Let $B(0,r)=\{x\in \mathbb R^n\mid \|x\| In other word, why
$$\frac{1}{|\partial B_1|}\int_{\partial B_1}f(R\sigma )d\sigma =\frac{1}{|\partial B_R|}\int_{\partial B_R}f(y)dy\ \ ?$$
If $B(0,r)\subset \mathbb R^n$ is a ball and $\sigma \in \partial B(0,1)$, if $y=r\sigma $, what is $dy$?
0
$\begingroup$
real-analysis
1 Answers
1
Integration in $\mathbb R^n$. Cartesian coordinates into polar coordinates for $B(0,R)$:
$$
\int_{|x| A convincing and rigorous explanation is given in Walter Rudin's, Real and Complex Analysis, Chapter 8, Exercise 6.
-
0I'm not sure to understand all. Why $$\frac{1}{|\partial B_1|}\int_{\partial B_1}f(R\sigma )d\sigma =\frac{1}{|\partial B_R|}\int_{\partial B_R}f(y)dy\ \ ?$$ I tryed to apply your formula, but it's not conclusive. – 2017-01-24
-
0In your formula, both Right and left hand side represent the average of $f$ on the surface $\partial B_R$. – 2017-01-25