calculus - Gauss Divergence Theorem in the Plane..

Let $F:\mathbb R^2\longrightarrow \mathbb R^2$ be a $C^1$ vector field, $F(x, y)=(P(x, y), Q(x, y))$. Consider the smooth curve $\gamma:[0, 2\pi]\longrightarrow \mathbb R^2$ given by $\gamma(t)=(\sin(t), \cos(t))$ and let $B=\{(x, y)\in\mathbb R^2: x^2+y^2\leq 1\}$. How can I show that

$$-\int_{\gamma} F\cdot \nu \ ds=\int\int_B\textrm{div}(F)\ dA,$$ where $\nu$ is the unit normal vector to $\gamma(t)$.

My Atempt: Firstly notice $\nu(\gamma(t))=(\sin(t), \cos(t))$ and $\|\gamma^{'}(t)\|=1$. 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?