ps6sol-2010 - 18.02A IAP 2010: Solutions Problem Set 6...

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18.02A IAP 2010: Solutions Problem Set 6 version January 15, 2010 Part II (35 points) Problem 1. F =3 xy 2 i y 3 j . We assume C 1 and C 2 do not cross. Let C x and C y be the horizontal and vertical curves joining the starting point and endpoint of C 1 and C 2 . Orient the closed simple curve C = C x + C 1 + C y C 2 so that it is positively oriented. Now we can apply the normal form of Green’s theorem on C : ° C F · n ds = ±± R ² ∂M ∂x + ∂N ∂y ³ dA = ±± R 3 y 2 3 y 2 dA =0 . But ´ C F · n ds = ´ C x F · n ds + ´ C 1 F · n ds + ´ C y F · n ds ´ C 2 F · n ds , and ° C x F · n ds = ° C x Mdy = ° C x 0 dy =0 , ° C y F · n ds = ° C y Ndx = ° C y 0 dx =0 . Thus 0 = ´ C F · n ds = ´ C 1 F · n ds ´ C 2 F · n ds as desired. Problem 2 F =( y + e x ) i +( x e y ) j . We do the conservative Feld test: curl F = N x M y = 1 1 = 0, thus F is conservative. We Fnd the scalar potential Feld f by the algebraic method: F = µ y + e x dx + C ( y )= yx + e x + C ( y ), and so F y = x + dC ( y ) dy should x e y ,so C ( y )= e y . Thus up to a constant
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ps6sol-2010 - 18.02A IAP 2010: Solutions Problem Set 6...

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