- #1
jostpuur
- 2,112
- 19
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
<br /> \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}<br />
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
<br /> B(x_0,r) = \{x\in\mathbb{R}^3\;|\;|x-x_0|<r\},<br />
and \partial is the boundary,
<br /> \partial B(x_0,r) = \{x\in\mathbb{R}^3\;|\;|x-x_0|=r\}.<br />
<br /> \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}<br />
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
<br /> B(x_0,r) = \{x\in\mathbb{R}^3\;|\;|x-x_0|<r\},<br />
and \partial is the boundary,
<br /> \partial B(x_0,r) = \{x\in\mathbb{R}^3\;|\;|x-x_0|=r\}.<br />
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