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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.
 Spring '13

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