Integration of functions with norm 
Let $\|\cdot\|$ be a norm on $\mathbb{R}^2$. Prove that $\exists c>0$ such that for every borelian function $f: \mathbb{R} \rightarrow \mathbb{R},$ $$\int_{\mathbb{R}^2}f(\|x\|)dx=c\int_{\mathbb{R}^2}f(|x|)dx,$$
where $|\cdot |$ is the Euclidian norm.
Prove the following extension to $\mathbb{R}^d$: $$\exists c'>0;\int_{\mathbb{R}^d}f(\|x\|)dx=c'\int_{0}^{+\infty}x^{d-1}f(x)dx$$

Trying the change $x=r\cos(\theta),y=r\sin(\theta)$ (from $]0,2\pi[ \times ]0,+\infty[ \to \mathbb{R}^2-([0,+\infty[ \times${$0$}$)$, so
$$\int_{0}^{2\pi}\int_{0}^{+\infty}rf(r\|(\cos(\theta),\sin(\theta))\|)drd\theta=\int_{\mathbb{R}^2}f(\|x\|)dx,$$
$$\int_{\mathbb{R}^2}f(|x|)dx=2\pi\int_0^{+\infty}rf(r)dr,$$
So how to obtain $c$? Is there another way to prove it? How can we prove the extension for this result?
 A: Using polar-coodrinates is not practicable, since the second norm is not necessarly invariant under rotations. For example, consider the $\infty$-norm. In fact, only the scale properties of the Lebesgue-measure are important:
Set $f_I(x) := 1_{I}(x)$ and note that $\int f_{[0,1)}(|x|) \, \mathrm{d} x = \omega_1$ is the volume of the unit ball with respect to the norm $| \cdot|$, thereby $\int f_{[0,1)}(\|x\|) \, \mathrm{d} x = \omega_2$ is the volume of the euclidean unit ball. Thus, the only possible choice for $c$ is $$c= \omega_2 / \omega_1.$$
Moreover, we have (scale properties of the Lebesgue-measure)
\begin{align}
\int f_{[0,b)}(\|x\|) \, \mathrm{d}x &= b^2 \int f_{[0,1)}(\|x\|) \, \mathrm{d}x \\
&= c b^2 \int f_{[0,1)}(|x|) \, \mathrm{d}x = c \int f_{[0,b)}(|x|) \, \mathrm{d}x.
\end{align}
Furthermore, we also find that
\begin{align}
\int f_{[a,b)}(\|x\|) \, \mathrm{d}x &= \int f_{[0,b)}(\|x\|) \, \mathrm{d}x - \int f_{[0,a)}(\|x\|) \, \mathrm{d}x \\
&= c \Big( \int f_{[0,b)}(|x|) \, \mathrm{d}x - c \int f_{[0,a)}(|x|) \, \mathrm{d}x \Big) = c \int f_{[a,b)}(|x|) \, \mathrm{d}x.
\end{align}
To establish this identity for all measurable functions, start with continuous functions $f \colon [0,\infty) \rightarrow \mathbb{R}$ with compact support: These functions can be approximated by functions of the type
$$f_n := \sum_{k=1}^{m} c_k 1_{[a_k,b_k)}$$
uniformly. By uniform convergence, we get that $\int f(\|x\|) \, \mathrm{d} x = c \int f(|x|) \, \mathrm{d} x$. If $f$ is continuous, but doesn't have compact support, we can multiple it with sequence $(g_n)$ of smooth bump function. Since $|fg_n|$ is monotone, we get that $f(\|x\|)$ is in $L^1$ if and only if $f(|x|)$ is in $L^1$. In this case an application of the dominated convergence theorem (note that $|f g_n| \leq |f|$) shows also that $$\int f(\|x\|) \, \mathrm{d} x = c \int f(|x|) \, \mathrm{d} x.$$
If we denote $\nu_1$ the pushforward measure induced by $x \rightarrow \|x\|$ and resp. $\nu_2$ the induced measure by $x \rightarrow |x|$, the latter identity says that
$$\tag{$*$} \int f(t) \, \mathrm{d} \nu_1(t) = c \int f(t) \, \mathrm{d} \nu_2(t)$$
for all continuous functions, which are integrable. Since $\nu_1$, $\nu_2$ are local-finite measures (all bounded open sets have finite measure), they are regular. For these measures it is well-known that ($*$) already implies that $\nu_1 = c \nu_2$.
There is also an alternative approach by considering the pushforward measure: Let $\nu_1$ be again the measure induced by $x \rightarrow \|x\|$. Then we have $\nu_1([0,1)) = \omega_1$ and a similiar calculation as above shows that $$\tag{$**$}\nu_1([a,b)) = (b^2-a^2) \omega_1.$$ If we restrict us on $[0,M]$ (i.e. $0 \leq a \leq b \leq M$), then we get that $\nu_1([a,b)) \leq 2 M (b-a) \omega_1$. By an appoximation argument (using that $\nu_1$ is regular), we see that $\nu_1(B) \leq 2M \omega_1 \lambda(B)$ for all Borel-sets $B \subset [0,M]$. This observation shows that $\nu_1$ is absolute continous with the respect to the Lebesgue-measure on $\mathbb{R}$, which I have denoted by $\lambda$. The identity ($**$) can be extended to all sets showing that
$$\nu_1(B) = \omega_1 \int_B 2 x \, \mathrm{d} x $$
and thus
$$ \int f(\|x\|) \, \mathrm{d} x = \int f(t) \, \mathrm{d} \nu_1( t) = \omega_1 \int_0^\infty 2x f(x) \, \mathrm{d} x.$$
The same argument applies to $\nu_2$ leading to
$$ \int f(|x|) \, \mathrm{d} x = \int f(t) \, \mathrm{d} \nu_2 (t) = \omega_2 \int_0^\infty 2x f(x) \, \mathrm{d} x.$$
