Worked examples — 3D shapes — cube, cuboid, cylinder, cone, sphere, prism, pyramid

We will lean on prerequisites you already met:

• Pythagorean Theorem — to find slant heights.
• Area of 2D Shapes and Circle Geometry — because every 3D volume is a stacked 2D shape.
• Units and Measurement — because a wrong unit is a wrong answer.
• Integration — for the sphere/cone slice arguments.
• Real-World Applications — for the word problems.

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The scenario matrix

Before solving anything, let us list every kind of situation a 3D-shapes question can be. Think of it as a checklist: if we solve one example from every row, nothing on an exam can surprise us.

Cell	Scenario class	What makes it tricky	Example
A	Clean forward calc	Just plug into a formula	Ex 1 (cuboid)
B	Reverse / solve-for-input	You are given the answer (volume) and must find a dimension	Ex 2 (cube edge)
C	Composite shape	Two solids glued together — add volumes	Ex 3 (cylinder + cone silo)
D	Subtracted / hollow shape	One solid removed from another — subtract	Ex 4 (pipe = tube)
E	Degenerate / zero input	A dimension goes to 0 → shape collapses	Ex 5 (flat box, zero-radius cone)
F	Partial fill / limiting value	Container only fraction full; check limits	Ex 6 (half-full tank)
G	Real-world word problem	Extract the shape and units from a story	Ex 7 (paint a silo)
H	Exam-style twist	Scaling: double a dimension, what happens to $V$ and $SA$?	Ex 8 (scaling law)

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Cell A — Clean forward calculation

Forecast: Before reading on — guess: is the number for volume bigger or smaller than the number for surface area? (Volume is a product of 3 lengths; surface area of 3 products of 2 lengths — so it depends. Guess, then check.)

Steps:

1. Volume $V=lwh=12\times 7\times 4=336$ cm³.
   • Why this step? Volume of a cuboid is stacking a $12\times 7$ rectangular floor ($=84$ cm²) up to height 4. Look at the figure: the shaded floor slab repeated 4 times.
2. Surface area — 3 pairs of faces: $SA=2(lw+lh+wh)=2(12\cdot 7+12\cdot 4+7\cdot 4)$.
   • Why this step? The skin has a top+bottom pair, a front+back pair, a left+right pair. Each pair is two identical rectangles.
3. $=2(84+48+28)=2(160)=320$ cm².

Verify: Units are right — cm³ for space, cm² for skin. Sanity: the floor alone is 84 cm², and there are 6 faces roughly this size, so a few hundred cm² is believable. ✓

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Cell B — Reverse: given the volume, find a dimension

Forecast: We normally do $V=a^3$ forwards. Here we have $V$ and want $a$ — so we must run the machine backwards. What operation undoes "cube"?

Steps:

1. Convert units first. $216$ litres $=216\times 1000=216000$ cm³.
   • Why this step? $a$ will come out in cm, so volume must be in cm³, not litres. See Units and Measurement.
2. Undo the cube. Since $V=a^3$, we take the cube root: $a=\sqrt[3]{V}$.
   • Why this step? The cube root is the exact inverse of "raise to the 3rd power" — it answers "which number, tripled-multiplied by itself, gives $V$?"
3. $a=\sqrt[3]{216000}=60$ cm.

Verify: Plug back forwards: $60^3=216000$ cm³ $=216$ litres. ✓ Matches the target exactly.

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Cell C — Composite shape (add two solids)

Forecast: Two solids stacked. Do we add or subtract? (They occupy different space, so we... guess.)

Steps:

1. Cylinder part: $V_{\text{cyl}}=\pi r^{2}h=\pi (3^2)(8)=72\pi$ m³.
   • Why this step? The straight body is a stack of circular floors of area $\pi (3^2)=9\pi$.
2. Cone cap: $V_{\text{cone}}=\frac{1}{3}\pi r^2{h}_c=\frac{1}{3}\pi (3^2)(4)=12\pi$ m³.
   • Why this step? A cone is exactly one-third of the cylinder with the same base and height.
3. Add (they sit in separate regions of space): $V=72\pi +12\pi =84\pi \approx 263.89$ m³.
   • Why this step? Total space = space in body + space in roof. Nothing overlaps, so we sum.

Verify: Cone volume should be smaller than a cylinder of the same $r$ and height 4, i.e. $<36\pi$; indeed $12\pi <36\pi$. ✓ Total $\approx 264$ m³ is a large tank — reasonable.

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Cell D — Subtracted / hollow shape (a pipe)

Forecast: A pipe is a solid cylinder with a smaller cylinder drilled out. So this time we... subtract. Guess whether the metal volume is closer to half or a tenth of the outer cylinder.

Steps:

1. Outer cylinder (as if solid): $V_{\text{out}}=\pi R^{2}L=\pi (5^2)(50)=1250\pi$ cm³.
2. Inner hole: $V_{\text{in}}=\pi r^{2}L=\pi (4^2)(50)=800\pi$ cm³.
   • Why these steps? The metal is everything between the two radii. The hole is empty air we must remove.
3. Subtract: $V_{\text{metal}}=1250\pi -800\pi =450\pi \approx 1413.72$ cm³.

Verify: The metal ring has cross-section area $\pi (R^2-r^2)=\pi (25-16)=9\pi$ cm²; times length 50 gives $450\pi$ cm³. ✓ Two routes agree. The metal is $\frac{450}{1250}=36\%$ of the outer volume — a thin-walled pipe, plausible.

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Cell E — Degenerate / zero input (shape collapses)

Forecast: With a length equal to zero, does the solid still hold anything? Predict $V$ before computing.

Steps:

1. Flat box, volume: $V=lwh=10\times 6\times 0=0$ cm³.
   • Why? A box with zero height is a flat sheet — it holds no space. This is the limiting case where a cuboid degenerates into a rectangle.
2. Flat box, surface area: $SA=2(lw+lh+wh)=2(60+0+0)=120$ cm².
   • Why? The four side faces have zero height ($lh=wh=0$), so only top + bottom survive — two $10\times 6$ rectangles. That is $2\times 60=120$ cm², exactly the two faces of a flat sheet. ✓ consistent.
3. Zero-radius cone, volume: $V=\frac{1}{3}\pi r^{2}h=\frac{1}{3}\pi (0)(5)=0$ cm³.
   • Why? With no radius, the cone shrinks to a line segment of length 5 — a 1D object with no volume.
4. Zero-radius cone, surface: slant $\ell =\sqrt{r^2+h^2}=\sqrt{0+25}=5$, so $SA=\pi r(r+\ell )=\pi \cdot 0\cdot (0+5)=0$ cm².

Verify: Both degenerate volumes are $0$ — correct, since a 3D formula fed a zero dimension must return $0$ space. The flat box keeps $120$ cm² of skin (it still has two faces), which is exactly what a physical sheet of card has. ✓

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Cell F — Partial fill / limiting value

Forecast: Half the height — is it exactly half the water? For a cylinder (constant cross-section) guess yes; for a cone it would not be. Why the difference?

Steps:

1. Water volume: the water is itself a shorter cylinder: $V=\pi r^{2}h=\pi (2^2)(2.5)=10\pi \approx 31.42$ m³.
   • Why this step? Water fills every circular slice from the floor up to height $h$; each slice has area $\pi (2^2)=4\pi$, so it is a genuine cylinder of height $h$.
2. Full tank for comparison: $V_{\text{full}}=\pi r^{2}H=\pi (4)(5)=20\pi \approx 62.83$ m³.
3. Fraction: $10\pi /20\pi =0.5$ — exactly half. Why? Because a cylinder's cross-section never changes with height, filling half the height fills half the volume.

Verify (limits):

• $h=0$: $V=\pi r^2(0)=0$ — empty tank. ✓
• $h=H=5$: $V=\pi (4)(5)=20\pi$ — full tank matches $V_{\text{full}}$. ✓

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Cell G — Real-world word problem

Forecast: Which surfaces are exposed? The top circle of the cylinder is hidden under the cone, so we do not count it. Guess: 4, 8, or 12 tins?

Steps:

1. Cone slant height (needed for its curved surface): $\ell =\sqrt{r^2+h_c^2}=\sqrt{3^2+4^2}=\sqrt{25}=5$ m. Uses Pythagorean Theorem.
2. Cone curved surface: $\pi r\ell =\pi (3)(5)=15\pi$ m².
   • Why? The unrolled cone is a sector; its area is $\pi r\ell$. We exclude the cone's base circle (it is hidden inside).
3. Cylinder curved side: $2\pi rh=2\pi (3)(8)=48\pi$ m².
   • Why? Unroll the side into a rectangle: width = circumference $2\pi r$, height $h$.
4. Bottom floor circle: $\pi r^2=\pi (3^2)=9\pi$ m².
5. Total paint area: $15\pi +48\pi +9\pi =72\pi \approx 226.19$ m².
6. Tins: $226.19/10=22.62$ → we cannot buy a fraction of a tin, so round up to 23 tins.
   • Why round up? Paint under-supply leaves bare metal; you must ceil.

Verify: Area breakdown adds correctly: $15+48+9=72$, times $\pi$. Ceiling of $22.62$ is $23$. ✓ Units: m² of surface ÷ m² per tin = dimensionless count. ✓

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Cell H — Exam twist: scaling law

Forecast: Doubling a length — does volume double? Quadruple? Or eight-fold? Guess before deriving. (Hint: volume has three length factors.)

Steps:

1. Original: $V_1=\frac{4}{3}\pi r^3=\frac{4}{3}\pi (6^3)=\frac{4}{3}\pi (216)=288\pi$ cm³.
2. Doubled: $V_2=\frac{4}{3}\pi (12^3)=\frac{4}{3}\pi (1728)=2304\pi$ cm³.
3. Volume factor: $V_2/V_1=2304\pi /288\pi =8$.
   • Why 8? Volume $\propto r^3$, and $2^3=8$. Every one of the three length-directions doubles.
4. Surface area factor: $SA\propto r^2$, so factor $=2^2=4$. Check: $S{A}_1=4\pi (6^2)=144\pi$; $S{A}_2=4\pi (12^2)=576\pi$; ratio $576/144=4$. ✓

Verify: $8=2^3$ and $4=2^2$ — consistent with the general scaling law: multiply every length by $k$ ⟹ area $\times k^2$, volume $\times k^3$. ✓

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Recall Quick self-test

Reverse a cube: it holds 27 litres, find edge in cm ::: $\sqrt[3]{27000}=30$ cm. Volume of water at depth $h$ in a cylinder radius $r$ ::: $\pi r^{2}h$. Painting a silo — why skip the top cylinder circle? ::: It is covered by the cone, so it is not an outer surface. Double every length of any solid: volume multiplies by ::: $2^3=8$. A cuboid with height 0 has volume ___ and surface area ___ ::: volume $0$; surface area $2lw$ (top+bottom survive).