# Translating integration domain

1. Jun 3, 2008

### jostpuur

If $$f:\mathbb{R}^3\to\mathbb{R}$$ is a continuous function, $$x_0\in\mathbb{R}^3$$ a fixed point, $$r>0$$ a fixed radius, and $$n\in\mathbb{R}^3$$ a fixed vector satisfying $$|n|=1$$, then is the equation

$$\int\limits_{B(x_0 + \alpha n, r)} d^3x\; f(x) \;=\; \int\limits_{B(x_0, r)} d^3x\; f(x) \;+\; \frac{\alpha}{r} \int\limits_{\partial B(x_0,r)} d^2x\; ((x-x_0)\cdot n) f(x) \;+\; O(\alpha^2),\quad\quad\alpha\in\mathbb{R}$$

true? I convinced myself of this somehow, but I'm still feeling unsure. I don't know how to deal with equations like this rigorously. There are other problems of similar nature, where the integration domain is changed a little bit, and then it is somehow possible to write the change as a functional of the restriction of the integrand onto the boundary.

The B notation means the ball

$$B(x_0,r) = \{x\in\mathbb{R}^3\;|\;|x-x_0|<r\},$$

and $$\partial$$ is the boundary,

$$\partial B(x_0,r) = \{x\in\mathbb{R}^3\;|\;|x-x_0|=r\}.$$

Last edited: Jun 3, 2008
2. Jun 4, 2008

### jostpuur

Later in my calculations I encountered a situation where some factor, that was supposed to be 1, was 3/2. I cannot yet know where the mistake is, but this formula could be one potential place for it.