# Area and Volume

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When working with areas and volumes, precise calculations are only possible by use of formulae and methods for combining and applying these formulae. Common examples of formulae are the area of a circle (given by $\ctext{Area} = \pi r^2$, where $\pi\approx 3.14159$ and $r$ is the circle's radius), the surface area of a semi-sphere (given by $\ctext{Surface area} =2\pi r^2$, where $r$ is the sphere's radius), and the volume of a cylinder (given by $\ctext{Volume} = \pi r^2h$, where $r$ is the base circle's radius and $h$ is the cylinder's height). Areas and volumes of less familiar shapes and solids can be calculated using the calculus-based techniques of numerical integration, but for most practical purposes this is done using formulae and methods behind the scenes' in programs such as Mathematica or Matlab. An example of combined use of formulae and a method is calculating the volume trapped' between a cylinder and a cube that is wholly contained within the cylinder. The method is to subtract the volume of the cube from that of the cylinder.

### Example problem

The North-facing wall of a house is to be built using a single layer of bricks (see the shaded region in the figure below). The wall's dimensions are $3400$ mm wide and $3400$ mm high. Two rectangular windows are to be cut into the wall whose dimensions are $700\;\ctext{mm} \times 280\;\ctext{mm}$ for the smaller one (width $\times$ height), and $1750\;\ctext{mm}\times 280\;\ctext{mm}$ for the larger one. What is the total area in square metres of bricks to be used in the wall construction? ### Solution

The method is to first calculate the area of the square wall surface, and then subtract from this the combined area of the two windows. Note that the first task is to convert the wall and window dimensions to metre measurements by dividing each of them by $1000$. We have

\begin{align*} \ctext{Brick area}&=\ctext{Area of wall}-\left(\ctext{area of small window}+\ctext{Area of large window}\right)\cr &=3.40\;\ctext{m}\times 3.40\;\ctext{m}-\left(0.70\;\ctext{m}\times 0.28\;\ctext{m}+ 1.75\;\ctext{m}\times 0.28\;\ctext{m}\right)\cr &\approx 11.56\;\ctext{m}^2-\left( 0.20\;\ctext{m}^2+ 0.49\;\ctext{m}^2\right)\cr &= 10.87\;\ctext{m}^2 \end{align*}