A disk of mass $M$ and radius $R$ is initially rotating at angular velocity of $\omega$. While rotating, it is placed on a horizontal surface whose coefficient of friction is $\mu=0.5$. How long will it take for the disk to stop rotating?

Someone had solved the problem following way

\begin{align*} \mathrm{df} &=\mu \mathrm{mg} \cdot \frac{2 \pi \mathrm{xdx}}{\pi \mathrm{R}^{2}}=\frac{2 \mu \mathrm{mgx}}{\mathrm{R}^{2}} \mathrm{dx} \\
\implies \mathrm{d} \tau &=\mathrm{xdf}=\frac{2 \mu \mathrm{mgx}^{2} \mathrm{dx}}{\mathrm{R}^{2}} \\ \implies\int \mathrm{d} \tau &=\frac{2 \mu \mathrm{mg}}{\mathrm{R}^{2}} \int_{0}^{R} \mathrm{x}^{2} \mathrm{~d} \mathrm{x} \end{align*} Again, \begin{align*} \tau &=\frac{2}{3} \mu \mathrm{mgR} \\ \tau &=\mathrm{I} \alpha=\frac{1}{2} \mathrm{~m} \mathrm{R}^{2} \alpha \\ \implies \alpha &=\frac{4}{3} \frac{\mu \mathrm{g}}{\mathrm{R}} \\ \alpha \mathrm{t}=\omega \implies \mathrm{t}&=\frac{\omega}{\alpha}=\frac{3 \omega \mathrm{R}}{4 \mathrm{\mu g}} \end{align*}

Unfortunately, I didn't understand first line of the math. Could you please explain it to me? Or, Is there any other way to solve the problem?

$$df=\mu mg\bullet \frac{2\pi xdx}{\pi R^2}$$