Volume of a Sphere
Geometry & MeasurementThe volume of a sphere is four-thirds times pi times the radius cubed.
Formula
V = \frac{4}{3}\pi r^3
Definition
The volume of a sphere (a ball shape) is $V = (4/3) \times \pi \times r^3$, where $r$ is the radius. You cube the radius (multiply it by itself three times), then multiply by $\pi$ and $4/3$.
Example
A ball with radius $3$ cm: $V = (4/3) \times \pi \times 27 = 36\pi = 113.1$ cm$^3$. A basketball with radius $12$ cm: $V = (4/3)\pi \cdot 1728 = 2304\pi = 7238$ cm$^3$.
Key Insight
A sphere fits perfectly inside a cylinder with the same diameter and height. The sphere takes up exactly $2/3$ of that cylinder's volume. Archimedes discovered this and was so delighted he asked for it to be carved on his tombstone.
Definition
A sphere of radius $r$ has volume $V = (4/3)\pi r^3$. The volume scales as $r^3$: doubling the radius multiplies volume by $8$. A sphere and its circumscribed cylinder (radius $r$, height $2r$) have volumes in ratio $2:3$. Hemisphere volume $= (2/3)\pi r^3$.
Example
Earth has radius $6371$ km. $V = (4/3)\pi(6371)^3 = 1.083 \times 10^{12}$ km$^3$. Jupiter has radius $69{,}911$ km, so $V_{Jupiter} / V_{Earth} = (69911/6371)^3 = (10.97)^3 = 1321$. Jupiter is $1321$ times Earth's volume.
Key Insight
Volume scales as the cube of linear dimensions, so small increases in radius create large increases in volume. A star $10$ times Earth's radius has $10^3 = 1000$ times Earth's volume. This cubic scaling explains why large planets and stars accumulate mass so rapidly as they grow.
Definition
$$V = \int_{-r}^{r} \pi(r^2 - z^2) \, dz = \pi\left[r^2 z - \frac{z^3}{3}\right]_{-r}^{r} = \pi\left(2r^3 - \frac{2r^3}{3}\right) = \frac{4}{3}\pi r^3.$$ Alternatively, in spherical coordinates: $$V = \int_0^{2\pi}\int_0^\pi\int_0^r \rho^2\sin\varphi \, d\rho \, d\varphi \, d\theta = \frac{4}{3}\pi r^3.$$
Example
The derivative of $V = (4/3)\pi r^3$ with respect to $r$ is $4\pi r^2 = SA$, the surface area. This is not coincidental: a thin shell of thickness $dr$ at radius $r$ contributes $SA \cdot dr$ to the volume. The same relationship holds for any dimension: $d/dr(\text{Volume}) = \text{Surface Area}$.
Key Insight
The relationship $V'(r) = SA(r)$ is a consequence of the fact that volume is the integral of surface area from $0$ to $r$. This generalizes: for an n-ball, the n-dimensional volume is the integral of the $(n-1)$-dimensional surface area, a beautiful unification of measurement in all dimensions.