can i write this $\int_{\mathbb{R}^2} \psi(\textbf{x}) \, d^{2}\textbf{x} = 2\pi \int_{0}^{\infty} \rho \, \phi(\rho)\, d\rho $?

Question

If the function $\psi$ is isotropic, (i.e. $\psi(\textbf{x}) = \phi(|\textbf{x}|)$, where $\phi \in L^{1}(\mathbb{R})$, then can i write \begin{equation} \label{eq:2.1} \int_{\mathbb{R}^2} \psi(\textbf{x}) \, d^{2}\textbf{x} = 2\pi \int_{0}^{\infty} \rho \, \phi(\rho)\, d\rho , \end{equation} where $\rho=|\textbf{x}|$ is the norm of $\textbf{x}$ in $\mathbb{R}^{2}?$

Answer 1

You can view it as a substation, $$\iint_{\mathbb R^2}\psi(x,y)dxdy$$ Let $x=r\cos(\theta)$, $y=r\sin(\theta)$.(i.e. polar coordinates) for which the jacobian is known to be $r$. $$\int_0^{2\pi}\int^\infty_{-\infty} \phi(r^2)r~dr~d\theta$$

$$2\int_0^{2\pi}\int^\infty_0 \phi(r^2)r~dr~d\theta$$
now, doing simple substitution $r^2=\rho$ $$\int_0^{2\pi}\int^\infty_0 \phi(\rho)d\rho d\theta$$ Because the integral inside isn't dependent on $\theta$ then it is equal to $$2\pi\int^\infty_0 \phi(\rho)d\rho$$

3D

$$\iiint_{\mathbb R^3}\psi(x,y,z)dxdydz$$ using spherical coordinates: $$\int^{2\pi}_0\int^{\pi}_0\int^\infty_0\phi(r^2)r\sin(\theta)drd\theta d\alpha$$ $$2\pi\int_0^\pi sin(\theta)d\theta \int_0^\infty\phi(r^2)rdr$$ $$4\pi\int^\infty_0 \phi(\rho)d\rho$$ I'm unsure of the existence of a general solution.