4}*2^4
= \frac{2^8*3^5}{35} - 12 = 1765 + \frac{13}{35}

Another Example

Problem: Calculate the following along the perimeter of the ellipse \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\,

\oint_{C} (y\, \mathrm{d}x + x\, \mathrm{d}y)

Solution: Using Green's Theorem, P=y and Q=x, and the above line integral equals:

=\iint_{R} \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right)\, \mathrm{d}x{d}y = \iint_{R} 2\, \mathrm{d}y{d}x.

This is just twice the area of the ellipse, or 2\pi{}ab\,.

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