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

$\mathbf{y}=\mathbf{x}+t\omega$

where

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

This system of coordinates can be seen via the picture below