I want to calculate the derivative $$F(r)=\underset{x_1^2+...+x_m^2\leq r^2}{\int ... \int} f(x_1,...,x_m) \, dx_1 \, dx_2 ... dx_m$$ with respect to $r$. $f:\mathbb{R}^m\rightarrow \mathbb{R}$ is a continous function.
My attempt: (heavily inspired by this question)
We can define the function
$$\phi: \mathbb{R}^m\rightarrow \mathbb{R}^m,\quad x\mapsto rx$$
$x_1^2+...+x_m^2 \leq r^2$ $\Rightarrow$ $\sqrt{x_1^2+...+x_m^2} \leq r$ $\overset{\phi^{-1}}{\Longrightarrow}$ $\sqrt{x_1^2+...+x_m^2} \leq \phi^{-1}(r) = 1$ So the region we're integrating is $B_{\mathbb{R}^m}(0,1)$, right? (I know my definition of $\phi$ can be misleading, maybe someone has a better suggestion?)
We can then change variables and see that... \begin{align*} \underset{x_1^2+...+x_m^2 \leq r^2}{\int ... \int} f(x_1,...,x_m) dx_1...dx_m= \underset{B_{\mathbb{R}^m}(0,1)}{\int ... \int} f(rx_1,...,rx_m) \cdot |\det D\phi(x_1,...,x_m)| dx_1...dx_m \end{align*}
We can easily calculate that... \begin{align*} \det \phi(x_1,...,x_m) = \underbrace{\begin{pmatrix} r & 0 & 0 & ... & 0 & 0\\ 0& r & 0 & ... & 0 & 0\\ ....\\ 0&0&0&...&r&0\\ 0&0&0&...&0&r\\ \end{pmatrix}}_{\in \mathbb{R}^{m\times m}} = r^m \end{align*}
\begin{align*} \underset{x_1^2+...+x_m^2 \leq r^2}{\int ... \int} f(x_1,...,x_m) dx_1...dx_m = \underset{B_{\mathbb{R}^m}(0,1)}{\int ... \int} r^m\cdot f(rx_1,...,rx_m) dx_1...dx_m \end{align*}
Now that the bounds are free from $r$, we can move $\frac{d}{dr}$ inside the integral. \begin{align*} F'(r)&=\frac{d}{dr} \underset{B_{\mathbb{R}^m}(0,1)}{\int ... \int} r^m\cdot f(rx_1,...,rx_m) dx_1...dx_m = \underset{B_{\mathbb{R}^m}(0,1)}{\int ... \int} \frac{d}{dr} r^m\cdot f(rx_1,...,rx_m) dx_1...dx_m\\ &\underset{B_{\mathbb{R}^m}(0,1)}{\int ... \int} f'(rx_1,...,rx_m)\cdot (x_1\cdot ... \cdot x_m)\cdot r^m + m\cdot r^{m-1}\cdot f(rx_1,...,rx_m) dx_1...dx_m \end{align*}
Any remarks? Is this correct?