Let F:R2⟶R2 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+y2≤1}. How can I show that −∫γF⋅ν 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,
−∫γF⋅ν ds=−∫2π0⟨F(γ(t)),ν(γ(t))⟩‖γ′(t)‖ dt=−∫2π0P(γ(t))sin(t)+Q(γ(t))cos(t) dt=−∫γP dx+Q dy=−∫∫B(∂Q∂x−∂P∂y) dA=∫∫B(∂P∂y+∂Q∂x) dA,
but I don't know how to finish the proof.. Can anyone help me?
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