Consider the region $R$ bounded by the closed curve $P$ in a simply connected space. $P$ can be given by a vector valued function $\vec{F}(x,y)=( f(x,y), g(x,y))$ . The region $R$ can then be described by $$\int\!\!\!\int_R \left(\frac{\partial g}{\partial x} - \frac{\partial f}{\partial y}\right)\;dA = \int\!\!\!\int_R \frac{\partial g}{\partial x}\;dA - \int\!\!\!\int_R \frac{\partial f}{\partial y}\;dA$$ The double integrals above can be evaluated separately. Let's look at $$\int\!\!\!\int_R \frac{\partial g}{\partial x}\;dA = \int_a^b\int_{A(y)}^{B(y)}\frac{\partial g}{\partial x}\;dxdy$$ Evaluating the above double integral, we get $$\int_a^b (g(A(y),y) - g(B(y),y))\;dy = \int_a^b g(A(y),y)\;dy - \int_a^b g(B(y),y)\;dy$$ According to the fundamental theorem of line integrals, the above equation is actually equivalent to the evaluation of the line integral of the function $\vec{F}_1(x,y)=( 0, g(x,y))$ over a path $P=P_1 + P_2$ , where $P_1=(A(y), y)$ and $P_2=(B(y), y)$ . $$\int_a^b g(A(y), y)\;dy - \int_a^b g(B(y), y)\;dy = \int_{P_1} \vec{F_1}\cdot d\vec{t} + \int_{P_2}\vec{F_1}\cdot d\vec{t} = \oint_P \vec{F_1}\cdot d\vec{t}$$ Thus we have $$\int\!\!\!\int_R \frac{\partial g}{\partial x}\;dA = \oint_P \vec{F_1}\cdot d\vec{t}$$ By a similar argument, we can show that $$\int\!\!\!\int_R \frac{\partial f}{\partial y}\;dA = -\oint_P \vec{F_2}\cdot d\vec{t}$$ where $\vec{F}_2=( f(x,y), 0)$ . Putting all of the above together, we can see that $$\int\!\!\!\int_R \left(\frac{\partial g}{\partial x} - \frac{\partial f}{\partial y}\right)\;dA = \oint_P \vec{F_1}\cdot d\vec{t} + \oint_P \vec{F_2}\cdot d\vec{t} = \oint_P (\vec{F}_1 + \vec{F}_2)\cdot d\vec{t}=\oint_P (f(x,y), g(x,y))\cdot d\vec{t}$$ which is Green's theorem.