integration of a 2-form over en sphere

**rayman** · May 12th 2011, 10:08 PM

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

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

**FernandoRevilla** · May 13th 2011, 09:15 PM

Originally Posted by rayman

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

That inequality has no sense.

**rayman** · May 13th 2011, 10:16 PM

Originally Posted by FernandoRevilla

That inequality has no sense.

of course it should be $x^2+y^2+z^2\ge R^2$

**FernandoRevilla** · May 13th 2011, 11:22 PM

Originally Posted by rayman

of course it should be $x^2+y^2+z^2\ge R^2$

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

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

So,

$\iint_Sw=-\iint_{S_r}w$

**rayman** · May 13th 2011, 11:29 PM

Thank you for your reply.

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

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