Surface area and volume
Many objects such as packing containers, buildings and components of machines have geometric representations that use two- and three-dimensional shapes. This leads to many practical scenarios that require the calculating the surface area or volume of shapes such as prisms, pyramids, cylinders and spheres.
Use this page to revise the concepts within surface area and volume of:
Prisms and Pyramids
Right rectangular prisms
Rectangular prisms are polygons with an two identical rectangle end sections and four identical rectangular faces along the length. When the faces of the prism are all at right-angles to each other, these are referred to as right rectangular prisms. Also referred to as ‘box-shaped’ objects, they occur in packaging, containers, building design and furniture.
The key dimensions of a rectangular prism are its width, \(w\), height, \(h\), and length,\(l\), as illustrated in the diagrams below, showing both the prism and corresponding net of the prism (the two dimensional layout of the faces of a prism, or, the flat shape that can be folded up to make the prism):
To calculate the surface area and volume of a right rectangular prism we can use the following formulae
\[ V = w \times h \times l\]
\[ S=2\times w\times h+2\times w\times l+2\times l\times h=2(wh+wl+lh) \]
Where:
- \(V\) is the volume
- \(S\) is the surface area
- \(w,\ h\) and \(l\) are the width, height and length of the prism
Worked Example
A storage box with a lid has dimensions \(100\text{ cm}\) by \(60\text{ cm}\) by \(40\text{ cm}\). Calculate the volume and surface area of the box.
The volume of the storage box is:
\[
V = 100 \times 60 \times 40 = 240\,000 \text{ cm}^3 = 0.24 \text{ m}^3.
\]
The surface area of the storage box is:
\[
S = 2(6000 + 4000 + 2400) = 24\,800 \text{ cm}^2 = 2.48 \text{ m}^2
\]
Right triangular prisms
Triangular prisms are polygons with two identical triangular end sections and three rectangular faces along the length. When the rectangular faces of the prism are at right-angles to the triangular end section, these are referred to as right triangular prisms. The end section is often an equilateral or isosceles triangle. Triangular prisms occur in packaging, containers, building design, house roofs, A-frame tents, and furniture.
The key dimensions of a triangular prism are its width, \(w\), height, \(h\), length,\(l\). The diagrams below, shows an isosceles right triangular prism and a corresponding net:
The volume of any right triangular prism can be calculated by:
\[ V = \frac{1}{2} \times w \times h \times l\]
The surface area, of a right triangular prism depends on the lengths of the other two sides of the end section triangle. If these side lengths are \(u\) and \(v\), then surface area of the right triangular prism can be calculated by:
\[ S = w \times h + w \times l + u \times l + v \times l \]
Worked Example
The attic of a house is in the shape of an isosceles right triangular prism. The end section has width \(8\text{ m}\) and height \(2\text{ m}\). The length of the attic is \(16\text{ m}\). Calculate the volume and surface area of the attic
The volume of the roof is:
\[
V = \frac{1}{2} \times 2 \times 8 \times 16 = 128 \text{ m}^3
\]
Use Pythagoras' theorem to calculate the length of the slant edge of the end section as:
\[
\sqrt{4^2 + 2^2} = \sqrt{20} \text{ m}
\]
The surface area of the attic, to the nearest square metre, is:
\[ S = 2 \times 8 + 2 \times \sqrt{20} \times 16 \approx 159 \text{ m}^2 \]
Right square-based pyramids
Square-based pyramids have a square base, sides that are isosceles triangles and a peak (also called the apex or vertex) that is vertically above the middle of their square base. They are at times used in for distinctive structural features in building and design.
The key dimensions of a square-based pyramid are the length of its base,\(l\), its height, \(h\), and the height of its triangular faces, called its slant height, \(s\). The diagrams below show a right square-based pyramid and a corresponding net:
To calculate the surface area and volume of a square-based pyramid we can use the following formulae
\[ V = \frac{1}{3} \times l^2 \times h \]
\[S=l^2+2\times s \times l\]
A right square based pyramid sculpture had base side length of \(10\text{ m}\), and height of \(4\text{ m}\).
The volume of the sculpture is:
\[V = \frac{1}{3} \times 10^2 \times 4 = \frac{400}{3} \approx 133.33 \text{ m}^3\]
Use Pythagoras theorem to calculate the length of the slant edge of the pyramid:
\[s = \sqrt{5^2 + 4^2} = \sqrt{41} \text{ m}\]
The surface area of the pyramid, including the base, is:
\[S = 10^2 + 2 \times 10 \times \sqrt{41} = 100 + 20\sqrt{41} \approx 228 \text{ m}^2\]
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Cylinders and Spheres
Cylinders
Cylinders have circular end sections joined by a single long curved face. They occur often in machinery, storage, packaging, construction and design, from cans to storage silos, pipes to batteries.
The length dimensions of a cylinder are the radius of its circular end, \(r\), and either its length, \(l\), or height, \(h\), depending on whether it has a horizontal or vertical orientation.
To calculate the surface area and volume of a cylinder we use the following formulae:
\[ V = \pi \times r^2 \times h \] \[ S = 2 \times \pi \times r^2 + 2 \times \pi \times r \times h \] \[= 2\pi r (r + h) \]
Worked Example
A tin of powdered chocolate drink, with lid, is a cylinder with diameter \(10\text{ cm}\) and height \(12\text{ cm}\).
Calculate the volume and surface area of the cylinder.
The volume of the tin is:
\[
V = \pi \times r^2 \times h = \pi \times 5^2 \times 12 = 300\pi \approx 942.5 \text{ cm}^3
\]
The surface area of the tin, including the lid, is:
\[ S = 10\pi (5 + 12) = 170\pi \approx 534 \text{ cm}^2 \]
Spheres
A sphere is a three-dimensional object that can be formed by rotating a circle around its diameter in three-dimensional space. Sometimes spheres are also referred to as balls or globes. Spheres, or approximate spheres, occur naturally as planets and moons, some fruits, and water droplets suspended in the atmosphere. They also occur in manufactured objects such as balls, globes and ball bearings.
The key dimension associated with a sphere is its radius, \(r\), as illustrated in the diagram below. Unlike the other objects considered on this page, a sphere does not have an exact net because it is curved everywhere and cannot be flattened out without disrupting its shape.
To calculate the surface area and volume of a sphere we use the following formulae:
\[ V = \frac{4}{3} \times \pi \times r^3 \]
\[ S = 4 \pi r^2 \]
Note that the surface area of a sphere is 4 times the area of the circle through its centre.
Worked Example
A soccer ball has a diameter of about \(22\text{ cm}\). Calculate its volume and surface area.
The volume of the soccer ball is:
\[ V = \frac{4}{3} \times \pi \times 11^3 \approx 5575 \text{ cm}^3 \]
The surface area of the soccer ball is:
\[ S = 4 \times \pi \times 11^2 = 484\pi \approx 1521 \text{ cm}^2 \]