Big picture
Surface areas and volumes studies the material needed to cover solids and the space they occupy or hold. A container's outside paint, inside lining and capacity are different quantities. A toy assembled from a cone and hemisphere also has hidden joining faces that do not belong to its exposed surface.
The main strategy is to identify the component solids, mark exposed surfaces and calculate volume separately. Before substituting numbers, decide whether dimensions are internal or external and convert all lengths to one unit. A clear sectional sketch often prevents more mistakes than a long calculation can correct afterwards.
Learning goals
Recall the surface and volume formulas for standard solids.
Distinguish curved, lateral, total and exposed surface areas.
Calculate cone slant height from vertical height and radius.
Add and subtract component volumes in composite shapes.
Solve capacity, recasting, displacement and rate problems.
Verify conservation assumptions, units and plausible results.
Core ideas
1. Standard solids and dimensions
A surface area measures covering and uses square units. Volume measures occupied space and uses cubic units. Capacity normally means the usable internal volume of a container. External volume includes wall material if the container has thickness.
Solid | Curved or lateral surface | Total surface | Volume |
|---|---|---|---|
Cube, edge a | 4a² lateral | 6a² | a³ |
Cuboid, length L, breadth b, height h | 2h(L + b) | 2(Lb + bh + hL) | Lbh |
Cylinder, radius r, height h | 2πrh | 2πr(h + r) | πr²h |
Cone, radius r, height h, slant l | πrl | πr(l + r) | ⅓πr²h |
Sphere, radius r | 4πr² | 4πr² | ⁴⁄₃πr³ |
Hemisphere, radius r | 2πr² | 3πr² | ⅔πr³ |
For a right circular cone, l = √(r² + h²). The vertical height runs from apex to base centre; the slant height runs to the base circumference. The word “right” means the apex is vertically above the base centre.
Worked example. A cuboid is 12 cm by 8 cm by 5 cm. Volume = 12 × 8 × 5 = 480 cm³. Total area = 2(96 + 40 + 60) = 392 cm². For a cube of edge 4 cm, volume = 64 cm³ and total area = 96 cm². The different units show why these results cannot be interchanged.
2. Open and closed surfaces
Curved surface area excludes flat bases. Total surface area includes every surface of an isolated solid. Exposed area counts only the surfaces visible or requiring treatment in a particular assembly. An open cylinder has one base rather than two; a tent without a floor may have no base area to include.
Worked example. A cylindrical container of radius 4 cm and height 10 cm is open at the top. Curved area = 2π × 4 × 10 = 80π cm². Bottom area = π × 4² = 16π cm². Material area, ignoring thickness and seams, = 96π cm². A closed cylinder of these dimensions has 112π cm². Capacity = π × 16 × 10 = 160π cm³.
At ₹2 per cm², covering the open container's counted area costs 2 × 96π = ₹192π. A rate per square unit must multiply area, not capacity. If the question asks for inner painting, use inner dimensions rather than automatically copying the external radius.
3. Cones, spheres and hemispheres
Worked cone example. A cone has radius 5 cm and vertical height 12 cm. Slant height = √(25 + 144) = 13 cm. Curved area = π × 5 × 13 = 65π cm². Total area = 65π + 25π = 90π cm². Volume = ⅓ × π × 25 × 12 = 100π cm³. Using 12 in πrl would wrongly substitute vertical height for slant height.
Worked sphere example. Radius = 3 cm. Surface area = 4π × 9 = 36π cm². Volume = ⁴⁄₃ × π × 27 = 36π cm³. The matching numerical coefficients here are coincidental; the dimensions and meanings remain different.
For a hemisphere of radius 3 cm, curved area = 18π cm², total area = 27π cm², and volume = 18π cm³. Total area includes the flat circular face. A hemispherical bowl's inner curved surface and a solid hemisphere's complete surface are different requests.
4. Joining solids: count shared faces correctly
When two solids join along complete circular faces, the joining faces become internal. Volumes add if the components overlap only at their boundary. Surface areas require subtracting the hidden faces rather than blindly adding both total areas.
Worked example. A toy consists of a cone mounted on a hemisphere, both radius 3 cm. The cone's vertical height is 4 cm, so slant height = √(9 + 16) = 5 cm. Exposed area = cone curved area + hemisphere curved area = π × 3 × 5 + 2π × 9 = 33π cm². The common circular base is not exposed.
Cone volume = ⅓ × π × 9 × 4 = 12π cm³. Hemisphere volume = ⅔ × π × 27 = 18π cm³. Total volume = 30π cm³. If one added both component total areas, the two copies of the shared base would need removal; each has area 9π cm².
5. Hollowing and subtracting volumes
A cavity removes volume, but its inside wall may add exposed surface. The volume of the remaining material is outer solid volume minus cavity volume. The surface calculation must reflect whether cavity openings remove flat regions or expose an annular rim.
Worked example. A cylinder of radius 5 cm and height 12 cm has a conical cavity of radius 5 cm and height 12 cm removed from its top. Cone slant height = 13 cm. Cylinder volume = 300π cm³. Removed cone volume = 100π cm³. Remaining material volume = 200π cm³.
Exposed area consists of the outside cylinder, bottom circle and inside conical wall: 2π × 5 × 12 + 25π + π × 5 × 13 = 120π + 25π + 65π = 210π cm². There is no top disk left because the cone opening covers the whole top. If the cavity radius were smaller, an annular top rim would remain and require a separate term.
6. Capacity and unit conversions
1 litre = 1,000 cm³, 1 m³ = 1,000 litres, and 1 cm³ = 1 mL. Since 1 m = 100 cm, squaring gives 1 m² = 10,000 cm² and cubing gives 1 m³ = 10,00,000 cm³. Do not apply a linear conversion factor directly to volume.
Worked example. A tank has internal dimensions 1.5 m × 0.8 m × 0.6 m. Capacity = 1.5 × 0.8 × 0.6 = 0.72 m³. Convert to litres: 0.72 × 1,000 = 720 litres. If initially filled to three-quarters capacity, water volume = ¾ × 720 = 540 litres. Empty space is 720 − 540 = 180 litres.
Dimensions labelled outside cannot determine exact capacity without wall thickness or inner dimensions. In school problems that neglect thickness, state that assumption rather than silently mixing internal and external measurements.
7. Recasting and conservation of volume
When a solid is melted and reshaped with no material loss, initial volume = total final volume. Surface area is not conserved. The new shape can have a much larger combined area even though the material volume is unchanged.
Worked example. A metal sphere of radius 6 cm is recast into spheres of radius 2 cm. Initial volume = ⁴⁄₃π × 6³. Each smaller volume = ⁴⁄₃π × 2³. Number = 6³/2³ = 216/8 = 27. This is a cubic ratio, not the radius ratio of three.
Different-shape example. A cylinder of radius 3 cm and height 8 cm is melted into cones of radius 3 cm and height 2 cm. Cylinder volume = 72π cm³. Each cone volume = ⅓ × π × 9 × 2 = 6π cm³. Number = 72π/6π = 12 cones. An exact whole-number answer is expected when all material forms identical complete objects without loss.
8. Displacement, flow and practical checks
A fully submerged solid displaces water equal to its volume, provided water does not enter it. In a uniform cylinder, water rise equals displaced volume divided by cross-sectional area. The vessel must have sufficient space and the solid must actually be submerged for this model to apply.
Worked example. A sphere of radius 3 cm is fully submerged in a cylindrical vessel of radius 6 cm. Sphere volume = 36π cm³. Vessel base area = 36π cm². Rise = 36π ÷ 36π = 1 cm. Dividing volume by area correctly produces a length.
Flow example. A pipe with internal radius 1 cm carries water at 50 cm/s for 60 s. Water advances 50 × 60 = 3,000 cm along the pipe. Volume = π × 1² × 3,000 = 3,000π cm³, or 3π litres. In general, volume = cross-sectional area × speed × time when flow is steady and the pipe is full.
The commonly taught frustum is outside this lesson's main rationalised Class 10 focus; confirm any additional school requirement with a teacher. For the solids covered here, end every solution by checking whether the required quantity is a surface, material volume, capacity, count or rise. Correct formulas cannot rescue a calculation of the wrong quantity.
A reliable approach to questions
Sketch component solids and label all radii and heights.
Identify internal or external dimensions and convert units.
Mark joining faces, openings and surfaces requiring treatment.
Calculate slant heights before cone areas where needed.
Add or subtract volumes; use conservation only when its assumptions apply.
Attach units and check capacity, whole-object counts and plausibility.
Connections that matter
Areas related to circles supplies bases, annuli and curved-boundary reasoning.
Pythagoras connects a cone's radius, vertical height and slant height.
Proportion and unit conversion support recasting, pricing and steady-flow calculations.
Common mistakes
Using cone height in curved area: calculate slant height first.
Counting joining disks as exposed: remove shared internal faces.
Conserving surface area in recasting: conserve material volume instead.
Converting cm³ with a linear factor: cube the length conversion or use litre relations.
Revision checklist
I can use standard solid formulas with consistent units.
I can distinguish open, closed, curved and exposed surfaces.
I can calculate composite and hollow-solid measurements.
I can convert internal volume into capacity.
I can solve recasting, displacement and steady-flow problems.
I can state the assumptions supporting each volume equation.