# 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

Thus

$\displaystyle |\mathbf{y}| = \sqrt {|\mathbf{y}|^2} = \sqrt {|\mathbf{x} + t\omega {|^2}} = \sqrt {{\mathbf{x}^2} + {t^2}{\omega ^2} + 2t\mathbf{x} \cdot \omega } = \sqrt {{r^2} + {t^2} + 2tr\cos \theta }$.

Let us remember once a surface $S$ can be parametrized by two variables $s,t$, we then have

$\displaystyle\iint\limits_{S} f(\mathbf{y})d\sigma_\mathbf{y} = \iint\limits_{T} f(\mathbf{y}(s, t)) \left|{\partial \mathbf{y} \over \partial s}\times {\partial \mathbf{y} \over \partial t}\right| dsdt$

Thus by the above coordinate

$\displaystyle\left|\frac{{\partial {\mathbf{y}}}}{{\partial \theta }} \times \frac{{\partial {\mathbf{y}}}}{{\partial \varphi }}\right| = \left| {\begin{array}{*{20}{c}}\vec i & \vec j & \vec k \\ {t\cos \theta \cos \varphi } & {t\cos \theta \sin \varphi } & { - t\sin \theta } \\ { - t\sin \theta \sin \varphi } & {t\sin \theta \cos \varphi } & 0 \\ \end{array} } \right| = t^2\sin\theta$

which helps us to write down

$\displaystyle\iint\limits_{|{\mathbf{y}} - {\mathbf{x}}| = t} {f(|{\mathbf{y}}|)d{\sigma _{\mathbf{y}}}} = \int_0^{2\pi } {\int_0^\pi {f(\sqrt {{r^2} + {t^2} + 2tr\cos \theta } ){t^2}\sin \theta d\theta d\varphi } }$.

The last integral can be estimated further as follows

$\displaystyle\begin{gathered} \int_0^{2\pi } {\int_0^\pi {f(\sqrt {{r^2} + {t^2} + 2tr\cos \theta } ){t^2}\sin \theta d\theta d\varphi } } \hfill \\ \qquad= 2\pi {t^2}\int_0^\pi {f(\sqrt {{r^2} + {t^2} + 2tr\cos \theta } )\sin \theta d\theta } \hfill \\ \qquad= 2\pi {t^2}\int_{ - 1}^1 {f(\sqrt {{r^2} + {t^2} + 2trs} )ds} \hfill \\ \qquad= \frac{{2\pi }}{r}\int\limits_{|r - t|}^{|r + t|} {sf(s)ds} . \hfill \\ \end{gathered}$

Thus

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

By a variable change

$\displaystyle{\mathbf{y}} - {\mathbf{x}} = {\mathbf{z}}$

we arrive at

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

Therefore if we choose $f$ to be $f(x)=\frac{1}{x}$ we obtain

Corollary. The following identity

$\displaystyle\iint\limits_{|{\mathbf{z}}| = t} {\frac{1}{{|{\mathbf{x}} + {\mathbf{z}}|}}d{\sigma _{\mathbf{z}}}} = 2\pi t\frac{{|{\mathbf{x}}| + t - ||{\mathbf{x}}| - t|}}{{|{\mathbf{x}}|}}$

holds.

It is well-known that

$\displaystyle\min \left\{ {a,b} \right\} = \frac{a + b - \left| {a - b} \right|}{2}$

which gives

$\displaystyle\frac{1}{{4\pi }}\iint\limits_{|{\mathbf{z}}| = t} {\frac{1}{{|{\mathbf{x}} + {\mathbf{z}}|}}d{\sigma _{\mathbf{z}}}} = {t^2}\min \left\{ {\frac{1}{{|{\mathbf{x}}|}},\frac{1}{t}} \right\}$.

Thus we have the following formula for the average of

$\displaystyle \frac{1}{{|{\mathbf{x}} + {\mathbf{z}}|}}$

over a sphere

$\displaystyle\frac{1}{{4\pi {t^2}}}\iint\limits_{|{\mathbf{z}}| = t} {\frac{1}{{|{\mathbf{x}} + {\mathbf{z}}|}}d{\sigma _{\mathbf{z}}}} = \min \left\{ {\frac{1}{{|{\mathbf{x}}|}},\frac{1}{t}} \right\}$.

This formula can also be found on page 249 of a book due to Lieb and Loss (AMS 2001).