Ngô Quốc Anh

April 27, 2010

Surface integral: The symmetric property, 2

Filed under: Các Bài Tập Nhỏ, Giải Tích 5 — Tags: — Ngô Quốc Anh @ 16:07

Today, we try to evaluate the following surface integral in \mathbb R^3. I found this result in a paper published in Math. Z. 198, 277-289 (1988). This topic can be considered as a continued part to the following topic.

Proposition. Let f(x) be a continuous function. Then we have

\displaystyle\frac{1}{t}\iint\limits_{|{\mathbf{y}} -  {\mathbf{x}}| = t} {f(|{\mathbf{y}}|)d{\sigma _{\mathbf{y}}}} =  \frac{{2\pi }}{r}\int\limits_{|r - t|}^{r + t} {sf(s)ds}

here we denote r=|\mathbf{x}|.

Proof. Remark that the integral only depends on |\mathbf{x}|, so that we may assume \mathbf{x} = (0, 0, r) and introduce the following spherical coordinate



\displaystyle \omega = (\sin \theta \cos \varphi ,\sin \theta \sin \varphi ,\cos \theta ).

This system of coordinates can be seen via the picture below


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