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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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The $\omega$ you give is a 1-form. Then the integral makes no sense. Probably a typo? Also, what did you try already? The formulation of the exercise is screaming for Stoke's theorem. –  ZulfiqarIII Aug 8 '12 at 19:33
    
Sorry,I correct it then. –  Takahiro Oba Aug 8 '12 at 23:54
    
Hint: Use polar coordinates. –  Alexander Thumm Aug 9 '12 at 9:38

2 Answers 2

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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"Let $\omega$ be a $2$-form on $\mathbb R^3 - \{ (1,0,0), (-1,0,0)\}$" –  Alexander Thumm Aug 9 '12 at 9:28
    
Yes, so what's the problem? –  ZulfiqarIII Aug 9 '12 at 9:37
    
Stokes is not applicable. –  Alexander Thumm Aug 9 '12 at 9:39
    
Can you elaborate? By definition $\int_S \omega = \int \iota^* \omega$, where $\iota : S \to \mathbf{R}^3 - \{(1,0,0),(-1,0,0)\}$ is the inclusion map. We can think of $\iota^* \omega$ as first restricting $\omega$ to $D$, and then restricting further to $S$. Where is the flaw in this? –  ZulfiqarIII Aug 9 '12 at 10:12
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@AlexanderThumm: Dear Alexander, I realize that it's not relevant to your point, with which I agree, but I just wanted to clarify that $D - \{(1,0,0),(-1,0,0)\}$ is a manifold (with boundary). Regards, –  Matt E Aug 9 '12 at 12:37

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