TSA of hemisphere = 3πr². The curved surface area is 2πr², plus the circular base πr².

Total Surface Area of a Hemisphere — Formula and Example

Question

A hemisphere has a radius of 7 cm. Find its total surface area.

(Take π = 22/7)

Solution — Step by Step

A solid hemisphere has two surfaces: the curved dome on top, and the flat circular base at the bottom. The TSA includes both. Students who only calculate the curved part lose half the marks — literally.

TSA = 2\pi r^2 + \pi r^2 = 3\pi r^2

The $2\pi r^2$ is the curved surface area (exactly half the sphere’s surface area of $4\pi r^2$ ). The $\pi r^2$ is the flat circular base.

TSA = 3 \times \frac{22}{7} \times 7^2

= 3 \times \frac{22}{7} \times 49

= 3 \times 22 \times 7

= 3 \times 154 = 462 \text{ cm}^2

Total Surface Area = 462 cm²

Why This Works

A sphere of radius $r$ has surface area $4\pi r^2$ . When we cut it exactly in half, the dome portion is $\frac{1}{2} \times 4\pi r^2 = 2\pi r^2$ . But now we’ve exposed a circular cross-section — that circle has area $\pi r^2$ .

So the hemisphere is like buying a bowl: you’re paying for the outside of the bowl ( $2\pi r^2$ ) plus the rim/base that seals it ( $\pi r^2$ ). Total: $3\pi r^2$ .

This formula is a high-frequency NCERT question. CBSE Class 9 and 10 boards both test it — sometimes directly, sometimes buried inside a composite solid problem.

Alternative Method

You can also arrive at the answer without memorising the combined formula. Calculate each part separately:

Curved Surface Area = $2\pi r^2 = 2 \times \frac{22}{7} \times 49 = 308 \text{ cm}^2$

Base Circle Area = $\pi r^2 = \frac{22}{7} \times 49 = 154 \text{ cm}^2$

TSA = $308 + 154 = \mathbf{462 \text{ cm}^2}$

Same answer, and this method makes it crystal clear what you’re actually adding. In exams where partial marks are given, showing the two components separately often earns you more credit.

If a question asks for CSA (curved surface area) only, stop at $2\pi r^2$ . Lots of questions ask for CSA of a hemispherical bowl — because a bowl has no base, so you don’t include $\pi r^2$ .

Common Mistake

The most common error is using $2\pi r^2$ as the TSA — forgetting to add the circular base. This happens because students confuse the formula for a hollow hemispherical shell (like a bowl open at the top) with a solid hemisphere. For a solid hemisphere, the base is a closed circle and must be counted. Check the word “solid” or “closed” in the problem — if it’s there, always use $3\pi r^2$ .

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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