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 ‖. Hence,
\begin{align*} \displaystyle -\int_{\gamma}F\cdot \nu\ ds&=-\int_{0}^{2\pi} \langle F(\gamma(t)), \nu(\gamma(t))\rangle\|\gamma^{'}(t)\|\ dt\\ &=-\int_{0}^{2\pi} P(\gamma(t))\sin(t)+Q(\gamma(t))\cos(t)\ dt\\ &=-\int_{\gamma} P\ dx+Q\ dy\\ &=-\int\int_{B} \left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)\ dA\\ &=\int\int_B\left(\frac{\partial P}{\partial y}+\frac{\partial Q}{\partial x}\right)\ dA, \end{align*}
but I don't know how to finish the proof.. Can anyone help me?
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