15_4_20 - 2 , x = f ( y ) , where y goes from d to c . Thus...

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20. Conjecture: If D is a domain in 2 whose boundary is a closed, non-self-intersecting curve , oriented counterclockwise, then I x dy = area of D , I y dx = - area of D . Proof for a domain D that is x -simple and y -simple: Since D is x -simple, it can be specified by the inequalities c y d , f ( y ) x g ( y ). Let consist of the four parts shown in the figure. On 1 and 3 , dy = 0. On 2 , x = g ( y ) , where y goes from c to d . On
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Unformatted text preview: 2 , x = f ( y ) , where y goes from d to c . Thus I x dy = Z 1 + Z 2 + Z 3 + Z 4 = + Z d c g ( y ) dy + + Z c d f ( y ) dy = ( g ( y )-f ( y ) ) dy = area of D . The proof that I y dx = -(area of D ) is similar, and uses the fact that D is y-simple. y x 2 3 4 1 x = g ( y ) D x = f ( y ) c d Fig. 4-20...
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