Maths with Lemon

Volume and Surface Area

IB Mathematics: Analysis and Approaches · SL 3.1 (SL/HL) · Lesson 6.2. Learn to calculate the volume and surface area of pyramids, cones, spheres, hemispheres and composite solids, using right-angled triangles to find missing dimensions.

Lesson 6.2 worksheet (PDF)AA Geometry & TrigonometryUnit Circle

1. Volume and Surface Area: The Big Picture

What you need to know

  • Volume measures the space inside a solid (cubic units).
  • Surface area measures its exposed outer area (square units).
  • Distinguish perpendicular height from slant height.
  • Identify hidden right-angled triangles in 3D shapes.

Key Points

For a composite solid, add the volumes of its constituent parts, or subtract a removed volume. For surface area, count only the exposed faces; joined faces inside the object are not included.

\[V_{\mathrm{composite}}=\sum V_{\mathrm{parts}}-\sum V_{\mathrm{removed}}\]
Always convert lengths to the same unit before calculating. Give area in cm² or m² and volume in cm³ or m³.
hr

Cone

Radius, vertical height, slant height

r

Sphere

All points are a radius from the centre

h

Pyramid

Base area and perpendicular height

2. Right Pyramids

What you need to know

  • Volume is one-third of the corresponding prism.
  • For a right pyramid, the apex is directly above the centre of the base.
  • Use Pythagoras to calculate a triangular face’s slant height.
  • Add the base area and all triangular faces for total surface area.

Worksheet: Examples 1–2, pages 1–2.

Key Formula

\[V=\frac13 A_{\mathrm{base}}h\]
Worked example · Square pyramid (worksheet Example 1)

Square base side = 12 cm; perpendicular height = 8 cm.

Step 1. Base area: \(12^2=144\text{ cm}^2\).

\[V=\frac13(144)(8)=384\text{ cm}^3\]

Step 2. Face slant height, using half the base side:

\[\ell=\sqrt{8^2+6^2}=10\text{ cm}\]

Step 3. Four triangular faces, each with area \(\frac12(12)(10)\):

\[A=144+4\left(\frac12\cdot12\cdot10\right)=384\text{ cm}^2\]

Notice that the numerical values of area and volume happen to be equal here, but their units are different.

For a regular hexagonal base, divide the base into six equilateral triangles to find its area.

3. Right Cones and Frustums

What you need to know

  • Use the perpendicular height for volume.
  • Use the slant height for curved surface area.
  • A frustum is a large cone with a smaller similar cone removed.
  • Check whether a question includes the circular base.

Worksheet: Examples 3–6 and 12.

Key Formulas

\[V=\frac13\pi r^2h\]
\[A_{\mathrm{curved}}=\pi r\ell,\qquad A_{\mathrm{total}}=\pi r\ell+\pi r^2\]
\[\ell=\sqrt{h^2+r^2}\]
Worked example · Cone (worksheet Example 4)

A cone has radius 3 cm and perpendicular height 7 cm. Find its exact volume.

\[V=\frac13\pi(3)^2(7)=21\pi\text{ cm}^3\]

Keep \(\pi\) in the answer when the question asks for an exact value.

Frustum · Piping nozzle (worksheet Example 3)

Outer radius = 15 mm; top radius = 3 mm. The large and removed cones have heights 36 mm and 7.2 mm.

\[\ell_1=\sqrt{36^2+15^2}=39,\quad \ell_2=\sqrt{7.2^2+3^2}=7.8\]

The nozzle has no base, so subtract the curved areas only:

\[A=\pi(15)(39)-\pi(3)(7.8)=561.6\pi\text{ mm}^2\]

4. Spheres and Hemispheres

What you need to know

  • A sphere has no flat faces.
  • A hemisphere has half the sphere’s volume.
  • The total surface area of a hemisphere includes its circular base.
  • Check whether a question asks for curved or total surface area.

Worksheet: Examples 7–8.

Formula Collection

Sphere
\[V=\frac43\pi r^3\]
\[A=4\pi r^2\]
Hemisphere
\[V=\frac23\pi r^3\]
\[A_{\mathrm{curved}}=2\pi r^2\]
\[A_{\mathrm{total}}=3\pi r^2\]
Worked example · Hemisphere (worksheet Example 8)

A hemisphere has diameter 20 cm, so its radius is 10 cm.

\[V=\frac23\pi(10)^3=\frac{2000\pi}{3}\text{ cm}^3\]
\[A_{\mathrm{total}}=3\pi(10)^2=300\pi\text{ cm}^2\]

5. Combined Solids

What you need to know

  • Split a solid into familiar shapes.
  • Add volumes of attached solids.
  • Subtract volumes of holes or cut-out sections.
  • For surface area, exclude shared internal faces and include newly exposed cut surfaces.

Worksheet: ice-cream model and Examples 9–11, page 4–5.

Key Strategy

Draw a quick labelled sketch and mark which surfaces are visible. The formula for volume and the formula for exposed area are often not built from the same set of pieces.

Worked example · Ice-cream model (worksheet page 4)

A cone of height 8 cm supports a hemisphere of radius 4 cm.

Volume: add cone and hemisphere.

\[V=\frac13\pi(4)^2(8)+\frac23\pi(4)^3=\frac{256\pi}{3}\text{ cm}^3\]

Surface area: only the cone’s curved surface and the hemisphere’s curved surface are exposed.

\[\ell=\sqrt{8^2+4^2}=4\sqrt5\]
\[A=\pi(4)(4\sqrt5)+2\pi(4)^2=16\pi\sqrt5+32\pi\text{ cm}^2\]
A shared circular face is hidden inside the combined solid, so it is not counted twice.

6. Formula Explorer

Try different dimensions

  • Choose a solid.
  • Adjust the radius or base side and perpendicular height.
  • Observe how the volume and total surface area change.
  • For a cone or pyramid, the slant height is calculated automatically.

The calculator assumes a closed solid, including its base.

Interactive calculation

Volume

Total surface area

7. Practice and IB-Style Challenges

From your worksheet

  • Examples 1–2: square and hexagonal pyramids.
  • Examples 3–6: cones, slant height and a piping nozzle.
  • Examples 7–8: sphere and hemisphere.
  • Examples 9–12: combined solids and a frustum.
  • Challenges 1–2: capsule redesign and fencepost manufacturing.

Revision checklist

Can you distinguish height and slant height, use Pythagoras in a 3D context, identify exposed surfaces, and explain your assumptions in an applied problem?

Quick check · Cone with radius 5 cm and height 12 cm

Slant height \(\ell=13\) cm.

\[V=100\pi\text{ cm}^3,\qquad A_{\mathrm{total}}=90\pi\text{ cm}^2\]
Quick check · Sphere with radius 3 cm
\[V=36\pi\text{ cm}^3,\qquad A=36\pi\text{ cm}^2\]
The worksheet’s Example 13 gives a pyramid height in cm but a volume in m³; check the intended units before attempting that problem.

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