Volume of a sphere

Pick up a cricket ball and ask a simple question: how much space does it actually take up? That is what volume measures, and for any ball-shaped solid — a sphere — the answer turns out to depend on just one number, the radius $r$ , raised to the third power. This article, part of the Class 9 chapter Surface Areas and Volumes, gives the volume formula for a sphere and for a hemisphere (half a sphere), shows why the formula has the shape it does using solids you already know — the cylinder and the cone — and works through the same numbers as the syllabus: a solid ball of radius $4.2$ cm and a hemispherical bowl of radius $9$ cm. By the end you should be able to plug a radius into $V = \tfrac43\pi r^3$ without hesitating over where the $\tfrac43$ comes from.

The core insight

A sphere of radius $r$ has volume $V = \tfrac43\pi r^3$ , exactly twice the volume of a hemisphere of the same radius, and the derivation follows from comparing the sphere to a cylinder and a cone of matching size.

The core idea

For a sphere of radius $r$ , the volume is:

V_{\text{sphere}} = \frac{4}{3}\pi r^3

where: $r$ is the radius of the sphere and $\pi \approx \tfrac{22}{7}$ or $3.14$ . A hemisphere is exactly half a sphere, cut through the centre, so its volume is exactly half of the sphere's:

V_{\text{hemisphere}} = \frac{1}{2}\times\frac43\pi r^3 = \frac23\pi r^3

Line the three mensuration solids of this chapter up side by side and a pattern jumps out: cylinder volume is $\pi r^2 h$ , cone volume is $\tfrac13\pi r^2 h$ , and sphere volume is $\tfrac43\pi r^3$ . The first two need a separate height $h$ because they have a flat base and rise up from it; a sphere has no separate height at all — its radius alone fixes its entire size, so the same $r$ does the job that $h$ does elsewhere. That is also why the exponent jumps from $r^2h$ to a clean $r^3$ : cube the one length that describes the whole solid. The figure below shows the classical way this formula is uncovered without calculus — by comparing a hemisphere to a cylinder with a cone hollowed out of it, all sharing the same radius $r$ and the same height $r$ .

Figure

A cylinder of radius r and height r with a cone of the same radius and height removed from its top, shown next to a solid hemisphere of the same radius, illustrating that the leftover cylinder volume equals the hemisphere volume

Derivation AVolume of a sphere from a cylinder and a cone

The goal is to obtain the sphere volume formula $V = \tfrac43\pi r^3$ without using calculus, by comparing a solid hemisphere of radius $r$ to a cylinder of radius $r$ and height $r$ that has a cone (same radius $r$ , same height $r$ ) hollowed out of its top, and showing that a horizontal slice through both solids at the same height always has the same area — so, by Cavalieri's principle, the two solids must have the same total volume.

Step 1. Set up both solids on the same base and slice them at height $h$ above that base, where $0 \le h \le r$ . For the hemisphere (a solid ball cut in half, flat side down), a horizontal slice at height $h$ is a circle. By Pythagoras' theorem applied to the right triangle formed by the sphere's centre, the slice's edge, and the axis, the radius of that circle is $\sqrt{r^2 - h^2}$ , so its area is $\pi(r^2 - h^2)$ .

Step 2. For the cylinder-with-cone-removed (radius $r$ , height $r$ , apex of the cone at the centre of the base), a horizontal slice at height $h$ is an annulus — a big circle of radius $r$ (the cylinder's cross-section) with a smaller circle removed (the cone's cross-section). Because the cone's radius grows in direct proportion to height and reaches $r$ at height $r$ , the cone's radius at height $h$ is exactly $h$ . So the annulus area is $\pi r^2 - \pi h^2 = \pi(r^2 - h^2)$ .

Step 3. Compare the two areas. Step 1 gave the hemisphere slice area as $\pi(r^2 - h^2)$ ; Step 2 gave the cylinder-minus-cone slice area as $\pi(r^2 - h^2)$ too — identical, for every single value of $h$ from $0$ to $r$ . Cavalieri's principle says that if two solids have matching cross-sectional areas at every height, they have the same volume. So the hemisphere's volume equals the cylinder's volume minus the cone's volume.

Step 4. Compute that right-hand side using formulas already known from this chapter. Cylinder volume is $\pi r^2 \times r = \pi r^3$ ; cone volume (same radius $r$ , same height $r$ ) is $\tfrac13\pi r^2 \times r = \tfrac13\pi r^3$ . So $V_{\text{hemisphere}} = \pi r^3 - \tfrac13\pi r^3 = \tfrac23\pi r^3$ . Doubling it for the full sphere gives $V_{\text{sphere}} = 2\times\tfrac23\pi r^3 = \tfrac43\pi r^3$ — the formula stated at the top of this article.

Worked example

Problem: Find the volume of a sphere of radius $4.2$ cm. (Use $\pi = \tfrac{22}{7}$ .)

Step 1. Write the formula and substitute the radius. $V = \tfrac43\pi r^3$ , with $r = 4.2$ cm.

Step 2. Compute $r^3$ . $(4.2)^3 = 4.2\times4.2\times4.2 = 74.088$ .

Step 3. Substitute and simplify. $V = \tfrac43\times\tfrac{22}{7}\times74.088$ . First, $\tfrac{22}{7}\times74.088 = 22\times10.584 = 232.848$ . Then $V = \tfrac43\times232.848 = \tfrac{931.392}{3} = 310.464$ .

The volume of the sphere is $V \approx 310.46 \text{ cm}^3$ .

Competitive stretch

Problem: A hemispherical bowl has radius $9$ cm. How many litres of water can it hold? (Use $\pi = \tfrac{22}{7}$ ; $1000\text{ cm}^3 = 1$ litre.)

Step 1. Apply the hemisphere formula. $V = \tfrac23\pi r^3 = \tfrac23\times\tfrac{22}{7}\times9^3$ , and $9^3 = 729$ .

Step 2. Simplify. $V = \dfrac{2\times22\times729}{3\times7} = \dfrac{32076}{21} \approx 1527.43 \text{ cm}^3$ .

Step 3. Convert to litres by dividing by $1000$ : $V \approx \dfrac{1527.43}{1000} \approx 1.53$ litres. The bowl holds about $1.53$ litres of water — notice this is roughly a fifth of the full sphere volume found above, even though the bowl's radius is more than twice as large, a reminder that the cubic power makes volume grow very fast with radius.

Common traps

Using $r^2$ instead of $r^3$ . Since the cone and cylinder formulas involve $r^2h$ , it is easy to slip and square the radius for a sphere instead of cubing it.
Forgetting the $\tfrac23$ factor for a hemisphere, or applying the full sphere formula $\tfrac43\pi r^3$ to a problem that is actually about half a sphere, like a bowl.
Diameter vs. radius mix-ups. If a problem states the diameter, halve it before cubing — cubing the diameter directly overstates the volume eightfold.
Skipping the unit conversion. Capacity questions expect litres or kilolitres, not raw cm³ or m³; forgetting to divide by $1000$ (or $10^6$ for kilolitres) is a common last-step slip.

Ask the tutor: Still shaky on why the sphere formula has a $\tfrac43$ in it, or want more practice converting between sphere and hemisphere problems? Ask the AI tutor for extra worked examples.

Why does the sphere formula use r cubed while the cone and cylinder use r squared times h?

Because a cylinder or cone has two independent dimensions — the size of its flat circular base (radius $r$ ) and how tall it stands (height $h$ ) — so its volume formula naturally carries $r^2$ from the base area and one separate factor of $h$ from the height. A sphere has only one dimension, the radius, since it looks the same from every direction and has no separate "height." To get a three-dimensional (cubic) quantity like volume out of a single length $r$ , that same $r$ has to be used three times, giving $r^3$ instead of $r^2h$ .

Do I need to remember the cylinder-and-cone derivation for exams, or just the final formula?

For most Class 9 exam purposes you need to state and correctly use $V = \tfrac43\pi r^3$ (and $\tfrac23\pi r^3$ for a hemisphere), not reproduce the full derivation — it is not typically asked to be proved at this level. That said, understanding the cylinder-minus-cone comparison is genuinely useful: it explains why the numbers $\tfrac13$ , $\tfrac23$ and $1$ (cone, hemisphere, cylinder of matching radius and height $r$ ) sit in that exact ratio, and recalling that relationship is a fast way to sanity-check whether an answer is roughly the right size.

How is the volume of a sphere different from its surface area, since both use only r?

Volume tells you how much space the solid ball fills — useful for "how much water/material does it hold or contain," measured in cubic units like cm³. Surface area tells you how much outer skin covers it — useful for "how much material to wrap or paint it," measured in square units like cm². Both formulas use only $r$ because a sphere has no other free dimension, but volume cubes $r$ (since it is a 3-D quantity) while surface area squares $r$ (since it is a 2-D quantity): $V = \tfrac43\pi r^3$ versus $A = 4\pi r^2$ . Reading the question's wording — "holds/contains" versus "covers/coats" — tells you which one to use.

Volume of a sphere

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads