20e-lecture25 - Lecture 25 Monday June 1st [email protected]

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Lecture 25 - Monday June 1st [email protected] Key words: Simple closed curve, Green’s theorem, Divergence, Divergence theorem, Incompressible fluid, Outward normal 25.1 Green’s Theorem. Green’s Theorem. Let γ be a simple closed curve in the plane and let D be the region it encloses. Suppose γ is oriented counterclockwise, and let g, h : D R are C 1 ( D ) functions. Then Z γ gdx + hdy = ZZ D ( h x - g y ) dA Green’s Theorem holds for more general regions than those bounded by closed simple curves. As long as we can cut up the given region into regions bounded by closed simple curves in such a way that all the integrals of pieces of curves inside the region cancel out, Green’s Theorem applies. For a general region D whose boundary consists of disjoint oriented closed simple curves γ 1 , γ 2 , . . . , γ n , Green’s Theorem applies provided that for each curve γ i , the outward normal to γ i is outside of the region D – in other words, walking around γ i in the given orientation, the region D must be on the left as we walk around γ . This is why we can apply Green’s Theorem with the inner curve in the above picture is oriented clockwise. If say γ 1 , γ 2 , . . . , γ m are oriented so that D is to the left of them, and γ m +1 , γ m +2 , . . . , γ n are oriented so that D is to the right of them as we walk around the curves, then by Green’s Theorem the general formula is Z γ f · dr = m X i =1 ZZ D i ( h x - g y ) dA - n X j = m +1 ZZ D j ( h x - g y ) dA where D i is the region enclosed by γ i . So we “subtract” the integrals over regions where the boundary curve has the “wrong” orientation. Example 1. Find R γ e y 2 dx +sin( x 2 ) dy where γ is the boundary of D = [0 , 3] × [0 , 3] \ [1 , 2] × [1 , 2] and the inner part of γ is oriented clockwise and the outer part of γ is oriented counterclockwise. 1
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Region D and boundary curve γ Solution. Clearly γ is not a simple curve: it consists of the boundaries of two squares, one inside the other. Nevertheless Green’s Theorem holds because D is always to the left in the given orientation of γ so we have a single formula: Z γ gdx + hdy = ZZ D ( g x - f y ) dA.
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