Volume of n-ball

From dimensional analysis, volume of a n-ball is proportional to $R^n$

$$ V_n=C_n R^n $$

Then from $\int_0^R S_n dr=V_n$

$$ \frac{dV_n}{dR}=V_n=n C_n R^{n-1} $$

The well-known Gaussion integration gives

$$ \int_{-\infty}^\infty e^{-x^2} = \sqrt{\pi} $$

For n-dimensional

$$ \int_{-\infty}^{\infty} e^{-(x_1^2+x_2^2+\cdots x_n^2)} dx_1 dx_2 \cdots dx_n = \pi^{n\over 2} $$

Also

$$ \begin{aligned} \int_{-\infty}^{\infty} e^{-(x_1^2+x_2^2+\cdots x_n^2)} dx_1 dx_2 \cdots dx_n&=\int_0^\infty e^{-r^2}S_n dr\\ &=nC_n \int_0^\infty e^{-r^2} r^{n-1} dr\\ &=\frac{1}{2}n C_n \int_0^\infty e^{-t} t^{\frac{n}{2}-1}dt\\ &=\frac{n}{2}C_n \Gamma(\frac{n}{2})=\pi^{\frac{n}{2}} \end{aligned} $$$$ C_n=\frac{\pi^{\frac{n}{2}}}{\frac{n}{2}\Gamma\left(\frac{n}{2}\right)} $$

Therefore the volume of n-ball is

$$ \color{red}{V_n=\frac{\pi^{\frac{n}{2}}}{\frac{n}{2}\Gamma\left(\frac{n}{2}\right)}R^n} $$