Volume and Surface Area

Volume and surface area extend area into three dimensions, and they are a dependable source of marks across OCR GCSE Mathematics (J560). Volume measures the space inside a solid; surface area measures the total area of all the faces that wrap around it. The recurring exam skills are choosing the correct formula, knowing which formulae the J560 sheet provides and which you must memorise, substituting carefully, and giving the answer in the right cubic or square units. Recalling and using a formula is AO1; combining solids or working backwards is AO2; and applying volume to a real context such as capacity or packing is AO3. Higher tier adds frustums and compound solids.

Key Vocabulary

Term	Meaning
Volume	The space inside a solid, in cubic units (cm$^{3}$3, m$^{3}$3)
Surface area	The total area of all the faces of a solid (cm$^{2}$2)
Cross-section	The 2D shape revealed by slicing straight through a solid
Prism	A solid with the same cross-section all the way through
Slant height	The distance up the sloping surface of a cone, from base edge to apex
Frustum	The solid left when the top of a cone or pyramid is cut off parallel to the base [H]

What the Formula Sheet Gives You

It pays to know exactly which formulae OCR provides and which you are expected to have learnt, because under exam pressure many candidates waste time searching the sheet for a formula that was never there. Given on the J560 formula sheet: the volume of a sphere $V = \dfrac{4}{3}\pi r^{3}$V=34​πr3, the surface area of a sphere $A = 4\pi r^{2}$A=4πr2, the volume of a cone $V = \dfrac{1}{3}\pi r^{2} h$V=31​πr2h, and the curved surface area of a cone $A = \pi r l$A=πrl (where $l$l is the slant height). You must memorise: the volume of a prism (area of cross-section $\times$× length), the volume and surface area of a cuboid, the volume of a cylinder $V = \pi r^{2} h$V=πr2h, the curved surface area of a cylinder $2\pi r h$2πrh, and the volume of a pyramid ($\dfrac{1}{3} \times$31​× base area $\times$× height). A sensible exam routine is to glance at the sheet at the start, remind yourself which solids are covered there, and trust your memory for the rest. Knowing where each formula lives — and roughly where it comes from — saves precious minutes and stops the panic of hunting for something that simply is not printed.

Prisms and Cylinders

A prism is any solid with a constant cross-section — meaning a slice taken anywhere along its length gives the same 2D shape — and its volume is beautifully simple: the area of the cross-section multiplied by the length (or height) of the prism. This one rule is far more powerful than it first looks, because it covers cuboids (rectangular cross-section), triangular prisms (triangular cross-section), L-shaped prisms (composite cross-section) and cylinders (circular cross-section) all at once. The strategy is always the same two steps: first work out the area of the cross-section using your 2D area skills, then multiply by how long the prism is. For a cylinder the cross-section is a circle of area $\pi r^{2}$πr2, so the volume becomes $\pi r^{2} h$πr2h — exactly "circle area times height". Surface area is handled separately by adding up every face: for a cylinder that is the curved tube ($2\pi r h$2πrh, the label of a tin can unrolled into a rectangle) plus the two circular ends. The figure shows a cylinder with its radius and height labelled.

Worked Example 1

A triangular prism has a right-angled triangular cross-section with base $6$6 cm and height $8$8 cm, and the prism is $15$15 cm long. Work out its volume.

Cross-section area $= \dfrac{1}{2} \times 6 \times 8 = 24$=21​×6×8=24 cm$^{2}$2. Volume $= 24 \times 15 = 360$=24×15=360 cm$^{3}$3.

Answer: $360$360 cm$^{3}$3.

Worked Example 2

A cylinder has radius $4$4 cm and height $10$10 cm. Work out (a) its volume and (b) its total surface area, each to $1$1 decimal place.

(a) Volume $= \pi r^{2} h = \pi \times 4^{2} \times 10 = 160\pi \approx 502.7$=πr2h=π×42×10=160π≈502.7 cm$^{3}$3.

(b) Curved surface $= 2\pi r h = 2\pi \times 4 \times 10 = 80\pi$=2πrh=2π×4×10=80π; the two circular ends $= 2 \times \pi \times 4^{2} = 32\pi$=2×π×42=32π; total $= 80\pi + 32\pi = 112\pi \approx 351.9$=80π+32π=112π≈351.9 cm$^{2}$2.

Answer: volume $502.7$502.7 cm$^{3}$3; surface area $351.9$351.9 cm$^{2}$2.

Common error: finding only the curved surface and forgetting to add the two circular ends for the total surface area.

Worked Example 3

A cuboid measures $5$5 cm by $4$4 cm by $3$3 cm. Work out its volume and its surface area.

Volume $= 5 \times 4 \times 3 = 60$=5×4×3=60 cm$^{3}$3. Surface area $= 2(5\times 4 + 5\times 3 + 4\times 3) = 2(20 + 15 + 12) = 2 \times 47 = 94$=2(5×4+5×3+4×3)=2(20+15+12)=2×47=94 cm$^{2}$2.

Answer: volume $60$60 cm$^{3}$3; surface area $94$94 cm$^{2}$2.

Cones and Pyramids

A pyramid is a solid that narrows from a flat polygon base up to a single point called the apex, and there is a lovely relationship that makes its volume easy to remember: a pyramid holds exactly one third of the prism that would share its base and height. So $V = \dfrac{1}{3} \times$V=31​× base area $\times$× height. A cone is simply a pyramid whose base is a circle, so the same one-third rule gives its volume as $\dfrac{1}{3}\pi r^{2} h$31​πr2h. The most important thing to get right with cones is that they involve two different lengths. The perpendicular height $h$h is the straight distance from the base up to the apex, measured at right angles to the base, and this is the length the volume formula uses. The slant height $l$l is the distance up the sloping outer surface from the rim of the base to the apex, and this is the length the curved-surface formula uses. The two are linked by Pythagoras, because $h$h, $r$r and $l$l form a right-angled triangle inside the cone: $l^{2} = r^{2} + h^{2}$l2=r2+h2. The curved surface area of a cone is $\pi r l$πrl, and adding the circular base $\pi r^{2}$πr2 gives the total surface area $\pi r l + \pi r^{2}$πrl+πr2. Keeping $h$h and $l$l straight in your head is the single biggest source of marks in this section. The figure shows a cone with $r$r, $h$h and $l$l all labelled.

Worked Example 4

A cone has radius $3$3 cm and perpendicular height $4$4 cm. Work out (a) its slant height, (b) its volume and (c) its total surface area. Give (b) and (c) to $1$1 decimal place.

(a) $l = \sqrt{r^{2} + h^{2}} = \sqrt{3^{2} + 4^{2}} = \sqrt{25} = 5$l=r2+h2​=32+42​=25​=5 cm.

(b) Volume $= \dfrac{1}{3}\pi r^{2} h = \dfrac{1}{3}\pi \times 3^{2} \times 4 = 12\pi \approx 37.7$=31​πr2h=31​π×32×4=12π≈37.7 cm$^{3}$3.

(c) Curved surface $= \pi r l = \pi \times 3 \times 5 = 15\pi$=πrl=π×3×5=15π; base $= \pi \times 3^{2} = 9\pi$=π×32=9π; total $= 15\pi + 9\pi = 24\pi \approx 75.4$=15π+9π=24π≈75.4 cm$^{2}$2.

Answer: $l = 5$l=5 cm; volume $37.7$37.7 cm$^{3}$3; surface area $75.4$75.4 cm$^{2}$2.

Common error: using the slant height in the volume formula. Volume needs the perpendicular height $h$h; only the curved surface area uses $l$l.

Worked Example 5

A square-based pyramid has base side $8$8 cm and perpendicular height $9$9 cm. Work out its volume.

Base area $= 8 \times 8 = 64$=8×8=64 cm$^{2}$2. Volume $= \dfrac{1}{3} \times 64 \times 9 = 192$=31​×64×9=192 cm$^{3}$3.

Answer: $192$192 cm$^{3}$3.

Worked Example 5b

A cylindrical water tank has radius $0.5$0.5 m and height $1.2$1.2 m. Work out its capacity in litres, given that $1$1 m$^{3} = 1000$3=1000 litres. Give your answer to the nearest litre.

Volume $= \pi r^{2} h = \pi \times 0.5^{2} \times 1.2 = \pi \times 0.25 \times 1.2 = 0.3\pi$=πr2h=π×0.52×1.2=π×0.25×1.2=0.3π m$^{3} \approx 0.9425$3≈0.9425 m$^{3}$3. Converting, $0.9425 \times 1000 \approx 942$0.9425×1000≈942 litres.

Answer: about $942$942 litres.