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AI HL · 12.10

Geometry & Trigonometry

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12.10.9

Volumes and surface areas of solids

12.10.9

Volumes and surface areas of solids

Target: I can find the volume and the curved and total surface area of a cone, using the slant height where the formula needs it. (+5 more)

By the end of this section

Key points

The formulae

  • These are in the formula booklet: cone \(V=\frac13\pi r^2h\), curved surface area \(\pi rl\); sphere \(V=\frac43\pi r^3\), surface area \(4\pi r^2\); pyramid \(V=\frac13\times\text{base area}\times h\).
  • \(h\) is the perpendicular height. \(l\) is the slant height, along the sloping side. They are linked by \(l^2=r^2+h^2\).
  • Volume uses \(h\). Curved surface area uses \(l\). Mixing them up is the commonest mistake.
  • A cone is one third of the cylinder with the same base and height. A pyramid is one third of the prism.

Curved or total surface area

  • Curved surface area leaves out flat faces. Total surface area adds them. Read the question: it says which.
  • A solid hemisphere has a curved surface \(2\pi r^2\) and a flat circular face \(\pi r^2\). Total: \(3\pi r^2\).
  • A bowl is open at the top: only the curved surface is outside, \(2\pi r^2\).

Solids made from other solids

  • Sketch the cross-section and mark where the solids join.
  • Volume: add the volumes of the parts.
  • Surface area: add only the faces that are on the outside. Where two solids join, the faces that touch disappear.
  • A hemisphere on a cylinder has the same radius as the cylinder. Say so in your working.

Volume and capacity

  • \(1\text{ cm}^3=1\text{ mL}\), \(1000\text{ cm}^3=1\text{ litre}\), \(1\text{ m}^3=1000\text{ litres}\).
  • When a solid is melted and recast, nothing is lost: volume before = volume after.
  • Liquid in a cone that stands on its point is a smaller cone of the same shape. Radius and depth are in the same ratio as in the full cone.
  • The volume of the smaller cone is \(k^3\) times the volume of the full one if its lengths are \(k\) times as long.

On the calculator

Solving for a radius or height

  • Write the volume equation with the unknown, for example \(\frac43\pi r^3=5600\).
  • Use the calculator's solver for the unknown. Give the positive answer.
  • In your working write the equation, then the solution: Solve for r: 11.0…
  • Check units first: change litres to cm³ before you solve.

Screenshot and video to come

Worked examples

Problems like the ones in class. Pick one, try it, then open the working.
A cone

A cone has base radius 6.5 cm and perpendicular height 11.2 cm. Find (a) its volume, (b) its total surface area, including the circular base.

11.2 cm6.5 cm
Working
  1. Volume

    \(V=\frac13\pi r^2h=\frac13\pi(6.5)^2(11.2)=495.5\) cm³

    One third of the cylinder with the same base and height.

  2. Slant height

    \(l=\sqrt{6.5^2+11.2^2}=12.95\) cm

    Pythagoras in the right-angled triangle made by the axis, the radius and the slant side.

  3. Curved surface

    \(\pi rl=\pi(6.5)(12.95)=264.4\) cm²

    The slant height goes in this formula, not the perpendicular height.

  4. Base

    \(\pi r^2=\pi(6.5)^2=132.7\) cm²

    The base is a circle.

  5. Total

    \(264.4+132.7=397.2\) cm², which is 397 cm²

    Give 3 significant figures.

Video walkthrough coming
A sphere and a hemisphere

A spherical water tank has radius 1.35 m. (a) Find the area of its outer surface. (b) Find the volume of water it holds when full, in litres. (c) A second tank is a solid hemisphere of the same radius. Find its total surface area, including the flat face.

Working
  1. Sphere surface

    \(4\pi r^2=4\pi(1.35)^2=22.90\) m², so 22.9 m²

    Four times the area of the circle through the centre.

  2. Volume in m³

    \(\frac43\pi r^3=\frac43\pi(1.35)^3=10.306\) m³

    Work in metres throughout.

  3. Litres

    \(1\text{ m}^3=1000\text{ litres}:\;10.306\times1000=10306\) litres, which is 10 300 litres

    Multiply by 1000 to go from m³ to litres.

  4. Hemisphere

    \(2\pi r^2+\pi r^2=3\pi r^2=3\pi(1.35)^2=17.18\) m², so 17.2 m²

    Curved half of the sphere plus the flat circular face.

Video walkthrough coming
A radius from a volume

A spherical balloon holds 5.6 litres of air. (a) Find its radius in cm. (b) A cone has volume 480 cm³ and base radius 7 cm. Find its perpendicular height.

Working
  1. Change the units

    \(5.6\text{ litres}=5600\text{ cm}^3\)

    1 litre is 1000 cm³.

  2. Equation

    \(\frac43\pi r^3=5600\)

    Set the volume formula equal to the volume.

  3. Solve

    \(\text{Solve for }r:\;r=11.016\), so 11.0 cm

    From the calculator, using the solver. Keep the positive answer.

  4. The cone

    \(\frac13\pi(7)^2h=480\;\Rightarrow\;h=\frac{3\times480}{\pi\times49}=9.354\), so 9.35 cm

    Here the unknown is to the first power, so rearranging is quick.

Video walkthrough coming
A cylinder with a hemisphere on top

A grain silo is a cylinder of radius 3.5 m and height 9 m, with a hemispherical roof of the same radius on top. Find (a) the volume of grain it holds when full, (b) the area of its curved outside surface, walls and roof together, not the floor.

7 m9 m
Working
  1. Split it

    cylinder + hemisphere

    Both have radius 3.5 m. The dashed line is where they join.

  2. Cylinder

    \(\pi r^2h=\pi(3.5)^2(9)=346.36\) m³

    Base area times height.

  3. Hemisphere

    \(\frac12\cdot\frac43\pi r^3=\frac23\pi(3.5)^3=89.80\) m³

    Half of the sphere.

  4. Volume

    \(346.36+89.80=436.2\) m³, so 436 m³

  5. Surface

    \(2\pi rh+2\pi r^2=197.92+76.97=274.9\) m², so 275 m²

    Wall plus roof. The circle where they meet is inside the solid, so it is not counted.

Video walkthrough coming
A cuboid with a pyramid on top

A monument is a cuboid with a square base of side 2.4 m and height 5 m. A square-based pyramid of height 1.8 m sits on top, with the same base. Find (a) the total volume, (b) the total area of the four triangular faces of the pyramid.

2.4 m5 m
Working
  1. Cuboid

    \(2.4\times2.4\times5=28.80\) m³

    Base area times height.

  2. Pyramid

    \(\frac13\times2.4^2\times1.8=3.456\) m³

    One third of base area times perpendicular height.

  3. Volume

    \(28.80+3.456=32.26\) m³, so 32.3 m³

  4. Slant height of a face

    \(\sqrt{1.8^2+1.2^2}=2.163\) m

    The height of one triangular face. It runs from the middle of a base edge up to the apex, so the horizontal distance is half the side: 1.2 m.

  5. Four faces

    \(4\times\frac12\times2.4\times2.163=10.38\) m², so 10.4 m²

    The slant height is the height of each triangle.

Video walkthrough coming
Melting and recasting

A solid metal sphere of radius 6.5 cm is melted and recast into small cones, each with base radius 2.1 cm and height 8 cm. No metal is lost. How many complete cones can be made?

Working
  1. Volume of the sphere

    \(\frac43\pi(6.5)^3=1150.3\) cm³

    The volume does not change when the metal is melted.

  2. Volume of one cone

    \(\frac13\pi(2.1)^2(8)=36.95\) cm³

  3. Divide

    \(\frac{1150.3}{36.95}=31.14\)

    Equal volumes: n × volume of one cone = volume of the sphere.

  4. Whole cones

    31 cones

    Round down, not to the nearest whole number.

Video walkthrough coming
A part-filled cone

A conical glass has internal radius 4.8 cm at the top and internal depth 12 cm. It stands on its point and holds juice to a depth of 7.5 cm. Find (a) the radius of the surface of the juice, (b) the volume of juice, (c) the juice as a percentage of the full glass.

9.6 cm7.5 cm
Working
  1. Radius of the surface

    \(\frac{r}{4.8}=\frac{7.5}{12}\;\Rightarrow\;r=3.00\) cm

    The juice is a smaller cone with the same shape, so its radius and depth are in the same ratio as the glass.

  2. Volume of juice

    \(\frac13\pi(3.0)^2(7.5)=70.69\) cm³, so 70.7 cm³

    Use the radius of the juice surface.

  3. Volume of the glass

    \(\frac13\pi(4.8)^2(12)=289.5\) cm³

  4. Percentage

    \(\frac{70.69}{289.5}\times100=24.4\)%, so 24.4%

    Depth is 62.5% of the full depth but the volume is a much smaller share: the volume scales as the cube.

Video walkthrough coming