# proof of Green’s theorem

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

 $\int\!\!\!\int_{R}\frac{\partial g}{\partial x}\;dA=\oint_{P}\vec{F_{1}}\cdot d% \vec{t}$
 $\int\!\!\!\int_{R}\frac{\partial f}{\partial y}\;dA=-\oint_{P}\vec{F_{2}}\cdot d% \vec{t}$

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.

Title proof of Green’s theorem ProofOfGreensTheorem 2013-03-22 12:28:47 2013-03-22 12:28:47 mathcam (2727) mathcam (2727) 11 mathcam (2727) Proof msc 26B12