Measurement - Surface Area and Volume

# Volume and surface area (mixed questions)

Lesson

A three-dimensional object can be thought of as an assembly of two-dimensional surface components or it can be thought of as a three-dimensional shape enclosed by its surface. Thus, we might be interested in the surface area of an object or we might be interested in the volume of the space it contains.

## Area

Many of the objects whose area we might wish calculate can be considered as separate pieces of simpler shape that together form the object of interest. If we can find the separate areas of the pieces, then we can sum them to get the whole surface area.

It can be helpful to consider the net formed by taking the shape apart and assembling it on a flat surface. The following diagram shows a triangular prism and its net.

Not all shapes can be flattened out in this way. Consider, for example, a sphere. It seems impossible to dissect a sphere into finitely many flat pieces and, in fact, we use methods of calculus to sum infinitely many very small flat pieces in this case. For a sphere, the volume is given by $4\pi r^2$4πr2 where $r$r is the radius.

The simple shapes whose areas will be needed are

• rectangle
• triangle
• circle
• parallelogram

These are shown below.

The surface area of a cone presents a more complicated problem. We can flatten a cone into a sector of a circle as shown below. (We have included only the curved surface of the cone.)

Evidently, the curved surface of the cone has area $\pi s^2$πs2 less the missing sector from the flattened out form. We need to find the fraction of the circle that is left after the missing sector is removed. This is the same proportion as the arc-length $k$k to the whole circumference $2\pi s$2πs of the flattened out circle.

The arc length $k$k is equal to the circumference $c$c of the base of the cone. So, we have $k=2\pi r$k=2πr. So, the fraction we want is $\frac{2\pi r}{2\pi s}=\frac{r}{s}$2πr2πs=rs. Then, the curved surface of the cone has area

$A=\frac{r}{s}\pi s^2=\pi rs$A=rsπs2=πrs.

If we now add the area of the circular base to this, we have the formula $A=\pi r(r+s)$A=πr(r+s) where $r$r is the radius of the base of the cone and $s$s is the length of its sloping side.

## Volume

If slices can be made through a solid shape perpendicular to an axis and the slices always have the same area, then the volume is the area of the slice$\times$× the full distance along the axis.

Thus, for any prism, its volume $V$V is given by $V=A_b\times h$V=Ab×h where $A_b$Ab is the area of the base (or any other cross-section) and $h$h is the height.

For shapes, including pyramids and cones, where there are cross-sections of similar shape but diminishing area as we move along an axis, it can be shown that the volume is given by $V=\frac{1}{3}A_b\times h$V=13Ab×h.

This fact has been known since ancient times but today it is usually demonstrated by calculus methods.

##### Example 1

Consider the triangular prism illustrated above. If the triangular cross-section is an equilateral triangle with edges $72\text{mm}$72mm , and the length of the prism is $190\text{mm}$190mm , what is the surface area and what is the volume?

Surface area:
The three rectangular faces have area $3\times72\times190\text{mm}^2$3×72×190mm2. The two triangular ends have area $2\times\frac{1}{2}\frac{\sqrt{3}}{2}72^2$2×1232722 (using the sine area formula and the fact that an equilateral triangle has $60^\circ$60° angles). In total, this is $45529.5\text{mm}^2$45529.5mm2. This is about $455.1\text{cm}^2$455.1cm2.

Volume:
We multiply the cross-section area by the length. This is $\frac{1}{2}\frac{\sqrt{3}}{2}\times72^2\times190=426500\text{mm}^3$1232×722×190=426500mm3  or $426.5\text{cm}^3$426.5cm3.

##### Example 2

In a physics experiment, a piece of equipment is constructed in the form of a hollow cylinder with hemispherical ends. The cylinder has length $100\text{cm}$100cm and inside diameter $3.6\text{cm}$3.6cm. The inside surface is to be coated with a film of scandium with a thickness of $0.1\text{mm}$0.1mm

What volume of scandium will be needed?

What is the volume of the space inside the piece of equipment?

The volume of scandium is the inside surface area multiplied by the thickness of the film. The circumference of the cylinder is $3.6\times\pi\text{cm}$3.6×πcm. So, the cylindrical part of the surface has area $3.6\pi\times100\text{cm}^2$3.6π×100cm2. The two hemispherical ends make a sphere with surface area $4\pi\times\left(\frac{3.6}{2}\right)^2\ \text{cm}^2$4π×(3.62)2 cm2

Thus, the surface area is $3.6\pi\times100+4\pi\times\left(\frac{3.6}{2}\right)^2=1171.688\ \text{cm}^2$3.6π×100+4π×(3.62)2=1171.688 cm2. This makes the volume of scandium required  $1171.688\times0.01\approx11.7\ \text{cm}^3$1171.688×0.0111.7 cm3.

The volume inside is $\frac{4}{3}\pi\times\left(\frac{3.6}{2}\right)^3+\pi\times3.6\times100=1155.4\ \text{cm}^3$43π×(3.62)3+π×3.6×100=1155.4 cm3.

We used the fact that the volume of a sphere is $V=\frac{4}{3}\pi r^3$V=43πr3 where $r$r is the radius.

##### Example 3

You have $9\ \text{m}^2$9 m2 of wrapping paper. What is the largest square prism that can be covered with this paper given that the length of the prism has to be $1.5$1.5 times the length of an edge of the square base?

We form an expression for the surface area of the object, using $x$x for the side of the square base.

$A=2\times x^2+4\times1.5x\times x=8x^2$A=2×x2+4×1.5x×x=8x2

We are given that $8x^2=9$8x2=9. Therefore, $x^2=\frac{9}{8}$x2=98 and so, $x=\sqrt{\frac{9}{8}}\approx1.06\text{m}$x=981.06m

This means the square prism should have dimensions $1\times1\times1.5$1×1×1.5 metres.

#### Worked Examples

##### Question 1

Consider the following rectangular prism with a width, length and height of $5$5 m, $7$7 m and $14$14 m respectively.

1. Find the total surface area of the rectangular prism.

2. Find the volume of the rectangular prism.

##### Question 2

Consider the following cylinder with a height of $20$20 mm and a base radius of $10$10 mm.

1. Find the surface area of the cylinder. Give your answer correct to $2$2 decimal places.

2. Find the volume of the cylinder. Give your answer correct to $2$2 decimal places.

##### Question 3

Given the following cone.

1. Find the total surface area correct to $2$2 decimal place.

2. Find the height of the cone correct to $2$2 decimal places.

3. Find the volume of the cone correct to $2$2 decimal places.

### Outcomes

#### 9D.MG2.06

Solve problems involving the surface areas and volumes of prisms, pyramids, cylinders, cones, and spheres, including composite figures