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These are a few steps I don't understand from my partial differential equations course. Let us be in $\mathbb{R}^3$. The average of a function $u(x,t)$ on the sphere $||x||=r$ of center $0$ and radius $r$ is denoted $\overline{u}(r,t) = \frac{1}{4\pi r^2} \int\int_{||x||=r} u(x,t) dS$.

My first question is, why is $\overline{u}(r,t) = \frac{1}{4\pi} \int_{0}^{2\pi}\int_{0}^{\pi}u(x,t) \sin (\theta) d \theta d \phi$? I know that we are changing to spherical coordinates, but I'm not sure about the intermediate details.

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In the integral $$ \overline{u}(r,t) = \frac{1}{4\pi r^2} \int\int_{||x||=r} u(x,t) dS $$ the area in the spharical polar coordinates is element $dS = r^2\sin(\theta)d\theta d\phi$, where $0\le \theta \le \pi$ and $0 \le \phi \le 2\pi$.

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  • $\begingroup$ Thanks for your answer. But how would your second comment imply $\Delta \overline{u} = \overline{\Delta u}$? $\endgroup$
    – Albert
    Commented Oct 7, 2020 at 0:07
  • $\begingroup$ I should have written "it might help to use it", but I don't know yet! I think you removed that part from the question $\endgroup$
    – Physor
    Commented Oct 7, 2020 at 9:07
  • $\begingroup$ I moved it to a different question : ) $\endgroup$
    – Albert
    Commented Oct 8, 2020 at 2:30

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