Ngô Quốc Anh

April 29, 2010

Surface integrals over a sphere when its radius varies

Filed under: Các Bài Tập Nhỏ, Giải Tích 5, Giải Tích Cổ Điển — Tags: — Ngô Quốc Anh @ 1:53

Let us consider the following integral in \mathbb R^3

\displaystyle\iint\limits_{|{\mathbf{x}}| = t} {f({\mathbf{x}})d{\sigma _{\mathbf{x}}}}

when t varies.

Obviously, the sphere |{\mathbf{x}}| = t can be parametrized as the following

\displaystyle {\mathbf{x}} = t\left( {\sin \theta \cos \varphi ,\sin \theta \sin \varphi ,\cos \theta } \right)

so

\displaystyle\begin{gathered} \iint\limits_{|{\mathbf{x}}| = t} {f({\mathbf{x}})d{\sigma _{\mathbf{x}}}} = \iint\limits_{\begin{subarray}{c} 0 \leqslant \theta \leqslant \pi \\ 0 \leqslant \varphi \leqslant 2\pi \end{subarray}} {f({\mathbf{x}}(\theta ,\varphi )){t^2}\sin \theta d\theta d\varphi } \hfill \\ \qquad= {t^2}\iint\limits_{\begin{subarray}{c} 0 \leqslant \theta \leqslant \pi \\ 0 \leqslant \varphi \leqslant 2\pi \end{subarray}} {f({\mathbf{x}}(\theta ,\varphi ))\sin \theta d\theta d\varphi } \hfill \\ \qquad= {t^2}\iint\limits_{|{\mathbf{x}}| = 1} {f \left(t\mathbf{x} \right)d{\sigma _{\mathbf{x}}}}. \hfill \\ \end{gathered}

If we wish to work on the average, the formula is much simpler than that, precisely

\displaystyle\frac{1}{{4\pi {t^2}}}\iint\limits_{|{\mathbf{x}}| = t} {f({\mathbf{x}})d{\sigma _{\mathbf{x}}}} = \frac{1}{{4\pi }}\iint\limits_{|{\mathbf{x}}| = 1} {f(t{\mathbf{x}})d{\sigma _{\mathbf{x}}}}

that means

\displaystyle \overline {\iint\limits_{|{\mathbf{x}}| = t} } f({\mathbf{x}})d{\sigma  _{\mathbf{x}}}=\overline {\iint\limits_{|{\mathbf{x}}| = 1} } f({t\mathbf{x}})d{\sigma  _{\mathbf{x}}}

where the bar means the average.

More general, we get

\displaystyle\iint\limits_{|{\mathbf{x}}| = {t_1}} {f({\mathbf{x}})d{\sigma _{\mathbf{x}}}} = {\left( {\frac{{{t_1}}}{{{t_2}}}} \right)^2}\iint\limits_{|{\mathbf{x}}| = {t_2}} {f\left( {\frac{{{t_1}}}{{{t_2}}}{\mathbf{x}}} \right)d{\sigma _{\mathbf{x}}}}.

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