Tutorial10-sol(1)

# Instead we just calculate as usual c can be

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Unformatted text preview: point (the origin) in the interior domain of C . Instead we just calculate as usual. C can be prametrised as = ε cos = ε sin ∈ [0 2π ] We get = −ε sin = ε cos and − 2+ C 2π 2 = 0 ε cos ε sin − ε sin (−ε sin ) ε2 2π = 1 = 2π 0 (c) Let Cε be a circle with radius ε oriented anticlockwise centred at the origin, where the radius ε is small enough for Cε to be contained in the interior region of C . Let D be the region inside C but outside Cε . The given vector ﬁeld satisﬁes the conditions of Green’s theorem on the region D, where we also have Q − P = 0, so by Green’s theorem we have C j− i − +2 2 Cε j− i = +2 2 (Q − P ) D = 0 =0 D But in part (b) we saw that Cε hence C j− i = 2π...
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## This document was uploaded on 02/10/2014.

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