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I'm studying the PDE textbook written by Evans. In the section about mean-value formulas for the Laplace equation, he asserts that $$\bar\int_{\partial B(x,r)}u(y)\mathrm{d}S(y) =\bar\int_{\partial B(0,1)}u(x+rz)\mathrm{d}S(z),$$ where $\bar\int$ signifies that the integration is taken for the average, that is, $$\bar\int_{\partial B(x,r)}u(y)\mathrm{d}S(y) =\frac{1}{n\alpha(n)r^{n-1}}\int_{\partial B(x,r)} u(y)\mathrm{d}S(y).$$ Here $\alpha(n)$ denotes the volume of the unit $n$-ball. I wonder why the assertion holds. Taking the Jacobian determinant into consideration, we should have $$\frac{1}{n\alpha(n)r^{n-1}}\int_{\partial B(x,r)} u(y)\mathrm{d}S(y)=\frac{1}{n\alpha(n)r^{n-1}}\int_{\partial B(0,1)} u(x+rz)r^n\mathrm{d}S(z).$$ Is there anything I missed? Thank you.

Boar
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    Since you are integrating on the surface of the sphere, making your change of variables you should have $r^{n-1}$ instead of $r^n$ inside the integral. – Duca_Conte Oct 27 '19 at 15:23
  • I found this is a duplicate of an old post. This might help: https://math.stackexchange.com/questions/673526/how-to-change-variables-in-a-surface-integral-without-parametrizing – Boar Oct 27 '19 at 22:39

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