Stokes formula

Submitted by Isoscel on Mon, 01/14/2008 - 17:19.

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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Stokes formula