Saturday, 30 April 2016

calculus - Gauss Divergence Theorem in the Plane..

Let F:R2R2 be a C1 vector field, F(x,y)=(P(x,y),Q(x,y)). Consider the smooth curve γ:[0,2π]R2 given by γ(t)=(sin(t),cos(t)) and let B={(x,y)R2:x2+y21}. How can I show that γFν ds=Bdiv(F) dA,

where ν is the unit normal vector to γ(t).



My Atempt: Firstly notice ν(γ(t))=(sin(t),cos(t)) and γ(t)=1. Hence,
γFν ds=2π0F(γ(t)),ν(γ(t))γ(t) dt=2π0P(γ(t))sin(t)+Q(γ(t))cos(t) dt=γP dx+Q dy=B(QxPy) dA=B(Py+Qx) dA,


but I don't know how to finish the proof.. Can anyone help me?

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