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Derivation of the volume of a sphere – method 2

In this blog here, I derived the expression for the volume of a sphere, $V= \frac{4}{3} \pi r^{3}$ using spherical polar coordinates. There is, however, another way to do it which some people may find easier, and that is to use something called the volume of rotation. Remember, a circle with radius $r$ which is centred on the origin has the equation $x^{2} + y^{2} = r^{2}$, but as I pointed out in my blog on deriving the area of a circle, because we can’t integrate $\sqrt{ \left( r^{2} - x^{2} \right) } \; dx$ we are stuck in trying to use conventional Cartesian coordinates, which is why we had to use polar coordinates instead.

However, if you are not comfortable with spherical polar coordinates to derive the volume of a sphere, there is a trick to get around using them. That is to use the volume of revolution. The idea is quite simple, imagine the strip which has area $dA = y \; dx$ as shown below.

Now, imagine rotating this trip about the x-axis, to produce a disk as shown in the figure below. The surface area of this disk is just the area of a circle with radius $y$, and so its area $dA = \pi y^{2}$. The volume element of the disk is then just $dV = dA \; \cdot dx = \pi y^{2} \; dx$, but as we can write $y^{2} = r^{2} - x^{2}$ this becomes $dV = \pi (r^{2} - x^{2}) \; dx$, which is easy to integrate.

This leads to the total volume $V$ being

$V = \int_{-r}^{+r} (r^{2} - x^{2}) dx = \left[ r^{2}x - \frac{x^{3}}{3} \right]_{-r}^{+r} = (r^{3} - \frac{r^{3}}{3}) - ( -r^{3} + \frac{r^{3}}{3})$

$V = 2r^{3} - \frac{2r^{3}}{3} = \; \boxed{ \; \frac{4}{3} \pi r^{3} }$

just as we had before. QED!