Stokes formula

Let $ \omega $ be a p-differential form and $ \lambda $ a p+1-chain, $ d\omega $ the exterior derivative of $ \omega $ and $ \partial\lambda $ the border of the chain $ \lambda $. Then

$$\int_{\partial\lambda}\omega=\int_{\lambda}d\omega$$

For a 1-differential form in $ R^3 $ we can write

$$\int_{\partial\lambda}A(x,y,z)dx+B(x,y,z)dy+C(x,y,z)dz=$$
$$=\int_{\lambda }(\frac{\partial C}{\partial y}-\frac{\partial B}{\partial z})dy\wedge dz+(\frac{\partial A}{\partial z}-\frac{\partial C}{\partial x})dz\wedge dx+(\frac{\partial B}{\partial x}-\frac{\partial A}{\partial y})dx\wedge d y<br />
$$

We need for $ \omega=A(x,y,z)dx+B(x,y,z)dy+C(x,y,z)dz $ the expression of $ d\omega $
We know $ d d\omega=0 $ so $ ddx=ddy=ddz=0 $ and
$ dz\wedge dy=-dy\wedge dz $ ,$ dx\wedge dz=-dz\wedge dx $,$ dy\wedge dx=-dx\wedge dy $
We can write

$$d\omega=dA(x,y,z)\wedge dx+dB(x,y,z)\wedge dy+dC(x,y,z)\wedge dz=$$
$$(\frac{\partial A}{\partial x}dx+\frac{\partial A}{\partial y}dy+\frac{\partial A}{\partial z}dz)\wedge dx+<br />
(\frac{\partial B}{\partial x}dx+\frac{\partial B}{\partial y}dy+\frac{\partial B}{\partial z}dz)\wedge dy+<br />
(\frac{\partial C}{\partial x}dx+\frac{\partial C}{\partial y}dy+\frac{\partial C}{\partial z}dz)\wedge dz=$$
$$(\frac{\partial C}{\partial y}-\frac{\partial B}{\partial z})dy\wedge dz+(\frac{\partial A}{\partial z}-\frac{\partial C}{\partial x})dz\wedge dx+(\frac{\partial B}{\partial x}-\frac{\partial A}{\partial y})dx\wedge d y<br />
$$

For a 2-differential form and a 3-chain in $ R^3 $ we have

$$\int_{\partial\lambda}A(x,y,z)dy\wedge dz+B(x,y,z)dz\wedge dx+C(x,y,z)dx\wedge dy=$$
$$=\int_{\lambda }(\frac{\partial A}{\partial x}+\frac{\partial B}{\partial y}+\frac{\partial C}{\partial z})dx\wedge dy\wedge dz$$
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