$\displaystyle f=\frac{E}{4\pi}(\frac{x}{(x^2+y^2+z^2)^{3/2}}dy\wedge dz+\frac{y}{(x^2+y^2+z^2)^{3/2}}dz\wedge dx+\frac{x}{(x^2+y^2+z^2)^{3/2}}dx\wedge dy)$

over a sphere $\displaystyle x^2+y^2+z^2 \ge \mathbb{R}^3$

Printable View

- May 12th 2011, 10:08 PMraymanintegration of a 2-form over en sphere
$\displaystyle f=\frac{E}{4\pi}(\frac{x}{(x^2+y^2+z^2)^{3/2}}dy\wedge dz+\frac{y}{(x^2+y^2+z^2)^{3/2}}dz\wedge dx+\frac{x}{(x^2+y^2+z^2)^{3/2}}dx\wedge dy)$

over a sphere $\displaystyle x^2+y^2+z^2 \ge \mathbb{R}^3$ - May 13th 2011, 09:15 PMFernandoRevilla
- May 13th 2011, 10:16 PMrayman
- May 13th 2011, 11:22 PMFernandoRevilla

I suppose you mean the sphere $\displaystyle S\equiv x^2+y^2+z^2=R^2$. In such case, apply the Divergence Theorem to $\displaystyle V-V_r$ where $\displaystyle 0<r<R$ . That is,

$\displaystyle \iint_Sw+\iint_{S_r}w=\iiint_{V-V_r}dw=0$

So,

$\displaystyle \iint_Sw=-\iint_{S_r}w$ - May 13th 2011, 11:29 PMrayman
Thank you for your reply.

actually it is supposed to be like I wrote earlier $\displaystyle x^2+y^2+z^2\ge R^2$ that is actually what my teacher says. I will apply Gauss-Otrogradski theorem to it and she what happens.