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Key concepts

  • Green’s theorem relates the integral over a connected region to an integral over the boundary of the region. Green’s theorem is a version of the Fundamental Theorem of Calculus in one higher dimension.
  • Green’s Theorem comes in two forms: a circulation form and a flux form. In the circulation form, the integrand is F · T . In the flux form, the integrand is F · N .
  • Green’s theorem can be used to transform a difficult line integral into an easier double integral, or to transform a difficult double integral into an easier line integral.
  • A vector field is source free if it has a stream function. The flux of a source-free vector field across a closed curve is zero, just as the circulation of a conservative vector field across a closed curve is zero.

Key equations

  • Green’s theorem, circulation form
    C P d x + Q d y = D Q x P y d A , where C is the boundary of D
  • Green’s theorem, flux form
    C F · d r = D Q x P y d A , where C is the boundary of D
  • Green’s theorem, extended version
    D F · d r = D Q x P y d A

For the following exercises, evaluate the line integrals by applying Green’s theorem.

C 2 x y d x + ( x + y ) d y , where C is the path from (0, 0) to (1, 1) along the graph of y = x 3 and from (1, 1) to (0, 0) along the graph of y = x oriented in the counterclockwise direction

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C 2 x y d x + ( x + y ) d y , where C is the boundary of the region lying between the graphs of y = 0 and y = 4 x 2 oriented in the counterclockwise direction

C 2 x y d x + ( x + y ) d y = 32 3

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C 2 arctan ( y x ) d x + ln ( x 2 + y 2 ) d y , where C is defined by x = 4 + 2 cos θ , y = 4 sin θ oriented in the counterclockwise direction

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C sin x cos y d x + ( x y + cos x sin y ) d y , where C is the boundary of the region lying between the graphs of y = x and y = x oriented in the counterclockwise direction

C sin x cos y d x + ( x y + cos x sin y ) d y = 1 12

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C x y d x + ( x + y ) d y , where C is the boundary of the region lying between the graphs of x 2 + y 2 = 1 and x 2 + y 2 = 9 oriented in the counterclockwise direction

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C ( y d x + x d y ) , where C consists of line segment C 1 from ( −1 , 0 ) to (1, 0), followed by the semicircular arc C 2 from (1, 0) back to (1, 0)

C ( y d x + x d y ) = π

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For the following exercises, use Green’s theorem.

Let C be the curve consisting of line segments from (0, 0) to (1, 1) to (0, 1) and back to (0, 0). Find the value of C x y d x + y 2 + 1 d y .

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Evaluate line integral C x e −2 x d x + ( x 4 + 2 x 2 y 2 ) d y , where C is the boundary of the region between circles x 2 + y 2 = 1 and x 2 + y 2 = 4 , and is a positively oriented curve.

C x e −2 x d x + ( x 4 + 2 x 2 y 2 ) d y = 0

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Find the counterclockwise circulation of field F ( x , y ) = x y i + y 2 j around and over the boundary of the region enclosed by curves y = x 2 and y = x in the first quadrant and oriented in the counterclockwise direction.

A vector field with focus on quadrant 1. A line is drawn from (0,0) to (1,1) according to function y=x, and a curve is also drawn according to the function y=x^2. The region enclosed between the two functions is shaded. Te=he arrows closer to the origin are much smaller than those further away, particularly vertically. The arrows point up and away from the origin to the right in the part of quadrant 1 shown.
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Evaluate C y 3 d x x 3 y 2 d y , where C is the positively oriented circle of radius 2 centered at the origin.

C y 3 d x x 3 d y = −24 π

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Evaluate C y 3 d x x 3 d y , where C includes the two circles of radius 2 and radius 1 centered at the origin, both with positive orientation.

A vector field in two dimensions. The arrows surround the origin in a clockwise circular motion. Those close to the origin are much smaller than those further away. A circle of radius 2 and a circle of radius with centers at the origin is drawn in, and the region between the two is shaded.
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Calculate C x 2 y d x + x y 2 d y , where C is a circle of radius 2 centered at the origin and oriented in the counterclockwise direction.

C x 2 y d x + x y 2 d y = 8 π

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Calculate integral C 2 [ y + x sin ( y ) ] d x + [ x 2 cos ( y ) 3 y 2 ] d y along triangle C with vertices (0, 0), (1, 0) and (1, 1), oriented counterclockwise, using Green’s theorem.

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Questions & Answers

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Source:  OpenStax, Calculus volume 3. OpenStax CNX. Feb 05, 2016 Download for free at http://legacy.cnx.org/content/col11966/1.2
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