Surface area of a sphere

A sphere is the set of all points in space at a fixed distance $r$ (the radius) from a fixed centre. Because that distance is the same in every direction, the surface has no flat patch anywhere on it — it is uniformly curved, like the outside of a perfectly round ball. This curved outer skin is the sphere's entire surface, so there is only one surface area to speak of: the total surface area (TSA).

where: $r$ is the radius of the sphere, and $\pi \approx \tfrac{22}{7}$ or $3.14$ as the problem specifies. Doubling the radius does not double the area — since $r$ is squared, doubling $r$ makes the surface area four times larger. A hemisphere is exactly half a sphere, produced by slicing straight through its centre. Cutting the ball this way exposes two very different kinds of surface: the curved half of the original sphere's skin, and a brand-new flat circular face where the cut was made (a circle of radius $r$, the same radius as the sphere).

where: the first term $2\pi r^2$ is half of the full sphere's $4\pi r^2$ (exactly half the ball's skin), and the second term $\pi r^2$ is the area of the flat circular disc left by the cut — an ordinary circle-area formula, added on because a hemisphere has one such face, not two. Which formula to use — CSA or TSA — depends entirely on whether the flat circular face is physically present in the problem. A solid hemispherical paperweight resting on a table shows both its curved dome and its flat base, so a "total surface area" question about it uses $3\pi r^2$. A hollow hemispherical bowl, painted only on the outside curved surface, uses just the curved part, $2\pi r^2$.

The goal is to show precisely why a hemisphere's total surface area is $3\pi r^2$ and not simply half the sphere's $4\pi r^2$, which is the single most common slip students make in this topic, so the derivation tracks each surface separately before adding them.

Step 1. Start from the whole sphere's curved skin, $4\pi r^2$. When the sphere is cut through its centre into two hemispheres, this curved skin is shared equally between the two halves, because the cut passes through the centre and splits the ball symmetrically.

Step 2. Halve the curved area to get one hemisphere's share. Each hemisphere keeps exactly half of $4\pi r^2$, giving the curved surface area $\text{CSA} = \tfrac{1}{2}\times 4\pi r^2 = 2\pi r^2$. Nothing has been added yet — this is purely the leftover curved skin, still bare.

Step 3. Account for the new flat face created by the cut. Slicing through the centre exposes a flat circular cross-section on each hemisphere, and this circle has the same radius $r$ as the sphere (because the cut passes through the centre, the widest point). Its area, by the ordinary circle-area formula, is $\pi r^2$.

Step 4. Add the two surfaces together for the total. $\text{TSA of hemisphere} = \text{CSA} + \text{flat face} = 2\pi r^2 + \pi r^2 = 3\pi r^2$. The flat face is added only once per hemisphere — each of the two pieces gets its own single cut face, not a shared one — which is why the coefficient is $3$, not $2$ (halving $4\pi r^2$ directly, ignoring the new face) and not $4$ (double-counting the face).

Problem: A hemispherical bowl made of steel is $0.25$ cm thick, and the inner radius of the bowl is $5$ cm. Find the total surface area of the outer surface of the bowl, treating it as a solid hemisphere of radius equal to the outer radius. (Use $\pi = \tfrac{22}{7}$.)

Step 1. Find the outer radius. The bowl's outer surface lies one thickness beyond the inner surface, so outer radius $=$ inner radius $+$ thickness $= 5 + 0.25 = 5.25$ cm.

Step 2. Write down the formula to apply. Since the problem asks for the total surface area treating the bowl as a solid hemisphere, use $\text{TSA} = 3\pi r^2$ with $r = 5.25$ cm.

Step 3. Compute the square of the radius. $(5.25)^2 = 27.5625$.

Step 4. Multiply through. $3\times\dfrac{22}{7}\times 27.5625 = \dfrac{66}{7}\times 27.5625 = 66\times 3.9375 = 259.875$.

So the outer total surface area of the bowl is about $\mathbf{259.9\ \text{cm}^2}$. Notice that the problem needed the outer radius throughout — using the given inner radius of $5$ cm directly, without adding the thickness, is the exact mistake examiners are testing for here.

• Halving $4\pi r^2$ for a hemisphere. A hemisphere's curved part is indeed half of the sphere's, $2\pi r^2$, but it also gains a brand-new flat face, so its TSA is $3\pi r^2$, not $2\pi r^2$.
• Adding the flat face twice, or not at all. A hemisphere has exactly one flat circular face, unlike a cylinder (two circular ends) or a plain sphere (none) — mixing this up is the single biggest source of error in hemisphere problems.
• Squaring the diameter instead of the radius. If a problem gives the diameter, halve it first to get $r$; forgetting to halve makes the computed area four times too large.
• Not matching "inner" and "outer" radius. In bowls, domes, or shells with a stated thickness, read carefully whether the question wants the inside surface, the outside surface, or both, and add or subtract the thickness accordingly.

Ask the tutor: Want to practice picking CSA versus TSA for hemisphere word problems, or work through a few more inner-outer radius bowl questions with the AI tutor?