Let $M = \{(x,y,z) \in \mathbb{R}^3 | x^2+y^2+z^2=1,z>0\}$, and let $\omega = xdy$. I would like to verify that $\int_M d\omega = \int_C \omega$, where $C = \partial M$ given by $C = \{(x,y,z) \in \mathbb{R}^3 | x^2+y^2 = 1, z=0\}$.

I first parametrize $M$ as the image of the set $A = \{(\theta, \phi) \in \mathbb{R}^2 | -\frac{\pi}{2} < \theta < \frac{\pi}{2}, 0 < \phi < 2\pi\}$ under the spherical transformation $\alpha(\theta, \phi) = (\sin\theta\cos\phi, \sin\theta\sin\phi, \cos\theta)$.

So we have that, $$\int_M d\omega = \int_M dx \wedge dy = \int_A \alpha^* (dx \wedge dy) = \int_A d\alpha_1 \wedge d\alpha_2.$$ Since $$d\alpha_1 = \cos\theta\cos\phi \ d\theta - \sin\theta\sin\phi \ d\phi,$$ and $$d\alpha_2 = \cos\theta\sin\phi \ d\theta + \sin\theta\cos\phi \ d\phi,$$ $d\alpha_1 \wedge d\alpha_2 = \sin\theta\cos^2\phi \ d\theta \wedge d\phi - \sin\theta\cos\theta\sin^2 \phi \ d\phi \wedge d\theta = \sin\theta\cos\theta \ d\theta \wedge d\phi$, so the integral becomes $$\int_A \sin\theta \cos\theta \ d\theta \wedge d\phi = \int_0^{2\pi} \int_{-\pi / 2}^{\pi / 2} \sin\theta \cos \theta \ d\theta d\phi = 0.$$

On the other hand, we can parametrize $C$ as the image of $U = \{\theta \in \mathbb{R} | 0 < \theta < 2\pi \}$ under $\gamma(\theta) = (\cos\theta, \sin\theta)$. Then $$\int_C \omega = \int_C xdy = \int_U \gamma^*(xdy) = \int_U \cos\theta \ d\gamma_2 = \int_U \cos^2\theta \ d\theta = \int_0^{2\pi} \frac{1+\cos2\theta}{2} \ d\theta = \pi.$$

Why do my answers not agree?

  • $\begingroup$ Did you mean to define $S$ and $C$ as the same thing? $\endgroup$ – DanielV Mar 23 '16 at 1:31
  • $\begingroup$ @DanielV My bad, I meant $M$ not $S$. $\endgroup$ – Simeon Mar 23 '16 at 1:33
  • $\begingroup$ @DanielRobert-Nicoud Another typo, sorry! You're right, that's a $>$ and $M$ is $2$-dimensional. $\endgroup$ – Simeon Mar 23 '16 at 1:48
  • $\begingroup$ Stokes', not Stoke's. $\endgroup$ – Akiva Weinberger Mar 23 '16 at 2:13

Your mistake is in the parametrization $\alpha$ you are using, or better in the definition of $A$. Your $\theta$ should only run from $0$ to $\frac{\pi}{2}$.

  • $\begingroup$ Why? That would give only a quarter of the sphere, won't it? $\endgroup$ – Simeon Mar 23 '16 at 2:32
  • $\begingroup$ @Simeon No, running $\theta$ from $0$ to $\frac{\pi}{2}$ gives you a quarter of circle, which spans your half sphere when you make it do a full turn running $\phi$ from $0$ to $2\pi$. You were actually covering your $M$ twice, and once with a negative sign picked up from the wrong parametrization, and that's why you got $0$. $\endgroup$ – Daniel Robert-Nicoud Mar 23 '16 at 2:38
  • $\begingroup$ Might be more accurate to limit $\phi < \pi/2$ (if using the coordinate system where $\phi$ measures angle from the north pole). (Anyone who doesn't specify which of a seemingly endless supply of different spherical coordinate systems one is using, should probably be forced to use one of the more awful ones. For instance bipolar cylindrical.) Why? $M$ is the upper hemisphere and $A$, as currently defined, is the eastern hemisphere. $\endgroup$ – Eric Towers Mar 23 '16 at 2:38
  • $\begingroup$ @DanielRobert-Nicoud You're right $\alpha$ is not injective on that set. Funny how I made the exact same mistake some time ago and did not at all learn from it. $\endgroup$ – Simeon Mar 23 '16 at 2:40
  • 1
    $\begingroup$ @Simeon Well, three times is a charm, so you'll only have to do the same mistake one more time in order to learn! ;) $\endgroup$ – Daniel Robert-Nicoud Mar 23 '16 at 2:44

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