Surface Area and Volume: Simple Examples and Solutions

The calculation of surface area and volume is fundamental in geometry and is frequently applied to a wide range of solid figures. Each solid has specific formulae that are derived from their respective geometric properties.

Geometric Approach to Surface Area and Volume Calculations

For a cube of edge length $a$, the total surface area is obtained by summing the areas of its six square faces. Each face has area $a^2$, so the total surface area is $6a^2$. The volume is given by $a^3$ because the cube occupies the space defined by its length, breadth, and height, each equal to $a$.

For a cuboid with length $l$, breadth $b$, and height $h$, the total surface area equals $2(lb + bh + lh)$, as it includes the pairs of opposite faces. The volume equals $lbh$. These formulae apply similarly to other common solids.

Stepwise Solution of Representative Problems on Surface Area and Volume

Example 1: Volume of a Hexagonal Prism

Given: The perimeter of the base of a regular hexagonal prism is $600\, \text{cm}$, and its height is $200\, \text{cm}$. The density of petroleum to be filled is $0.8\, \text{g}/\text{cm}^3$. Take $\sqrt{3} = 1.73$.

First, the length of one side of the hexagon is found as $\dfrac{600}{6} = 100\, \text{cm}$.

The area $A$ of a regular hexagon with side $a$ is $A = \dfrac{3\sqrt{3}}{2}a^2$.

Substituting $a = 100\, \text{cm}$, the area is:

$A = \dfrac{3 \times 1.73}{2} \times 100^2 = \dfrac{5.19}{2} \times 10^4 = 2.595 \times 10^4\, \text{cm}^2 = 25,950\, \text{cm}^2$

The volume $V$ of the prism is $V = \text{Base area} \times \text{height} = 25,950 \times 200 = 5,190,000\, \text{cm}^3$.

The weight of petroleum it can hold is $5,190,000 \times 0.8 = 4,152,000\, \text{g} = 4,152\, \text{kg}$.

Examples On Surface Area And Volume

Example 2: Volume of a Cone Inscribed in a Cube

Given: The maximal right circular cone is to be carved from a cube of edge $a = 42\, \text{cm}$.

The radius $r$ of the cone will be half the edge ($r = 21\, \text{cm}$). The height $h$ of the cone equals the cube’s edge: $h = 42\, \text{cm}$.

Volume of the cone $= \dfrac{1}{3}\pi r^2 h$.

Substituting, $= \dfrac{1}{3} \times \dfrac{22}{7} \times 21^2 \times 42$.

First, calculate $21^2 = 441$.

Thus, $= \dfrac{1}{3} \times \dfrac{22}{7} \times 441 \times 42$.

$= \dfrac{1}{3} \times \dfrac{22}{7} \times 441 \times 42$

$= \dfrac{1}{3} \times \dfrac{22}{7} \times 18,522$

First, $441 \times 42 = 18,522$

Now, $18,522 \div 7 = 2,646$

Then, $2,646 \times 22 = 58,212$

Finally, $58,212 \div 3 = 19,404\, \text{cm}^3$.

Result: The maximum cone that can be carved from the cube has volume $19,404\, \text{cm}^3$.

Volume Of Parallelepiped

Example 3: Rise in Water Level Due to a Solid Sphere

Given: An iron sphere of diameter $6\, \text{inches}$ is submerged in a cylindrical vessel of diameter $12\, \text{inches}$ filled with water. Find the rise $h$ in water level.

Radius of sphere $= 3\, \text{inches}$.

Radius of vessel $= 6\, \text{inches}$.

Volume of sphere $= \dfrac{4}{3}\pi r^3 = \dfrac{4}{3}\pi (3)^3 = \dfrac{4}{3}\pi \times 27 = 36\pi\, \text{in}^3$.

Let $h$ be the rise in water level. Volume of water displaced $= \pi R^2 h = 36\pi$.

$\pi (6)^2 h = 36\pi$

$36\pi h = 36\pi$

$h = 1\, \text{inch}$

Result: The water rises by $1\, \text{inch}$.

Example 4: Recasting a Sphere into a Cylindrical Wire

Given: A sphere of diameter $9\, \text{cm}$ is melted and drawn into a wire of radius $1.5\, \text{cm}$. Find the length of the wire.

Radius of sphere $= 4.5\, \text{cm}$.

Volume of sphere $= \dfrac{4}{3}\pi r^3 = \dfrac{4}{3}\pi (4.5)^3$.

Calculate $(4.5)^3$:

$4.5 \times 4.5 = 20.25$

$20.25 \times 4.5 = 91.125$

So, volume $= \dfrac{4}{3}\pi \times 91.125 = 121.5\pi\, \text{cm}^3$.

Let length of wire be $H$.

Volume of cylinder (wire) $= \pi r^2 H = \pi (1.5)^2 H = \pi \times 2.25 H$

By conservation of volume: $121.5\pi = 2.25\pi H$

$H = \dfrac{121.5}{2.25} = 54\, \text{cm}$

Result: The length of the wire is $54\, \text{cm}$.

Area Of A Circle Formula

Example 5: Height of a Square Pyramid Given Surface Area and Volume

Given: A square pyramid has base area $36\, \text{m}^2$ and volume $288\, \text{m}^3$. Find the height ($H$) of the pyramid.

Volume of a pyramid $= \dfrac{1}{3} \text{Base area} \times \text{Height}$

$288 = \dfrac{1}{3} \times 36 \times H$

$288 = 12H$

$H = \dfrac{288}{12} = 24\, \text{m}$

Result: The height of the pyramid is $24\, \text{m}$.

Example 6: Surface Area and Volume of a Triangular Prism

Given: A prism with an equilateral triangle base of side $12\, \text{cm}$ and height $17\, \text{cm}$.

Perimeter $P = 12 + 12 + 12 = 36\, \text{cm}$

Area of base $= \dfrac{\sqrt{3}}{4} a^2 = \dfrac{1.73}{4} \times 12^2 = 0.4325 \times 144 = 62.28\, \text{cm}^2$ (use $\sqrt{3}=1.73$)

Lateral Surface Area (CSA): $= \text{Perimeter} \times \text{Height} = 36 \times 17 = 612\, \text{cm}^2$

Total Surface Area (TSA): $= \text{Lateral Surface Area} + 2 \times (\text{Area of Base}) = 612 + 2 \times 62.28 = 612 + 124.56 = 736.56\, \text{cm}^2$

Volume: $= \text{Area of Base} \times \text{Height} = 62.28 \times 17 = 1,058.76\, \text{cm}^3$

Result: The lateral surface area is $612\, \text{cm}^2$, total surface area is $736.56\, \text{cm}^2$, and volume is $1,058.76\, \text{cm}^3$.

Analysis of Surface Area and Volume Ratio in a Cube

For a cube with edge $s$, the surface area is $6s^2$ and the volume is $s^3$. The ratio $\dfrac{\text{surface area}}{\text{volume}} = \dfrac{6s^2}{s^3} = \dfrac{6}{s}$, which shows that as the edge increases, the ratio decreases.

Remarks on Real-Life Relevance of Surface Area and Volume Calculations

Calculations of surface area are crucial for estimating materials needed to cover solids, such as paint, tiles, or fabric. Volume calculations are essential when determining the capacity of containers, the amount of substances transferred between different geometric forms, or understanding physical phenomena governed by displacement. Common real-life examples include determining the quantity of liquid a tank holds or the material required to manufacture hollow cylinders.