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Volume of More Complex Shapes

Real objects are rarely one neat solid. The trick never changes: break the shape into simple pieces you already know, then add the volumes together or take one away from another.

MYP 5Standard MathsMensurationCriteria A · C · D~10 min read

Written by Fayek Jawad Haider · MYP student

The add-or-subtract idea

A compound, or composite, solid is one built from two or more basic solids. Because volume simply measures space, the volume of the whole is the sum of the parts, or the leftover after a part is removed. So every compound question comes down to a plan: name the simple pieces, find each volume with its own formula, then combine.

Compound solid
A single object made by joining simpler solids together, or by cutting one solid out of another.

Building up by adding

When one solid sits on top of another, such as a cone on a cylinder or a hemisphere capping a tube, the total volume is just the pieces added:

\[ V_{\text{total}} = V_1 + V_2 \]

Work each piece out fully on its own, keeping answers as multiples of \(\pi\) where you can, then add at the end. This keeps rounding out of the middle of your working, where it can drift.

Sketch and label the pieces

Before calculating, draw the solid and mark where you split it. Note the radius and height of each piece next to it. A clear diagram stops you from reusing a length in the wrong formula and shows the marker exactly how you saw the shape.

Hollowing out by subtracting

When material is removed, such as a hole drilled through a block or a scoop taken from a solid, find the volume of the outer solid and subtract the volume of the gap:

\[ V_{\text{solid}} = V_{\text{outer}} - V_{\text{removed}} \]

A tube (a pipe) is the classic case: a large cylinder with a smaller cylinder removed down its centre.

A worked example

Worked example

A grain silo is a cylinder of radius 3 m and height 10 m with a cone of the same radius and perpendicular height 4 m on top. Find its total volume, to 1 decimal place.

1
Cylinder: \(V = \pi r^2 h = \pi \times 3^2 \times 10 = 90\pi\) m\(^3\).
2
Cone: \(V = \tfrac13 \pi r^2 h = \tfrac13 \pi \times 3^2 \times 4 = \tfrac13 \pi \times 36 = 12\pi\) m\(^3\).
3
Add the two: \(90\pi + 12\pi = 102\pi\) m\(^3\).
4
Evaluate: \(102\pi = 320.44\ldots\)
Total volume \(\approx 320.4\) m\(^3\)

Where this is assessed

Breaking the solid into known pieces and applying each formula is Criterion A (Knowing and understanding). A labelled diagram and clear add or subtract working is Criterion C (Communicating). Modelling a real object such as a silo, a bottle or a machined part is Criterion D (Applying maths in context). Decide up front whether pieces are added or one is subtracted.

Check yourself

1. A solid cuboid 8 cm by 8 cm by 10 cm has a cylindrical hole of radius 2 cm drilled straight through its 10 cm height. Find the remaining volume (to 1 d.p.). +

Cuboid: \(8 \times 8 \times 10 = 640\) cm\(^3\). Hole (a cylinder): \(\pi r^2 h = \pi \times 2^2 \times 10 = 40\pi = 125.66\ldots\) cm\(^3\). Remaining \(= 640 - 40\pi = 640 - 125.66\ldots = \mathbf{514.3}\) cm\(^3\).

2. A cube of edge 6 cm has a hemisphere of radius 3 cm scooped out of one face. Find the remaining volume (to 1 d.p.). +

Cube: \(6^3 = 216\) cm\(^3\). Hemisphere is half a sphere: \(\tfrac12 \times \tfrac43 \pi r^3 = \tfrac23 \pi \times 3^3 = \tfrac23 \pi \times 27 = 18\pi = 56.54\ldots\) cm\(^3\). Remaining \(= 216 - 18\pi = 216 - 56.54\ldots = \mathbf{159.5}\) cm\(^3\).

3. A solid is a cuboid 5 cm by 5 cm by 4 cm with a smaller cuboid 3 cm by 3 cm by 6 cm stacked on top. Find its total volume. +

Lower cuboid: \(5 \times 5 \times 4 = 100\) cm\(^3\). Upper cuboid: \(3 \times 3 \times 6 = 54\) cm\(^3\). Total \(= 100 + 54 = \mathbf{154}\) cm\(^3\).

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