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Let $\omega$ be a $2$-form on $\mathbb{R}^3-\{(1,0,0),(-1,0,0)\}$, $\omega=((x-1)^2+y^2+z^2)^{-3/2}((x-1)dy\wedge dz+ydz\wedge dx+zdx \wedge dy)+ ((x+1)^2+y^2+z^2)^{-3/2}((x+1)dy\wedge dz+ydz\wedge dx+zdx \wedge dy)$ and $S=\{(x,y,z)\in \mathbb{R^3}: x^2+y^2+z^2=5 \}$.

In this condition, we calculate $\int_{S}\omega$, where the orientation of $S$ is the natural orientation induced by $D=\{(x,y,z)\in \mathbb{R^3}: x^2+y^2+z^2 \leq 5 \}$.

I can't calculate this, so if you solve this, please teach me the answer for this.

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    Hint: Use polar coordinates.2012-08-09

2 Answers 2

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Let $\omega_\pm = ((x\mp 1)^2 + y^2 + z^2)^{-3/2} ( (x \mp 1)dy \wedge dz + ydz\wedge dx + z dx \wedge dy)$.

Now $\omega_\pm$ is defined on $\mathbb R^3 - \{ (\pm 1,0,0)\}$ and $\omega = \omega_+ + \omega_-$. Using polar coordinates centered at $(\pm 1,0,0)$: $x = r \sin \theta \cos \phi \pm 1 \\ y = r \sin \theta \sin \phi \\ z = r\cos \theta$ we can now calculate $\omega_\pm = \sin\theta \;d\theta\wedge d\phi.$ Since $d\omega_\pm = 0$ we have by Stokes theorem, that $\int_S \omega + \int_{-S_+} \omega_+ + \int_{-S_-} \omega_- = 0$ where $S_\pm = \{(x,y,z) \in \mathbb R^3 \;|\; (x \mp 1)^2 + y^2 + z^2 = \epsilon\}$, thus

$\int_s \omega = 2\int_0^{2\pi}\int_o^\pi \sin\theta \;d\theta \wedge d\phi = 8\pi.$

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This is a closed form, i.e. $d \omega = 0$. Since the orientation of $S = \partial D$ is induced by the orientation of $D$, by Stoke's theorem $ \int_S \omega = \int_D d \omega = 0. $ I leave checking that $d \omega = 0$ up to you.

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    @MattE: Yes, you are absolutely right. I am probably so used to thinking compact manifold, that I have missed clarifying. So the correct statement would be $\omega$ does not have compact support on $D - \{ (\pm 1,0,0)\}$. Thank you for pointing out my mistake. Regards2012-08-09