Show that $$\int\limits_{0}^{2\pi} \int\limits_{0}^{\pi} \frac{u_{\phi \phi}}{\sin\theta} d\theta\,d\phi = 0 $$ Where $u$ is a function of $\theta$ and $\phi$.

I am unable to show that this integral is in fact $0$. This is because the of the $\frac{1}{\sin\phi}$, which ultimately yields undefined values over that interval for $\phi$.

Is it possible to show that this integral is undefined or is it really $0$?

  • $\begingroup$ Hint: replace the inner integral with two integrals, one from $t$ to whatever and the other from whatever to $\r$ and then take limits of your variables $\endgroup$ – Brevan Ellefsen Feb 28 '16 at 4:43
  • $\begingroup$ Is it really $u_{\phi\phi}$ in the numerator ? $\endgroup$ – Jean Marie Feb 28 '16 at 5:38
  • 2
    $\begingroup$ $$\int_0^{2\pi }u_{\phi \phi}\,d\phi =u_{\phi}(\theta,2\pi)-u_{\phi}(\theta,0)=0$$assuming that $u_{\phi}$ is single-valued. $\endgroup$ – Mark Viola Feb 28 '16 at 5:55

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