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## Derivation of the volume of a sphere

In this blog, I derived the expression for the surface area of a sphere, $A = 4 \pi r^{2}$. In today’s blog, I will derive the expression for the volume of a sphere. Actually, once one has understood how to derive the surface area of a sphere using spherical polar coordinates, deriving the volume is pretty straight forward. It only involves one extra step, and that is to create a volume element $dV$ with the same surface area $dA$ that we had before, but with a thickness $dr$, and to integrate over $r$ in addition to integrating over $\theta \text{ and } \phi$.

As we can see from the figure below, the volume element $dV$ is given by $dr \cdot dA$ where $dA$ is the same surface element we derived before, namely $dA = r^{2} \; \cos \theta d \theta \; d \phi$. So, the expression for our volume element is

$\boxed{ dV = dr \cdot dA = r^{2} \; \cos \theta d \theta \; d \phi \; dr = r^{2} dr \; \cos \theta d \theta \; d \phi }$

The volume element $dV$ is given by $dA \cdot dr = r^{2}dr \; \cos \theta d \theta \; d \phi$

We need to integrate this volume element over all three variables $r, \theta \text{ and } \phi$ so we have

$\text{total volume of a sphere} = \int d V = \int_{0}^{r} r^{2} dr \; \int_{-\pi/2}^{+ \pi/2} \cos \theta d \theta \; \int_{0}^{2 \pi} d \phi$

$\text{total volume of a sphere} = \left[ \frac{ r^{3} }{ 3 } \right]_{0}^{r} \left[ \sin \theta \right]_{-\pi / 2 }^{ + \pi / 2 } \left[ \phi \right]_{0}^{2 \pi} = \frac{ r^{3} }{ 3 } \cdot ( 1 + 1 ) \cdot 2 \pi = \boxed{ \frac{ 4 }{ 3} \pi r^{3} }$

as required.