Chapter 10.5 — Exercise 10.4 — Sphere and Hemisphere

Four Formulas Built From One Radius

A hemisphere's curved surface area is exactly half a full sphere's, 2πr² instead of 4πr², which makes sense since it genuinely is half a sphere's outer shell. Its total surface area, though, adds the flat circular face's own area, πr², on top of that half-shell — giving 2πr² + πr² = 3πr², not simply half of the sphere's 4πr². Volume halves cleanly with no such correction needed, since slicing a sphere in half through its center produces exactly two equal halves of space with nothing left over or added. This single-radius simplicity is what makes this exercise, in some ways, the most arithmetically direct in the entire chapter — with only one measurement to track throughout, most problems here reduce to a single substitution or a single reversal, rather than the multi-step chains that a cylinder's or cone's extra dimension usually demands elsewhere.

Radius In, Both Measurements Out

A sphere with radius 3.5 cm: surface area = 4πr² = 4(22/7)(3.5)² = 154 cm², and volume = (4/3)πr³ = (4/3)(22/7)(3.5)³ ≈ 179.67 cm³. A second problem runs the identical pair of formulas in reverse: given surface area 1018 2/7 cm², solve 4πr² = 7128/7 for r² = 81 exactly, giving r = 9 cm, then substitute that same recovered radius into the volume formula to get 3052.08 cm³ — the same "recover one quantity, reuse it in the next formula" pattern seen constantly throughout this chapter's cylinder and cone exercises. What makes the sphere's version of this pattern noticeably cleaner than either of those two is that a sphere never needs a second measurement recovered from a right-triangle relationship the way a cone's slant height does — solving 4πr² for r hands back the one and only unknown the volume formula will ever need, with nothing further to untangle before substituting it straight in.

Scaling Two Spheres Against Each Other

Two spheres have radii in ratio 2:3; find the ratio of their surface areas and volumes. Surface area ratio = r₁²:r₂² = 2²:3² = 4:9. Volume ratio = r₁³:r₂³ = 2³:3³ = 8:27. This is exactly the identical scaling pattern already seen back in the cube-and-cuboid exercise, now applied to spheres instead of boxes: surface area, built from two multiplied lengths, scales with the square of the radius ratio; volume, built from three multiplied lengths, scales with the cube of it. A related balloon problem confirms this exact same rule numerically, with real measured figures rather than a plain ratio: a balloon's diameter doubling from 14 cm to 28 cm doubles its radius too, so its surface area scales by 2² = 4 — confirmed directly, since 616 cm² becomes 2464 cm², exactly four times as much. It's worth noticing that this scaling rule applies just as cleanly to a hemisphere as to a full sphere, since a hemisphere's surface area and volume formulas are both still built from powers of the same single radius — 2πr² or 3πr² still scales with r², and (2/3)πr³ still scales with r³, no matter that the constant multiplying r² or r³ has changed from the full sphere's own 4π or (4/3)π.

A Hemisphere's Total Surface Area, Directly

A hemisphere of radius 10 cm: total surface area = 3πr² = 3(3.14)(100) = 942 cm² in total. This single substitution is the most direct problem in the whole exercise — no recovery step, no ratio, no reversed formula — precisely because both the shape and the one measurement it needs are handed over already complete. A companion problem asks the same question about a hemispherical bowl instead, but with a genuinely different twist: the bowl has a real physical thickness, 0.25 cm, so its outer radius (5.25 cm) and inner radius (5 cm) are two slightly different numbers rather than one shared value, and the question asks for the ratio between the outer curved surface area and the inner curved surface area — 4π(5.25)² to 4π(5)², which simplifies to 441:400 once the common 4π factor cancels out of both sides.

Density Converts Volume Into Weight

A lead ball with diameter 2.1 cm, made from lead with density 11.34 g/cm³; find its weight. Radius = 1.05 cm exactly, so volume = (4/3)πr³ ≈ 4.851 cm³. Weight = volume × density = 4.851 × 11.34 ≈ 55 grams total. Density is the bridge this problem needs between a purely geometric quantity, volume, and a physical one, weight — multiplying cubic centimetres by grams-per-cubic-centimetre cancels the volume units entirely, leaving a plain weight in grams, the same unit-cancellation logic used whenever a geometric formula gets connected to a real physical measurement outside pure mathematics. This is really the only problem in the entire chapter that steps outside geometry proper — every other formula here answers a question about space alone, but density brings in an entirely separate physical property of the material itself, one that has nothing to do with the sphere's shape and would be completely different for a lead ball versus, say, one made of wood or aluminium of the exact same size.

Melting a Cylinder Into a Sphere

A metallic cylinder, diameter 5 cm and height 10/3 cm, is melted down and recast into a sphere; find the resulting sphere's diameter. Since melting preserves volume, πr²h = (4/3)πR³, giving (2.5)²(10/3) = (4/3)R³. Solving this out carefully, R³ = 125/8, so R = 5/2 cm, meaning the sphere's own diameter comes out to 5 cm — coincidentally identical to the cylinder's own diameter, though its own height and the resulting sphere's radius turn out to be genuinely different numbers entirely. This is the same melting-and-recasting principle from the cylinder exercise, now connecting two entirely different solid shapes rather than a cuboid and a cylinder — volume is still the one quantity guaranteed to survive the transformation unchanged, regardless of which two shapes are involved on either side of it. Across all three of this chapter's melting-and-recasting problems — cuboid into cylinder, and now cylinder into sphere — the governing equation is always "old volume formula equals new volume formula," never anything involving surface area, since only the volume, the actual amount of material present, is a physically conserved quantity through the melting process.

The Chapter's Closing Connection

Every solid across this chapter — cuboid, cylinder, cone, sphere, hemisphere — has been measured by extending a flat 2D area formula into three dimensions. Chapter 11, Areas returns to those flat figures directly, revisiting rectangles, triangles, and other plane shapes in genuinely far more depth than this chapter's brief opening review ever covered.