Sphere: Volume and Surface Area

Insert the radius into $V =\frac{ 4}{3}\cdot{}\pi{}\cdot{}r^{3}$: multiply the radius by itself three times, times π, times $\frac{4}{3}$. Example: $r = 3 cm$ gives $V =\frac{ 4}{3}\cdot{}\pi{}\cdot{}27 = 36\pi{} \approx{} 113.1 cm^{3}$. If the diameter is given, halve it first: $r =\frac{ d}{2}$. Watch the order: only r is cubed, π and $\frac{4}{3}$ remain factors. The unit is always a cubic unit like $cm^{3}$ or $m^{3}$, because three lengths are multiplied. For litres use $1000 cm^{3} = 1 l$. A quick sanity check: the sphere must hold less than the enclosing cube with edge 2r, here $216 cm^{3}$.

Rearrange the formula: $r = \sqrt[3]{(\frac{3V}{(4\pi{})})}$. So multiply the volume by 3, divide by 4π and take the cube root. Example: $V = 500 cm^{3}$ gives $\frac{3\cdot{}500}{(4\pi{})} =\frac{ 1500}{12.566} \approx{} 119.4$ and thus $r = \sqrt[3]{119.4} \approx{} 4.92 cm$. Check: $\frac{4}{3}\cdot{}\pi{}\cdot{}4.92^{3} \approx{} 499 cm^{3}$ ✓. The most common mistake is taking the square root instead of the cube root; but the volume depends on $r^{3}$, so the cube root is needed. On the calculator use the ∛ key or raise to the power $\frac{1}{3}$. Analogously the surface gives $r = \sqrt{(\frac{O}{(4\pi{})})}$, there with a square root.

The volume $V =\frac{ 4}{3}\cdot{}\pi{}r^{3}$ measures the capacity, i.e. how much fits inside the sphere; the surface $O = 4\pi{}r^{2}$ measures the skin of the sphere, i.e. how much material is needed to wrap or paint it. You can tell them apart by exponent and unit: volume has $r^{3}$ and $cm^{3}$, surface has $r^{2}$ and $cm^{2}$. For radius $r = 3 cm$ both numerical values happen to be $36\pi{} \approx{} 113.1$, but with different units; that is a peculiarity of $r = 3$. Also note the scaling: doubling the radius quadruples the surface and multiplies the volume by eight. That is why small bodies cool faster: they have relatively much surface per volume.

The classical route is due to Archimedes: he compared the sphere with the circumscribed cylinder (radius r, height 2r) and proved that the sphere fills exactly $\frac{2}{3}$ of its volume. The cylinder holds $\pi{}r^{2}\cdot{}2r = 2\pi{}r^{3}$, two thirds of that is $\frac{4}{3}\cdot{}\pi{}r^{3}$. In upper school you can verify this with the volume of revolution: the semicircle $f(x) = \sqrt{(r^{2} - x^{2})}$ rotates over [−r; r] around the x-axis, and $V = \pi{}\cdot{}\int{}(r^{2} - x^{2}) dx$ gives exactly $\frac{4}{3}\cdot{}\pi{}r^{3}$. The surface formula is tied to it: $O = 4\pi{}r^{2}$ is the derivative of the volume with respect to r, because a sphere grows in shells like an onion.

The volume is simply half: $V =\frac{ 2}{3}\cdot{}\pi{}r^{3}$. With the surface you must be careful, because the circular cut face $\pi{}r^{2}$ is added to the half shell $2\pi{}r^{2}$: $O = 2\pi{}r^{2} + \pi{}r^{2} = 3\pi{}r^{2}$. Example with $r = 3 cm$: $V =\frac{ 2}{3}\cdot{}\pi{}\cdot{}27 = 18\pi{} \approx{} 56.5 cm^{3}$ and $O = 3\pi{}\cdot{}9 = 27\pi{} \approx{} 84.8 cm^{2}$. The typical mistake is to give the surface of the hemisphere as half the sphere surface $2\pi{}r^{2}$ and forget the cut face; that is only correct if the open shell is explicitly meant (e.g. a bowl without a lid). So read carefully whether the solid is closed.