Green’s
Theorem
Here P(x,y) = y2
, Q(x,y) = xy and the closed
curve, C, is the Now ∂Q/∂x =
y and ∂P/∂y =
2y so ∂Q/∂x -
∂P/∂y = - y y = 2 x =
3√[1-(y/2)2] Thus the area integral
becomes ∫ ∫ -y dA = -2 ∫
∫ y dx dy
R y = -2 x
= 0 y = 2 x = 3√[1-(y/2)2] y = 2 I
= -2 ∫
y x| dy = - 6 ∫ y √[1-(y/2)2] dy y = -2 x = 0
y = -2 let
u = 1- (y/2)2 , du =
- (1/2) y dy,
y dy
= - 2 du 2 I
= 12 ∫ √u du = 8 [1-(y/2)2 ]3/2 | = 0 (result)
-2 Note this result makes
sense since ∫ ∫ -y dA should equal
zero when encompassing R the region, R, by going
completing around the ellipse (the areas on both sides of the y-axis cancel out). |
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Example: Evaluate the line
integral: ∫ (7y – esinx)dx + (15x –
sin(y3 + 8y) dy C where C is
a circle of radius 3 centered at (5, -7).
Note that this is a very difficult integral to evaluate. So try simplifying the calculation using the
RHS of Green’s Theorem. ∂Q/∂x = 15,
∂P/∂y = 7 ; ∫
∫[∂Q/∂x - ∂P/∂y] dA =
∫ ∫ 8 dA = 8
(π 32 ) =
72π (result) R R Note the simplification that results in using Green’s Theorem. Click here for more examples
using Green’s Theorem. |
Copyright © 2017 Richard C. Coddington
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