Surface area and volume of all above 3D shapes: Worked examples

This page is the practice ground for the parent topic. Before we solve, we map out every kind of problem this topic can throw at you, then hit each one on purpose. If a scenario exists, it lives in the table below and gets solved.

The scenario matrix

Cell	Scenario class	What makes it tricky	Solved in
A	Direct plug-in	Nothing — build confidence	Ex 1
B	Diameter given, not radius	Must halve first (Mistake 1)	Ex 2
C	Slant vs vertical height	Must pick the right length (Mistake 2)	Ex 3
D	Composite solid (glued shapes)	Add volumes, but subtract hidden faces	Ex 4
E	Degenerate / limiting input ( $h \to 0$ , $r \to 0$ )	Shape collapses — does formula still make sense?	Ex 5
F	Reverse problem (given SA or V, find a length)	Solve an equation, not just substitute	Ex 6
G	Unit change / scaling	Volume scales as length³, area as length²	Ex 7
H	Real-world word problem	Translate English into a shape	Ex 8
I	Exam twist (hollow / two radii, volume and surface)	Outer minus inner, and two walls	Ex 9

Every numeric answer below is machine-checked. Let's go.

The tools you'll reuse

Recall Formulas we lean on (from the parent)

Cube: $V = a^{3}$ , $S A = 6 a^{2}$
Cylinder: $V = π r^{2} h$ , $T S A = 2 π r (r + h)$ , curved $C S A = 2 π r h$
Cone: $V = \frac{1}{3} π r^{2} h$ , $T S A = π r (r + l)$ , $l = r^{2} + h^{2}$
Sphere: $V = \frac{4}{3} π r^{3}$ , $S A = 4 π r^{2}$
Hemisphere: $V = \frac{2}{3} π r^{3}$ , $T S A = 3 π r^{2}$

Here, $r$ = radius (centre to edge), $h$ = the vertical height (straight up), and $l$ = the slant height (along the sloped surface). Keep those three words apart in your head — most errors are a swap of two of them.

Cell A — Direct plug-in

Cell B — Diameter given, not radius

Figure — Surface area and volume of all above 3D shapes

Worked example Ex 2 · The diameter trap (sphere)

A ball has diameter $14 cm$ . Find its surface area. Take $π = \frac{22}{7}$ .

Forecast: If you blindly plug $14$ into $4 π r^{2}$ , your answer will be four times too big. Why four? Because $r$ is squared, and you'd have doubled $r$ . Watch for it.

Step 1 — Convert diameter to radius. $r = \frac{d}{2} = \frac{14}{2} = 7 cm$ . Why this step? Every formula uses $r$ , the centre-to-edge distance; but we were handed the diameter $d$ (across the whole circle). Since $d = 2 r$ , we halve it. Look at the red radius in the figure — it is half the black diameter.

Step 2 — Apply the surface-area formula. $S A = 4 π r^{2} = 4 \times \frac{22}{7} \times 49$ . Why this step? $S A = 4 π r^{2}$ for a sphere; now $r$ is correct, so substitution is safe.

Step 3 — Arithmetic. $\frac{22}{7} \times 49 = 22 \times 7 = 154$ , then $4 \times 154 = 616 cm^{2}$ . Why this step? We chose $π = \frac{22}{7}$ precisely because $49$ is a multiple of $7$ , so the $7$ cancels and leaves clean whole numbers instead of a messy decimal.

Verify: Had we wrongly used $r = 14$ : $4 π (196) = 2464$ , exactly $4 \times 616$ . That confirms the "square doubles into ×4" warning. Units: $cm^{2}$ ✓.

Cell C — Slant height vs vertical height

Worked example Ex 3 · Which height? (cone)

A cone has base radius $r = 9 cm$ and vertical height $h = 12 cm$ . Find its total surface area.

Forecast: The curved-surface part needs the slant height $l$ , not $h$ . If you use $h = 12$ in $π r l$ , you'll under-count the sloped skin. Predict whether $l$ is bigger or smaller than $h$ — then check.

Step 1 — Find the slant height with the Pythagorean theorem. $l = r^{2} + h^{2} = 9^{2} + 1 2^{2} = 81 + 144 = 225 = 15 cm .$ Why this step? Drop straight down from the apex: $r$ , $h$ , and $l$ form a right triangle where $l$ is the hypotenuse (the longest, sloped side). See the red slant in the figure hugging the outside. Because the hypotenuse is always the longest side, $l = 15 > h = 12$ — your forecast should have said "bigger." More on this in Pythagorean Theorem.

Step 2 — Total surface area. $T S A = π r (r + l) = π \times 9 \times (9 + 15) = π \times 9 \times 24 = 216 π cm^{2}$ . Why this step? $T S A$ = base circle $π r^{2}$ plus curved surface $π r l$ , which factors to $π r (r + l)$ .

Step 3 — Decimal. $216 π \approx 678.6 cm^{2}$ . Why this step? The exact form $216 π$ is best for further algebra, but a decimal tells us the physical size ("about $679$ square centimetres of surface"), so we convert at the very end.

Verify: If we had wrongly used $h = 12$ as slant: $π \times 9 \times (9 + 12) = 189 π$ , smaller than $216 π$ . Under-counting confirmed. Units $cm^{2}$ ✓.

Slant vs height

The slant

l = r^{2} + h^{2}

is the hypotenuse, always longer than the vertical

h

; use

l

for curved surface,

h

for volume.

Cell D — Composite solid

The figure below shows the toy in side view: a black cone sitting on a black hemisphere, with the red dashed circle marking exactly where they meet. That red circle is the pedagogical star of this example — it is the surface that disappears into the join and must NOT be painted. Keep your eye on it through Step 5.

Worked example Ex 4 · Ice-cream shape (cone + hemisphere)

A toy is a cone mounted on a hemisphere. Both share radius $r = 6 cm$ . The cone's vertical height is $h = 8 cm$ . Find the total volume and the outer surface area (the flat circle where they join is inside, so it is not painted).

Forecast: For volume we simply add. For surface area, the tricky bit: the cone's base circle and the hemisphere's flat circle both vanish into the join — neither gets counted. Guess: will the two circular bases cancel exactly? (They will — same $r$ .)

Step 1 — Cone volume. $V_{cone} = \frac{1}{3} π r^{2} h = \frac{1}{3} π (36) (8) = 96 π$ . Why this step? A cone is one-third of the cylinder with the same base and height.

Step 2 — Hemisphere volume. $V_{hemi} = \frac{2}{3} π r^{3} = \frac{2}{3} π (216) = 144 π$ . Why this step? Half of the sphere's $\frac{4}{3} π r^{3}$ .

Step 3 — Total volume. $V = 96 π + 144 π = 240 π \approx 754.0 cm^{3}$ . Why this step? The solids fill separate regions of space, so volumes simply add.

Step 4 — Slant height for the cone's curved surface. $l = 6^{2} + 8^{2} = 36 + 64 = 100 = 10 cm$ . Why this step? The cone's curved surface uses the slant height $l$ , not the vertical $h$ (same trap as Cell C). Since only $r$ and $h$ are given, we rebuild $l$ from the right triangle with the Pythagorean theorem before we can find any curved area.

Step 5 — Outer surface = cone's curved surface + hemisphere's curved surface. $S A = π r l + 2 π r^{2} = π (6) (10) + 2 π (36) = 60 π + 72 π = 132 π \approx 414.7 cm^{2} .$ Why this step? We drop both flat circles because they sit hidden at the join (the red dashed circle in the figure). Cone base $π r^{2}$ and hemisphere base $π r^{2}$ are both excluded.

Verify: If someone forgot and added the two hidden circles, they'd add $2 π r^{2} = 72 π$ of paint that doesn't exist. Excluding them is the whole point of a "composite" question. Units: $cm^{3}$ and $cm^{2}$ ✓. See Real-World Applications.

Cell E — Degenerate / limiting inputs

The next figure has two panels of the same idea from two directions: on the left, a can loses height until it flattens onto a single red disc ( $h \to 0$ ); on the right, a can loses radius until it collapses onto a single red vertical line ( $r \to 0$ ). Both red objects are what remains after the collapse — watch how each has no interior volume.

Worked example Ex 5 · What happens as a dimension shrinks to zero? (both directions)

Take a cylinder of radius $r = 5 cm$ . (a) Compute its volume for heights $h = 4, 1,$ and $0 cm$ . (b) Now fix height $h = 4 cm$ and shrink the radius: $r = 5, 1,$ and $0 cm$ . Describe what the shape becomes in each case.

Forecast: As $h \to 0$ the can gets flatter; as $r \to 0$ the can gets thinner (a needle). In both directions, does volume go to a positive number, or to zero? Guess before computing.

Step 1 — Height collapse, $h = 4$ : $V = π (25) (4) = 100 π \approx 314.2 cm^{3}$ . Why this step? This is our full-size reference can; we compute it first so we have a baseline to watch shrink as we lower the height in the next steps.

Step 2 — $h = 1$ : $V = π (25) (1) = 25 π \approx 78.5 cm^{3}$ . Why this step? Cutting the height from $4$ to $1$ (a quarter) cuts the volume to a quarter too — because volume is linear in height. Seeing this middle value proves the shrink is smooth, not a sudden jump.

Step 3 — $h = 0$ : $V = π (25) (0) = 0 cm^{3}$ . Why this step? Volume is base area × height. Height zero means no stacking at all — the solid degenerates into a flat disc, which has area but no volume. The left panel shows the can flattening onto the red circle.

Step 4 — Radial collapse, $r = 5$ : $V = π (25) (4) = 100 π$ ; $r = 1$ : $V = π (1) (4) = 4 π \approx 12.6 cm^{3}$ ; $r = 0$ : $V = π (0) (4) = 0 cm^{3}$ . Why this step? Now the base circle shrinks. When $r = 0$ the base has no area, so no matter how tall the can is, there is nothing to stack — it degenerates into a red vertical line segment (a "needle") with no volume. The right panel shows this. Notice the drop is faster than for height ( $r$ is squared), which is why $r = 1$ gives only $4 π$ while $h = 1$ gave $25 π$ .

Step 5 — Cross-check the surface areas at the two limits.

$h \to 0$ : curved $C S A = 2 π r h \to 0$ , but the two end-circles remain: $T S A = 2 π r (r + h) \to 2 π r^{2} = 50 π \approx 157.1 cm^{2}$ .
$r \to 0$ : every term has a factor of $r$ , so $T S A = 2 π r (r + h) \to 0$ . A needle has essentially no surface either. Why this step? A degenerate input should always give a sensible limit, never nonsense. Notice the beautiful asymmetry: a flattened can keeps its two circles ( $50 π$ ), but a needle keeps nothing ( $0$ ) — because $r = 0$ kills area, while $h = 0$ only kills the wall.

Verify: Height trend $314.2 \to 78.5 \to 0$ and radius trend $314.2 \to 12.6 \to 0$ both fall smoothly to zero, matching "collapsed solid has no interior." Both edge cases give finite, meaningful limits — that's the test we're teaching here.

Cell F — Reverse problem (given the answer, find a length)

Worked example Ex 6 · Work backwards (sphere volume → radius)

A spherical balloon has volume $V = 288 π cm^{3}$ . Find its radius, then its surface area.

Forecast: This is an equation to solve, not a formula to plug into. We'll isolate $r^{3}$ , then take a cube root. Guess: will $r$ be a whole number? (It's designed to be.)

Step 1 — Set up the equation. $\frac{4}{3} π r^{3} = 288 π$ . Why this step? We know the output (volume) and want the input (radius), so we write the formula and treat $r$ as the unknown.

Step 2 — Cancel $π$ and isolate $r^{3}$ . $\frac{4}{3} r^{3} = 288 ⟹ r^{3} = 288 \times \frac{3}{4} = 216.$ Why this step? Dividing both sides by $π$ removes it cleanly; multiplying by $\frac{3}{4}$ undoes the $\frac{4}{3}$ .

Step 3 — Cube root. $r = 3216 = 6 cm$ . Why this step? Cube root undoes the cube: we want the length that, cubed, gives $216$ . Since $6^{3} = 216$ , $r = 6$ .

Step 4 — Surface area. $S A = 4 π r^{2} = 4 π (36) = 144 π \approx 452.4 cm^{2}$ . Why this step? With $r$ now known we substitute into the sphere's surface-area formula to answer the second half.

Verify: Plug $r = 6$ back into volume: $\frac{4}{3} π (216) = 288 π$ ✓ — matches the given. That round-trip is the whole point of a reverse problem.

Cell G — Scaling / unit change

Worked example Ex 7 · Doubling every side (scaling law)

A cube has edge $a = 3 cm$ . A second cube has double the edge, $a^{'} = 6 cm$ . How many times bigger are its surface area and volume?

Forecast: Guess the two multipliers. Most people say "double" for both — wrong. Area and volume scale differently.

Step 1 — Small cube. $V = 27$ , $S A = 6 (9) = 54$ . Why this step? We compute the original's volume and area so we have baseline numbers to compare against.

Step 2 — Big cube. $V^{'} = 216$ , $S A^{'} = 6 (36) = 216$ . Why this step? Same formulas with the doubled edge $a^{'} = 6$ , giving the enlarged solid's numbers.

Step 3 — Ratios. $\frac{V ^{'}}{V} = \frac{216}{27} = 8 = 2^{3}, \frac{S A ^{'}}{S A} = \frac{216}{54} = 4 = 2^{2} .$ Why this step? Volume is a product of three lengths, so scaling each by $2$ multiplies $V$ by $2^{3} = 8$ . Surface area is a product of two lengths, so it scales by $2^{2} = 4$ . This is the general rule for any Similar Solids: linear scale $k$ gives area $\times k^{2}$ , volume $\times k^{3}$ .

Verify: $2^{3} = 8$ ✓ and $2^{2} = 4$ ✓. The exponents equal the dimension of the quantity — the deep reason lives in Dimensional Analysis.

Linear scale

k

area multiplies by

k^{2}

, volume by

k^{3}

Cell H — Real-world word problem

Worked example Ex 8 · Painting a water tank

A cylindrical water tank (no lid, closed base) has radius $r = 7 m$ and height $h = 3 m$ . Paint covers $10 m^{2}$ per litre. How many litres are needed for the outside curved wall plus the bottom? Use $π = \frac{22}{7}$ .

Forecast: "No lid" means we skip the top circle. Which surfaces stay? Curved wall + one bottom circle. Guess whether we need more or fewer litres than a fully closed tank.

Step 1 — Translate English into surfaces. Paint area $= wall (C S A) 2 π r h + bottom π r^{2}$ . Why this step? "Outside curved wall" is the curved surface $2 π r h$ ; "bottom" is one circle $π r^{2}$ ; "no lid" drops the top circle. Turning words into the right pieces is the whole skill of a word problem.

Step 2 — Wall. $2 \times \frac{22}{7} \times 7 \times 3 = 2 \times 22 \times 3 = 132 m^{2}$ . Why this step? We plug into $2 π r h$ ; picking $π = \frac{22}{7}$ lets the $7$ cancel the radius $7$ , leaving clean integers.

Step 3 — Bottom. $\frac{22}{7} \times 49 = 22 \times 7 = 154 m^{2}$ . Why this step? The single base circle is $π r^{2}$ ; again the $7$ divides into $49$ to keep the arithmetic exact.

Step 4 — Total area and litres. $132 + 154 = 286 m^{2}$ . Litres $= \frac{286}{10} = 28.6 L$ . Why this step? Add the two surfaces for total paintable area, then divide by coverage ( $10 m^{2}$ per litre) to convert area into litres.

Verify: A fully closed tank would need one more circle ( $154 m^{2}$ ), i.e. more paint — so "no lid" giving fewer litres matches the forecast. Units: $m^{2} \div (m^{2} / L) = L$ ✓.

Cell I — Exam twist (hollow solid, two radii)

The final figure is the pipe's cross-section — the flat slice you'd see if you sawed straight through it. The red shaded ring (an annulus) is the actual metal; the white hole in the middle is empty. Everything we compute below — both the metal volume and the two walls of paint — is read off this red ring.

Worked example Ex 9 · Hollow pipe (volume AND surface area)

A hollow cylindrical pipe has outer radius $R = 5 cm$ , inner radius $ρ = 3 cm$ , and length $h = 10 cm$ . (a) Find the volume of the metal (material only, not the hole). (b) Find the total surface area to be painted, counting the outer wall, the inner wall, and the two ring-shaped ends. Give exact multiples of $π$ .

Forecast: For volume, the metal is the big cylinder minus the empty tube inside — subtract two volumes. For surface area, a hollow pipe surprisingly has more surface than a solid one, because you must also paint the inside wall. Guess: is the painted area bigger or smaller than a solid rod's?

Step 1 — Outer solid cylinder volume. $V_{outer} = π R^{2} h = π (25) (10) = 250 π$ . Why this step? We first imagine the pipe as completely solid, then carve the hole out.

Step 2 — Inner hole cylinder volume. $V_{inner} = π ρ^{2} h = π (9) (10) = 90 π$ . Why this step? The hollow interior is itself a smaller cylinder of the same length; because it is empty, we must remove it from the solid rod.

Step 3 — Metal volume. $V = 250 π - 90 π = 160 π \approx 502.7 cm^{3}$ . Why this step? Metal = outer cylinder − inner void. Equivalently $π (R^{2} - ρ^{2}) h = π (25 - 9) (10) = 160 π$ , which reads straight off the red ring: ring area $π (R^{2} - ρ^{2})$ times length $h$ .

Step 4 — Outer wall area. $A_{out} = 2 π R h = 2 π (5) (10) = 100 π$ . Why this step? The outside is the curved surface ( $C S A$ ) of the big cylinder, $2 π R h$ .

Step 5 — Inner wall area. $A_{in} = 2 π ρ h = 2 π (3) (10) = 60 π$ . Why this step? Looking down the hole, the inside is also a curved cylinder wall, of the smaller radius $ρ$ — this is the surface a solid rod never had.

Step 6 — Two ring-shaped ends. Each end is an annulus of area $π (R^{2} - ρ^{2}) = π (25 - 9) = 16 π$ ; two of them give $2 \times 16 π = 32 π$ . Why this step? The top and bottom are not full discs but flat rings (the red cross-section itself), one at each end of the pipe.

Step 7 — Total painted area. $S A = A_{out} + A_{in} + 2 π (R^{2} - ρ^{2}) = 100 π + 60 π + 32 π = 192 π \approx 603.2 cm^{2}$ . Why this step? Sum the four surfaces: outer wall, inner wall, and the two ring ends.

Verify: Metal volume via factored form $π (R^{2} - ρ^{2}) h = π (16) (10) = 160 π$ matches Step 3 ✓. Surface: a solid rod of radius $5$ would have $2 π R h + 2 π R^{2} = 100 π + 50 π = 150 π$ ; our hollow pipe's $192 π$ is larger — matching the forecast that the inner wall adds surface. Units: $cm^{3}$ and $cm^{2}$ ✓. Relates to Density and Mass if you next want the pipe's weight.

Recall Self-test

A cone's curved surface uses which height? ::: the slant $l = r^{2} + h^{2}$ , not the vertical $h$ . When gluing a cone onto a hemisphere, which faces are NOT painted? ::: the two coinciding flat circles at the join. Doubling every edge multiplies volume by what factor? ::: $2^{3} = 8$ . A cylinder's volume as $h \to 0$ ? ::: $0$ — it degenerates to a flat disc with no interior. A cylinder's volume as $r \to 0$ ? ::: $0$ — it degenerates to a needle (a line segment) with no interior. What is the difference between $C S A$ and $T S A$ ? ::: $C S A$ is only the curved/wrapped side; $T S A$ adds the flat ends.