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Surface area and volume of 3D solidsIB MYP Maths Extended: Revision notes

Section 1

Units and conversions

Volume is the space inside a solid, in cubic units such as cm3^3. Surface area is the total area of all the faces, in square units such as cm2^2. When you convert units, remember that lengths scale by the conversion factor, areas by its square and volumes by its cube:

  • 11 cm =10=10 mm, so 11 cm2=100^2=100 mm2^2 and 11 cm3=1000^3=1000 mm3^3.
  • 11 m =100=100 cm, so 11 m3=1 000 000^3=1\,000\,000 cm3^3.
  • 11 litre =1000=1000 cm3^3, and 11 m3=1000^3=1000 litres. Make sure every length is in the same unit before you substitute into a formula.
Key termsvolumesurface areacapacity
Common mistake

Converting a volume by multiplying by 1010 or 100100 instead of 10001000 or 1 000 0001\,000\,000. Cube the length conversion factor.

Section 2

Prisms and cylinders

A prism has the same cross-section all the way along its length. For any prism: V=area of cross-section×length.V=\text{area of cross-section}\times\text{length}. For a cylinder with radius rr and height hh: V=πr2h,curved surface area=2πrh,total surface area=2πr2+2πrh.V=\pi r^2h,\qquad \text{curved surface area}=2\pi rh,\qquad \text{total surface area}=2\pi r^2+2\pi rh. For the surface area of a prism, find the area of each face and add. Example: a triangular prism with a right-angled triangle (66 cm, 88 cm, 1010 cm) of length 1515 cm has V=24×15=360V=24\times15=360 cm3^3 and surface area 48+360=40848+360=408 cm2^2.

Key termsprismcross-sectioncylinder
Exam tip

For a net-based surface area, list every face (including both ends) and tick them off as you add them.

Section 3

Pyramids and cones

A pyramid has a polygon base and triangular faces meeting at an apex. A cone has a circular base. Both have volume that is one third of the matching prism or cylinder: Vpyramid=13×base area×h,Vcone=13πr2h.V_{\text{pyramid}}=\frac13\times\text{base area}\times h,\qquad V_{\text{cone}}=\frac13\pi r^2h. Here hh is the perpendicular height. The slant height ll runs along a sloping face. For a cone, curved surface area =πrl=\pi rl. For a pyramid, each triangular face is 12×base×slant height\frac12\times\text{base}\times\text{slant height}. Example: base 1010 cm square, h=12h=12 cm gives V=13(100)(12)=400V=\frac13(100)(12)=400 cm3^3.

Key termspyramidconeslant heightperpendicular height
Common mistake

Using the perpendicular height instead of the slant height for the area of a sloping face (or the other way round in the volume formula).

Section 4

Spheres

A sphere of radius rr has V=43πr3,surface area=4πr2.V=\frac43\pi r^3,\qquad \text{surface area}=4\pi r^2. A hemisphere is half a sphere. Its curved surface is 2πr22\pi r^2 and its total surface area, with the flat circle, is 3πr23\pi r^2. Example: r=3r=3 cm gives V=36π=113V=36\pi=113 cm3^3 and A=36π=113A=36\pi=113 cm2^2 (the numbers match here, but the units differ). The formulae are given in the formula booklet, so focus on substituting carefully.

Key termsspherehemisphere
Common mistake

Cubing the wrong thing. In 43πr3\frac43\pi r^3 you cube only the radius, not 43πr\frac43\pi r.

Section 5

Solving 3D problems

Many problems combine formulae. Plan the steps:

  1. Pick the formula and write it down.
  2. Check all lengths are in the same unit.
  3. Substitute, keeping π\pi on your calculator until the end.
  4. Round to 3 significant figures and give the correct units (cm3^3 or cm2^2). For liquids poured between containers, the volume stays the same. If the cup holds 100π100\pi cm3^3 and is poured into a cylinder of radius 44 cm, then 16πh=100π16\pi h=100\pi and h=6.25h=6.25 cm. In real-life problems, round down for 'complete' items (servings, boxes) and round up for 'enough to buy'.
Key termssignificant figures
Exam tip

Write the formula first, then substitute. Even if you slip on the arithmetic, you still earn the method mark.

That's the notes covered.

Carry on to the next subtopic.

Exam questions on Surface area and volume of 3D solids

  1. A closed cylindrical water tank has radius 0.80.8 m and height 1.51.5 m. Volume of a cylinder =πr2h=\pi r^2h. Curved surface area of a cylinder =2πrh=2\pi rh.
    Find the total outer surface area of the closed tank.2 marks
  2. A solid metal prism has a right-angled triangle as its cross-section, with perpendicular sides 66 cm and 88 cm and hypotenuse 1010 cm. The prism is 1515 cm long.
    Write the volume of the prism in mm3^3.2 marks
  3. A square-based pyramid has a base of side 1010 cm and a perpendicular height of 1212 cm. Each of its four triangular faces has a slant height of 1313 cm. Volume of a pyramid =13×base area×height=\frac13\times\text{base area}\times\text{height}.
    Find the volume of the pyramid. A cuboid box has the same base and the same height. What fraction of the box does the pyramid fill?3 marks
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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).