Sphere and hemisphere

A football, a marble, a ball bearing, the Earth itself to a first approximation — all are spheres, the most symmetric solid there is, with every surface point the same distance $r$ from the centre. Slice a sphere exactly through its centre and each half is a hemisphere: a mixing bowl, a dome, half a globe. This is the third and last solid of the Class 10 Mensuration unit, after the cylinder and the cone, and it closes the set with the cleanest formulas of all — a sphere needs only one surface measure, its total surface area, while a hemisphere splits into a curved dome and a flat circular base. Both formulas trace back to a single elegant idea that Archimedes proved over two thousand years ago: fit the sphere exactly inside a cylinder, and the cylinder's own volume and surface area hand you the sphere's for free.

The core insight

A sphere of radius $r$ has total surface area $4\pi r^2$ and volume $\dfrac{4}{3}\pi r^3$ , and a hemisphere — exactly half a sphere — has curved surface area $2\pi r^2$ , total surface area $3\pi r^2$ (curved plus the flat circular base), and volume $\dfrac{2}{3}\pi r^3$ .

The core idea

A sphere has no edges, no vertices, and no separate "flat" and "curved" parts — its entire boundary is one smooth surface, so it has a total surface area (TSA) but no curved surface area (CSA) in the usual sense.

\text{TSA}_{\text{sphere}} = 4\pi r^2

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

where: $r$ is the radius of the sphere, the constant distance from centre to surface. Cut that sphere exactly through its centre and you get a hemisphere with two genuinely different surfaces: the curved dome (half the sphere's original surface) and a flat circular base of radius $r$ sealing the cut.

\text{CSA}_{\text{hemi}} = 2\pi r^2, \qquad \text{TSA}_{\text{hemi}} = 2\pi r^2 + \pi r^2 = 3\pi r^2, \qquad V_{\text{hemi}} = \frac{2}{3}\pi r^3

where: the extra $\pi r^2$ in the TSA is the area of the flat circular base ( $\pi r^2$ , the ordinary area of a circle of radius $r$ ), and the volume is exactly half the full sphere's volume since a hemisphere is literally half a sphere. For a hollow spherical shell, such as a thick metal ball with a cavity, bounded by an outer radius $R$ and inner radius $r$ , the volume of material used is the difference of two sphere volumes:

V_{\text{shell}} = \frac{4}{3}\pi\left(R^3 - r^3\right)

where: $R$ is the outer radius and $r$ the inner (cavity) radius, both measured from the common centre. Reading a problem correctly matters as much as the formula itself. A solid sphere or a solid hemisphere includes every bit of material, so "total surface area" means all of it — for the hemisphere, curved plus flat base. An open object like a bowl, dome, or tent has nothing on the open side, so it typically needs only the curved surface, never a flat-base term for a surface that is not actually there.

Figure

A sphere of radius r fitted exactly inside a cylinder of the same radius and height 2r touching it at the top bottom and around the equator

Derivation AArchimedes's cylinder: where the sphere formulas come from

The goal of this derivation is to reconstruct both headline sphere formulas, the total surface area $4\pi r^2$ and the volume $\dfrac{4}{3}\pi r^3$ , from nothing more than the volume and curved-surface-area formulas of an ordinary cylinder that a Class 10 student already knows, using the exact geometric relationship Archimedes discovered between a sphere and the cylinder that fits tightly around it.

Step 1. Build the circumscribing cylinder. Take a sphere of radius $r$ and enclose it in the smallest cylinder that will hold it: a cylinder of the same radius $r$ , with height $2r$ so that the sphere touches the top, the bottom, and the curved wall all the way round the middle (the equator).

Step 2. Compute the cylinder's own volume and curved surface using formulas from the previous mensuration topics. Its volume is $V_{\text{cyl}} = \pi r^2 h = \pi r^2 (2r) = 2\pi r^3$ , and its curved surface area is $\text{CSA}_{\text{cyl}} = 2\pi r h = 2\pi r(2r) = 4\pi r^2$ .

Step 3. State Archimedes's result, which links the two solids exactly. The sphere's volume is precisely two-thirds of this circumscribing cylinder's volume, and the sphere's total surface area exactly equals the cylinder's curved surface area (with no extra factor at all).

Step 4. Substitute the Step 2 values into Archimedes's relations. For the volume, $V_{\text{sphere}} = \dfrac{2}{3} \times V_{\text{cyl}} = \dfrac{2}{3} \times 2\pi r^3 = \dfrac{4}{3}\pi r^3$ . For the surface area, $\text{TSA}_{\text{sphere}} = \text{CSA}_{\text{cyl}} = 4\pi r^2$ directly — both of the standard sphere formulas are recovered without any calculus.

Step 5. Extend to the hemisphere by halving. Since a hemisphere is exactly half a sphere sliced through the centre, its volume is half of $\dfrac{4}{3}\pi r^3$ , giving $\dfrac{2}{3}\pi r^3$ , and its curved surface is half of $4\pi r^2$ , giving $2\pi r^2$ ; adding the flat circular base $\pi r^2$ gives the hemisphere's total surface area, $3\pi r^2$ .

Worked example

Find the total surface area and volume of a solid sphere of radius $10.5\text{ cm}$ . Take $\pi = \dfrac{22}{7}$ .

Step 1. Apply the TSA formula. $\text{TSA} = 4\pi r^2 = 4 \times \dfrac{22}{7} \times 10.5 \times 10.5 = 4 \times 22 \times 1.5 \times 10.5 = 88 \times 15.75 = 1386\text{ cm}^2$ .

Step 2. Apply the volume formula. $V = \dfrac{4}{3}\pi r^3 = \dfrac{4}{3} \times \dfrac{22}{7} \times 10.5 \times 10.5 \times 10.5 = \dfrac{4}{3} \times 22 \times 1.5 \times 165.375 = \dfrac{4}{3} \times 22 \times 165.375$ .

Step 3. Simplify carefully. $\dfrac{22}{7} \times 10.5 = 33$ , so $V = \dfrac{4}{3} \times 33 \times 10.5 \times 10.5 = \dfrac{4}{3} \times 33 \times 110.25 = \dfrac{4}{3} \times 3638.25 = 4851\text{ cm}^3$ .

The sphere has total surface area $1386\text{ cm}^2$ and volume $4851\text{ cm}^3$ .

Competitive stretch

A hemispherical bowl of internal radius $9\text{ cm}$ is full of liquid, and the liquid is poured into a cylindrical vessel of radius $6\text{ cm}$ . Find the height to which the liquid rises.

Step 1. Since the liquid is only transferred, not created or destroyed, its volume is conserved: volume of the hemisphere equals volume of liquid in the cylinder, $\dfrac{2}{3}\pi r_{\text{hemi}}^3 = \pi r_{\text{cyl}}^2 h$ .

Step 2. Substitute the given radii and cancel $\pi$ . $\dfrac{2}{3} \times 9^3 = 6^2 \times h \;\Rightarrow\; \dfrac{2}{3} \times 729 = 36h \;\Rightarrow\; 486 = 36h$ .

Step 3. Solve for $h$ . $h = \dfrac{486}{36} = 13.5\text{ cm}$ . The liquid rises to a height of $13.5\text{ cm}$ in the cylinder.

Common traps

Using the full sphere's volume $\dfrac{4}{3}\pi r^3$ for a hemisphere instead of halving it to $\dfrac{2}{3}\pi r^3$ .
Forgetting the flat circular base when a "solid hemisphere" asks for total surface area, writing $2\pi r^2$ (curved only) instead of $3\pi r^2$ .
Adding a flat-base area for a hollow bowl or dome that has no solid base to include — check whether the object is solid or an open shell before choosing CSA or TSA.
Confusing radius with diameter: a sphere described as "14 cm in diameter" has $r = 7\text{ cm}$ , and skipping the halving throws the volume off by a factor of $8$ since $r$ is cubed.

Ask the tutor: Want to work through a melting-and-recasting problem or a combined solid — like a cone topped with a hemisphere — step by step?

Why does a sphere have only a TSA and no separate CSA?

CSA and TSA are different measures only when a solid has distinct flat and curved parts, like a cylinder's two circular ends plus its curved wall. A sphere's boundary is one continuous smooth surface with no flat part anywhere, so there is nothing to separate out — the total surface area, $4\pi r^2$ , is the whole story. The moment you cut it into a hemisphere, a genuine flat base appears, and CSA and TSA become different numbers.

How do I know if a hemisphere problem wants CSA or TSA?

Read what the object actually is. A solid hemisphere, like a paperweight or a dome sitting on solid ground, has material everywhere including the flat circular face, so it needs the TSA, $3\pi r^2$ . An open object like a bowl, a tent, or a mixing bowl has nothing filling the flat circular opening, so only the curved surface is physically present, and you use the CSA, $2\pi r^2$ . When in doubt, ask whether you could actually touch a flat surface on that face.

Does Archimedes's cylinder trick work for any curved solid, or just the sphere?

It is special to the sphere. Archimedes proved that the volume ratio sphere to circumscribing cylinder is always exactly $2:3$ , and the sphere's TSA exactly equals the cylinder's curved surface area, regardless of the radius — it is a fixed, radius-independent relationship unique to this pairing of shapes. A cone inscribed in the same cylinder has a different ratio ( $1:3$ for volume), so the trick does not carry over automatically; each solid pair needs its own proof, and this one happens to be unusually clean, which is why Archimedes asked for a sphere-and-cylinder diagram on his own tombstone.

Sphere and hemisphere

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads