Grade 7-12 Math Review · Grade 11-12 · Solid Mensuration
Surface Area and Volume of a Sphere
Two formulas, no heights, no slants. The difficulty in this topic is never the sphere — it is the half-sphere and the tank built out of two solids at once.
Find the volume
A ball has radius 3 cm. Leave your answer in terms of π.
V = ⅔πr³ = ⅔π(27) = 36π cm³. Each wrong answer is one slip: 9π is πr², the flat circle through the middle rather than the solid; 27π is πr³ with the fraction dropped; and 12π comes from ⅔πr², squaring the radius where it should be cubed. Note that 4πr² also comes to 36 here, so the units are what separate a volume from a surface area.
1
Two formulas, and a way to keep them straight.
V = ⅔πr³ SA = 4πr²
Count the r’s. Volume is cubic, so r is cubed; surface is square, so r is squared.
That single check tells you which formula you have written down, without remembering which
fraction goes where.
The three solids on the right share a radius, and the cylinder’s height is 2r so the
sphere fits inside it exactly. Their volumes are in the ratio 1 : 2 : 3
— Archimedes was proud enough of this to have it carved on his tomb, and it is a
quick way to check any sphere answer against a cylinder you can compute easily.
2
A hemisphere has a flat face.
curved = ½(4πr²) = 2π(25) = 50π
half the sphere’s skin
flat = πr² = 25π
the circle you cut across
TSA = 50π + 25π = 75π cm²
≈ 235.6 cm²
Halving the sphere does not halve its surface area
Cutting a sphere in two creates a new face that was not there before. The curved part
halves to 2πr², but the flat circle πr² is added, so a solid hemisphere
has 3πr² of surface — three quarters of the whole
sphere’s, not half. A hollow bowl is a different question again: no flat face, so
2πr². Read which one is meant.
3
A composite solid is an addition problem.
cylinder = π(4)(5) = 20π m³
πr²h
hemisphere = ½(⅔π)(8) = 16π3 m³
half of ⅔πr³
V = 20π + 16π3 = 76π3 m³
≈ 79.6 m³
≈ 79 600 litres
1 m³ = 1 000 L
The radius is shared, which is what makes the two pieces fit. Take that as the signal:
wherever two solids meet, the matching dimension is given once and used twice.
4
Surface area of a composite: the join disappears.
Curved cylinder2πrhcounted in full
Curved hemisphere2πr²counted in full
Bottom of the cylinderπr²counted, if the tank has a base
The circle where they meet—not counted; it is inside the solid
Volumes add without thinking. Surface areas do not: any face that ends up buried between
the two pieces is no longer surface. Sketch the solid and ask of each face, could you
paint it?
5
Working back to a radius.
288π = ⅔πr³
π cancels on both sides
288 = ⅔r³ → r³ = 288(34) = 216
multiply by the reciprocal
r = 3√216 = 6 cm
check: ⅔π(216) = 288π ✓
Leaving π as a symbol rather than turning it into 3.14 early is what keeps this tidy.
It cancels, the cube root is exact, and no rounding has crept into the answer.
6
Variations: back from the surface, a diameter given, melting and recasting.
Still the two formulas; the radius just arrives by a different road.
A sphere has surface area 100π cm². Find its volume.
4πr² = 100π, so r² = 25
π cancels, divide by 4
r = 5 cm
a square root this time, not a cube root
V = 43 π(125) = 500π3 cm³
≈ 523.6 cm³
From an area, undo a square; from a volume, undo a cube.
A ball has diameter 14 cm. Find its volume.
r = 14 ÷ 2 = 7 cm
halve the diameter first
V = 43 π(343) = 1 372π3 cm³
≈ 1 436.8 cm³
Cubing 14 instead of 7 makes the answer 2³ = 8 times too big.
A lead sphere of radius 9 cm is melted and recast into balls of radius 3 cm. How many balls?
big: 43 π(729) = 972π
the metal available
small: 43 π(27) = 36π
metal in each ball
972π ÷ 36π = 27 balls
check: (9 ÷ 3)³ = 27 ✓
Melting keeps the volume, not the surface. A radius three times smaller means 3³ times as many balls, not 3.
On the exam
Sphere items give a radius, a diameter or one measure and ask for the other: volume from surface area, radius from volume, a hemisphere’s total surface, or how many small balls one large one makes. Halve any diameter, keep π as a symbol so it cancels, and pick the formula by counting r’s. The 10-second check: put your radius back into the formula you started from.
Now you try
Practice problem
Find the total surface area
A solid hemisphere has radius 4 cm. Leave your answer in terms of π.
Curved: 2πr² = 32π. Flat circle: πr² = 16π. Total 3πr² = 48π cm². A is the curved part only — correct for a hollow bowl, but this one is solid and has a flat top. B is the whole sphere, 4πr². D is the flat circle alone.
Hint: the word solid is doing the work. Cutting a sphere in half
exposes a circle, and that circle is part of the surface you would have to paint.
