*This could be seen as a second-year university-level post.*

Even though the well-known Archimedes has derived the formula for the inside of a sphere long before we were born, its derivation obtained through the use of spherical coordinates and a volume integral is not often seen in undergraduate textbooks.

In this post, we will derive the following formula for the volume of a ball:

\begin{equation}

V = \frac{4}{3}\pi r^3,

\end{equation}

where $r$ is the radius.

Note the use of the word ball as opposed to sphere; the latter denotes the infinitely thin shell, or, surface, of a perfectly round geometrical object in three-dimensional space. A surface has no volume, hence, we prefer to refer to it as a ball.

### Spherical coordinates

The volume of a cuboid $\delta V$ with length $a$, width $b$, height $c$ is given by $\delta V = a \times b \times c$.

In Figure 1, you see a sketch of a volume element of a ball. Although its edges are curved, to calculate its volume, here too, we can use

\begin{equation}

\delta V \approx a \times b \times c,

\end{equation}

even though it is only an approximation.

To use spherical coordinates, we can define $a$, $b$, and $c$ as follows:

\begin{align}

a &= PQ\delta\phi = r\sin\theta \, \delta\phi, \\

b &= r\delta\theta, \\

c &= \delta r.

\end{align}

So, equation (2) becomes

\begin{align}

\delta V &\approx r\sin\theta \, \delta\phi \times r\delta\theta \times \delta r, \nonumber \\

&\approx r^2\sin\theta \, \delta\phi \, \delta\theta \, \delta r.

\end{align}

### Volume integral

Note that the relation becomes more precise when $\delta\phi$, $\delta\theta$, and $\delta r$ tend to zero. So, we can now write the volume integral for our ball $B$ as follows:

\begin{equation*}

V_B = \int_B dV_B = \int_\phi \int_\theta \int_r r^2\sin\theta \, dr \, d\theta \, d\phi.

\end{equation*}

To set the upper and lower bounds for our integrals, we note that a ball has rotational symmetry about the $z$-axis (besides infinitely many others through the centre too). We will exploit this. We refer to Figure 2.

Firstly, to integrate over infinitely many points between $0$ and $r$, the lower bound is $0$ and the upper bound is $r$:

\begin{equation*} V_B = \int_B dV_B = \int_\phi \int_\theta \int_{r=0}^r r^2\sin\theta \, dr \, d\theta \, d\phi.

\end{equation*}

Secondly, to integrate over infinitely many points in the plane of angle $\theta$, we only need to regard the angles between $0$ and $\pi$,

\begin{equation*}

V_B = \int_B dV_B = \int_\phi \int_{\theta=0}^{\theta=\pi} \int_{r=0}^r r^2\sin\theta \, dr \, d\theta \, d\phi,

\end{equation*}

as we will proceed to, thirdly, rotate this plane, as it were, about the $z$-axis to integrate over infinitely many planes about said axis, which complete the shape of our ball. Hence, $\phi$ varies between $0$ and $2\pi$.

And so, we calculate

\begin{align}

V_B = \int_B dV_B &= \int_{\phi=0}^{\phi=2\pi} \int_{\theta=0}^{\theta=\pi} \int_{r=0}^r r^2\sin\theta \, dr \, d\theta \, d\phi, \\

&= \int_{\phi=0}^{\phi=2\pi} \int_{\theta=0}^{\theta=\pi} \left(\frac{1}{3} r^3\sin\theta \Big|_0^r\right) d\theta \, d\phi, \nonumber \\

&= \frac{1}{3} \int_{\phi=0}^{\phi=2\pi} \int_{\theta=0}^{\theta=\pi} r^3\sin\theta \, d\theta \, d\phi, \nonumber \\

&= -\frac{1}{3} \int_{\phi=0}^{\phi=2\pi} \left( r^3\cos\theta \Big|_0^{\pi} \right) \, d\phi, \nonumber \\

&= \frac{2}{3} \int_{\phi=0}^{\phi=2\pi} r^3 \, d\phi, \nonumber \\

&= \frac{2}{3} \left( \phi r^3 \Big|_0^{2\pi} \right), \nonumber \\

&= \frac{4}{3}\pi r^3,

\end{align}

which is the desired result equal to equation (1).

*@kjrunia** is reading mathematics and theoretical physics (final year) in England, at The Open University, Walton Hall, Milton Keynes. He also works on coding for the Mars Rover of the university’s Planetary Robotics Team.*