Prove by integration Volume Sphere is $V=\displaystyle {4\pi R^3\over 3}$

Volume of a Sphere by integration

Prove by integration Volume Sphere of radius $R$ is $$V=\displaystyle {4\pi R^3\over 3}$$

Volume of revolution of the semicircle about the $x$ axis gives the volume

$$\begin{array}{ll}
V&=\displaystyle \pi \int^R_0 y^2\,dx\\
&=\displaystyle 2\pi \int^R_0 y^2\,dx\\
&=\displaystyle 2\pi \int^R_0 R^2-x^2\,dx\\
&=\displaystyle 2\pi \left.\left(R^2x-{x^3\over 3} \right)\right|^R_0\\
&=\displaystyle 2\pi \left(R^2-{R^3\over 3} \right)\\
&=\displaystyle 2\pi\times {2R^3\over 3}\\
&=\displaystyle {4\pi R^3\over 3}
\end{array}$$