Maths · Suggested Grade 10
Sphere and Hemisphere: Surface Area & Volume | Class 10
A hemispherical bowl has a curved surface and an open circular rim. Adding a flat circular face is correct for a closed solid hemisphere—but not for every bowl question.
Original explanations and practice by BananaLearn · Published 1 November 2026
On a small screen, swipe across the illustration to read every label.
Name the object and the measured quantity
A sphere is the surface of a ball; in school volume problems a solid sphere means the enclosed solid. A hemisphere is half a sphere cut through its centre. Radius r runs from the centre to the surface; diameter d = 2r. Surface area uses square units, and volume uses cubic units.
Before choosing a formula, decide whether the question asks for a surface to cover, a capacity to hold, or a solid amount of material. An opening does not need a circular cover unless the question explicitly supplies one. Our diagrams are idealised geometry, not measurements of a manufactured object.
Sphere formulas and a worked example
A sphere has surface area 4πr² and volume (4/3)πr³. For radius 3 cm, surface area = 36π cm² and volume = 36π cm³. The numerical coefficients happen to agree here, but the units and meanings differ. Use the value of π specified by the question or leave an exact multiple of π.
If the diameter is 14 cm, the radius is 7 cm. Surface area is 196π cm², or 616 cm² when π = 22/7. Substituting 14 as r would multiply surface area by four. Write r = d/2 as a visible first step.
Hemisphere: curved area versus total area
Halving the sphere’s curved surface gives 2πr². A closed solid hemisphere also has a circular flat face of area πr², so its total surface area is 3πr². Its volume is half the sphere’s volume: (2/3)πr³. Cutting creates a new face; it does not create more volume.
For radius 7 cm with π = 22/7, curved area is 308 cm² and total area is 462 cm². The difference, 154 cm², is the flat circle. A thin ideal open bowl’s inner curved area uses its inner radius; real wall thickness is not automatically zero.
Worked question: paint or fill a bowl?
An ideal hemispherical bowl has inner radius 3 cm. Its inner curved area is 2π × 9 = 18π cm². Its capacity is (2/3)π × 27 = 18π cm³. At π ≈ 3.14 the capacity is about 56.52 cm³, equivalent to 56.52 mL. These quantities answer different questions despite the matching coefficients.
If asked to paint the outer curved surface, an outer radius is needed when thickness matters. If asked for the material volume of a hemispherical shell with outer radius R and inner radius r, subtract capacities: (2/3)π(R³ − r³). Do not subtract surface areas to find a volume.
Scaling gives a useful prediction
When every radius doubles, surface area multiplies by 2² = 4 and volume by 2³ = 8. A sphere of radius 6 cm therefore has four times the area and eight times the volume of one with radius 3 cm. This helps detect errors before calculating exact values.
For similar hemispheres, the same scaling rules apply separately to curved area, total area and volume. Doubling the radius does not merely double capacity. If a task compares volumes, use the cube of the scale factor rather than the square.
Joining solids changes exposed surface area
Two hemispheres joined along matching flat circles form a sphere. Their volumes add, but the joined flat circles are internal and do not contribute to the outside surface area. Adding two total hemisphere areas would count these hidden circles. Instead add their curved areas: 2πr² + 2πr² = 4πr².
For a hemisphere on a cylinder of the same radius, identify exposed curved surfaces and any remaining base. The shared circular interface is hidden. Draw or shade the surfaces actually included before inserting dimensions; the phrase total area always belongs to a specified object.
Choose and justify the formula
Use the surface checklist to label radius, diameter, opening and any hidden join. Then choose the formula and state units. A sound explanation says why a flat circle is included or excluded, rather than only quoting 2πr² or 3πr².
Parents can vary one feature at a time: turn a closed hemisphere into an open ideal bowl, or replace a given radius with a diameter. Keep π exact until the final rounding step when possible. None of these geometric examples establishes the capacity of a real bowl without measured inner dimensions.
Choose the surface before the formula
Copy the record into a notebook or use it to organise your own answer. Blank spaces are for your work.
| Object or quantity | Formula | Include flat circle? |
|---|---|---|
| Sphere outside | 4πr² | No separate flat face |
| Hemisphere curved area | 2πr² | No |
| Closed hemisphere total area | 3πr² | Yes |
| Hemisphere volume | (2/3)πr³ | Volume, not a surface |
| Your question | ___ | ___ |
12 questions with explained answers
Try the questions before opening the answers. Explain the reason, then try a changed example to check your understanding.
Sphere radius 3 cm: area?
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36π cm²
4π × 3².Same sphere: volume?
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36π cm³
(4/3)π × 3³.Sphere diameter 14 cm: radius?
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7 cm
Divide diameter by 2.Radius 7 sphere: area using π = 22/7?
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616 cm²
4 × 22/7 × 49.Radius 7 hemisphere: curved area with π = 22/7?
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308 cm²
2 × 22/7 × 49.Same closed hemisphere: total area?
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462 cm²
Add the 154 cm² flat circular face.Ideal inner hemispherical bowl radius 3 cm: capacity?
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18π cm³
(2/3)π × 27.56.52 cm³ equals how many mL?
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56.52 mL
1 cm³ = 1 mL.Radius doubles: area factor?
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4
Area scales with r².Radius doubles: volume factor?
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8
Volume scales with r³.Do two joined hemispheres expose their shared flat faces?
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No
The common faces become internal.Hemispherical shell R = 4, r = 3: material volume?
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(74/3)π cubic units
(2/3)π(64 − 27) = 74π/3.
Keep the ideas connected
- Compare cylinder surfaces and volume
- Choose the right height for a cone
- Keep area and volume units clear
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The examples, diagrams and practice are original BananaLearn work. Grade labels suggest a learning level; teaching sequences and school requirements can differ. Official sources below support the stated facts and resource links. These activities are independent resources, not official exam questions or endorsed campaign materials.
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