Volume of Standard Solids
Volume: Definition and Concept for Solids
The volume of a three-dimensional (3D) object, or solid, is a fundamental property that quantifies the amount of space it occupies. It measures the capacity of the solid, or how much substance (like liquid, gas, or solid material) it can hold if it's a container, or how much space it displaces.
Unlike surface area, which is a measure of the two-dimensional boundary of the solid, volume is a measure of the space contained within that boundary. While surface area is measured in square units, volume is measured in cubic units because it involves three dimensions: length, breadth, and height.
Concept of Volume using Unit Cubes
The most intuitive way to understand volume is by thinking about filling the solid object with unit cubes. A unit cube is a cube with side lengths of 1 unit (e.g., 1 cm × 1 cm × 1 cm, or 1 m × 1 m × 1 m). The volume of a unit cube is $1 \times 1 \times 1 = 1$ cubic unit.
The volume of a solid is equal to the total number of unit cubes that can fit perfectly inside it without overlapping.
For example, a cuboid that is 3 cm long, 2 cm wide, and 1 cm high can be filled with $3 \times 2 \times 1 = 6$ unit cubes of size 1 cm³. Therefore, its volume is 6 cm³.
Units of Measurement for Volume
Volume is measured in cubic units, denoted by the symbol of the linear unit raised to the power of 3 (e.g., $\text{cm}^3$, $\text{m}^3$).
- Metric System: Cubic millimetre ($\text{mm}^3$), Cubic centimetre ($\text{cm}^3$), Cubic metre ($\text{m}^3$), Cubic kilometre ($\text{km}^3$).
- Imperial System: Cubic inch ($\text{in}^3$ or cu in), Cubic foot ($\text{ft}^3$ or cu ft), Cubic yard ($\text{yd}^3$ or cu yd), Cubic mile ($\text{mi}^3$ or cu mi).
Basic Metric Conversions (Volume):
These conversions are derived by cubing the linear conversions:
- Since $1 \$ \text{cm} = 10 \$ \text{mm}$, then $1 \$ \text{cm}^3 = (10 \$ \text{mm})^3 = 1000 \$ \text{mm}^3$.
- Since $1 \$ \text{m} = 100 \$ \text{cm}$, then $1 \$ \text{m}^3 = (100 \$ \text{cm})^3 = 1,000,000 \$ \text{cm}^3$.
- Since $1 \$ \text{km} = 1000 \$ \text{m}$, then $1 \$ \text{km}^3 = (1000 \$ \text{m})^3 = 1,000,000,000 \$ \text{m}^3$.
Capacity and its Relationship with Volume
The term capacity is often used specifically for the volume of fluid or gas that a container can hold. Capacity is typically measured in units like litres (L) and millilitres (mL) in the metric system. There is a direct relationship between cubic volume units and capacity units:
- $\mathbf{1 \$ \text{millilitre (mL)} = 1 \$ \text{cubic centimetre (cm)}^3}$ (also known as cc). This is a fundamental conversion.
- Since $1 \$ \text{L} = 1000 \$ \text{mL}$, then $\mathbf{1 \$ \text{L} = 1000 \$ \text{cm}^3}$.
- Since $1 \$ \text{m}^3 = 1,000,000 \$ \text{cm}^3$, and $1000 \$ \text{cm}^3 = 1 \$ \text{L}$, then $1 \$ \text{m}^3 = \frac{1,000,000}{1000} \$ \text{L} = 1000 \$ \text{L}$. So, $\mathbf{1 \$ \text{cubic metre (m)}^3 = 1000 \$ \text{Litres (L)}}$. A kilolitre (kL) is defined as 1000 litres, so $1 \$ \text{m}^3 = 1 \$ \text{kL}$.
Summary of Common Volume and Capacity Conversions (Metric):
$\mathbf{1 \$ \text{mL} = 1 \$ \text{cm}^3}$
$\mathbf{1 \$ \text{L} = 1000 \$ \text{cm}^3}$
$\mathbf{1 \$ \text{m}^3 = 1000 \$ \text{L} = 1 \$ \text{kL}}$
Understanding volume and its units is crucial for calculations involving how much material a solid contains or displaces, or the capacity of containers.
Example (Conceptual)
Example 1. A small box measures $4$ cm in length, $3$ cm in breadth, and $2$ cm in height. How many unit cubes of side $1$ cm can fit inside it?
Answer:
Given:
Box dimensions: Length $= 4$ cm, Breadth $= 3$ cm, Height $= 2$ cm.
Unit cube side length $= 1$ cm.
To Find:
Number of unit cubes of side 1 cm that fit inside the box.
Solution:
A unit cube of side 1 cm has a volume of $1 \$ \text{cm} \times 1 \$ \text{cm} \times 1 \$ \text{cm} = 1 \$ \text{cm}^3$.
The number of unit cubes that fit along the length is 4.
The number of unit cubes that fit along the breadth is 3.
The number of unit cubes that fit along the height is 2.
The total number of unit cubes that can fill the box is the product of the numbers along each dimension:
Number of unit cubes $= \text{Length} \times \text{Breadth} \times \text{Height}$
"$= 4 \times 3 \times 2$"
"$= 12 \times 2$"
"$\mathbf{= 24}$"
Thus, 24 unit cubes of side 1 cm can fit inside the box. This number also represents the volume of the box in cubic centimetres, i.e., Volume $= 24 \$ \text{cm}^3$.
Volume Formulas for Cubes and Cuboids
Cubes and cuboids are the simplest solid shapes for understanding volume. Their volume formulas are derived directly from the concept of filling the space with unit cubes.
Cuboid
A cuboid is a three-dimensional solid bounded by six rectangular faces. It has three pairs of identical parallel faces. Its dimensions are typically referred to as length, breadth, and height.
Let the dimensions of the cuboid be:
- Length = $l$
- Breadth (or Width) = $b$
- Height = $h$
Derivation of the Volume Formula:
Imagine the base of the cuboid, which is a rectangle with dimensions $l \times b$. The area of this base is $A_{\text{base}} = l \times b$.
Now, imagine stacking layers of this base area up to the height $h$. If we consider thin slices of height $\Delta h$, the volume of each slice is approximately $A_{\text{base}} \times \Delta h$. Summing these infinitesimally thin slices from height 0 to $h$ gives the total volume.
Alternatively, based on the unit cube concept: If the dimensions $l, b, h$ are integers, we can fit $l$ unit cubes along the length, $b$ unit cubes along the breadth, and $h$ unit cubes along the height. The total number of unit cubes filling the cuboid is the product of these three numbers.
Volume of Cuboid $= (\text{Number of unit cubes along length}) \times (\text{Number along breadth}) \times (\text{Number along height})$
$= l \times b \times h$
... (1)
This concept extends to non-integer dimensions as well.
A more general way to think about it, applicable to many solids that have the same cross-sectional area throughout their height (like prisms and cylinders), is:
Volume $= \text{Area of Base} \times \text{Height}$
... (2)
For a cuboid with length $l$ and breadth $b$, the area of the base is $l \times b$. Substituting this and the height $h$ into (2):
$\mathbf{Volume_{\text{cuboid}} = l \times b \times h}$
... (3)
Formula:
The formula for the Volume of a cuboid is:
$\textbf{Volume of Cuboid} = \mathbf{Length \times Breadth \times Height}$
Or simply, $\mathbf{V = lbh}$. The unit is cubic units (e.g., $\text{cm}^3$, $\text{m}^3$).
Cube
A cube is a special type of cuboid where all three dimensions (length, breadth, and height) are equal. It is bounded by six congruent square faces. Let the length of each edge (or side) of the cube be '$a$'.
So, for a cube: $l = a$, $b = a$, and $h = a$.
Derivation of the Volume Formula:
We can use the volume formula for a cuboid, substituting $l=a$, $b=a$, and $h=a$ into formula (3):
$\text{Volume}_{\text{cube}} = l \times b \times h = a \times a \times a$
[Using (3) with $l=b=h=a$]
$\mathbf{Volume_{\text{cube}} = a^3}$
... (4)
Alternatively, using the general formula Volume = Area of Base $\times$ Height (2):
The base of a cube is a square with side length $a$. Area of Base $= a \times a = a^2$.
The height of the cube is $h = a$.
Volume $= a^2 \times a = a^3$
[Using (2) with base area $a^2$ and height $a$]
Formula:
The formula for the Volume of a cube is:
$\textbf{Volume of Cube} = \mathbf{(Side \$\$ Length)^3}$
Or simply, $\mathbf{V = a^3}$. The unit is cubic units (e.g., $\text{cm}^3$, $\text{m}^3$).
Examples
Example 1. Find the volume of a cuboid with length $8$ cm, breadth $5$ cm, and height $4$ cm.
Answer:
Given:
Cuboid with dimensions: Length, $l = 8$ cm; Breadth, $b = 5$ cm; Height, $h = 4$ cm.
To Find:
Volume of the cuboid, $V$.
Solution:
Using the formula for the volume of a cuboid, $V = l \times b \times h$. Using formula (3) derived above:
"$V = l \times b \times h$"
Substitute the given values:
"$V = 8 \$ \text{cm} \times 5 \$ \text{cm} \times 4 \$ \text{cm}$"
[Substituting values]
Perform the multiplication:
"$V = (8 \times 5) \times 4 \$ \text{cm}^3$"
"$V = 40 \times 4 \$ \text{cm}^3$"
"$\mathbf{V = 160 \$\$ cm^3}$"
The volume of the cuboid is 160 cubic centimetres ($\text{cm}^3$).
Example 2. A cubical box has an edge length of $1.5$ metres. Find its volume in cubic metres.
Answer:
Given:
Cubical box (cube).
Edge length, $a = 1.5$ m.
To Find:
Volume of the cube, $V$, in $\text{m}^3$.
Solution:
Using the formula for the volume of a cube, $V = a^3$. Using formula (4) derived above:
"$V = a^3$"
Substitute the given edge length:
"$V = (1.5 \$ \text{m})^3$"
[Substituting $a=1.5$ m]
"$V = 1.5 \$ \text{m} \times 1.5 \$ \text{m} \times 1.5 \$ \text{m}$"
Perform the multiplication:
"$1.5 \times 1.5 = 2.25$"
"$V = 2.25 \$ \text{m}^2 \times 1.5 \$ \text{m}$"
"$\mathbf{V = 3.375 \$\$ m^3}$"
The volume of the cubical box is 3.375 cubic metres ($\text{m}^3$).
Volume Formulas for Cylinders
Similar to prisms, the volume of a cylinder is found by multiplying the area of its base by its height. This is because a cylinder can be thought of as a stack of congruent circular layers.
We primarily consider a Right Circular Cylinder with:
- Radius ($r$): The radius of the circular base.
- Height ($h$): The perpendicular distance between the two circular bases.
Derivation of the Volume Formula:
The general principle for finding the volume of solids that have a constant cross-sectional area from base to top (prisms, cylinders) is:
Volume $= \text{Area of Base} \times \text{Height}$
... (1)
For a right circular cylinder, the base is a circle with radius $r$. The area of this circular base is $A_{\text{base}} = \pi r^2$.
The height of the cylinder is $h$.
Substitute the area of the base and the height into formula (1):
$\text{Volume}_{\text{cylinder}} = (\pi r^2) \times h$
[Substituting base area and height into (1)]
$\mathbf{Volume_{\text{cylinder}} = \pi r^2 h}$
... (2)
This formula represents the idea of stacking up circles of area $\pi r^2$ to a height of $h$.
Formula:
The formula for the Volume of a right circular cylinder is:
$\textbf{Volume of Cylinder} = \mathbf{\pi \times (Radius)^2 \times Height}$
Or simply, $\mathbf{V = \pi r^2 h}$. The unit is cubic units (e.g., $\text{cm}^3$, $\text{m}^3$).
Examples
Example 1. Find the volume of a cylinder whose base radius is $7$ cm and height is $15$ cm. (Use $\pi = \frac{22}{7}$)
Answer:
Given:
Cylinder.
Base radius, $r = 7$ cm.
Height, $h = 15$ cm.
Value of $\pi = \frac{22}{7}$.
To Find:
Volume of the cylinder, $V$.
Solution:
Using the formula for the volume of a cylinder, $V = \pi r^2 h$. Using formula (2) derived above:
"$V = \pi r^2 h$"
Substitute the given values for $\pi$, $r$, and $h$:
"$V = \frac{22}{7} \times (7 \$ \text{cm})^2 \times 15 \$ \text{cm}$"
[Substituting values]
"$V = \frac{22}{7} \times 49 \$ \text{cm}^2 \times 15 \$ \text{cm}$"
Simplify by cancelling the common factor 7:
"$V = \frac{22}{\cancel{7}_1} \times \cancel{49}^7 \$ \text{cm}^2 \times 15 \$ \text{cm}$"
"$V = 22 \times 7 \times 15 \$ \text{cm}^3$"
Perform the multiplication:
"$V = (22 \times 7) \times 15 \$ \text{cm}^3 = 154 \times 15 \$ \text{cm}^3$"
"$\mathbf{V = 2310 \$\$ cm^3}$"
The volume of the cylinder is 2310 cubic centimetres ($\text{cm}^3$).
Example 2. The volume of a cylinder is $308$ $\text{m}^3$. If its height is $8$ m, find the radius of its base. (Use $\pi = \frac{22}{7}$)
Answer:
Given:
Cylinder.
Volume, $V = 308 \$ \text{m}^3$.
Height, $h = 8$ m.
Value of $\pi = \frac{22}{7}$.
To Find:
Radius of the base, $r$.
Solution:
Using the formula for the volume of a cylinder, $V = \pi r^2 h$. Using formula (2):
"$V = \pi r^2 h$"
We are given $V$ and $h$, and need to find $r$. Substitute the given values:
"$308 \$ \text{m}^3 = \frac{22}{7} \times r^2 \times 8 \$ \text{m}$"
[Substituting values]
Rearrange the equation to solve for $r^2$. Divide both sides by $\frac{22}{7} \times 8$:
"$r^2 = \frac{308}{\frac{22}{7} \times 8} \$ \text{m}^2$"
"$r^2 = \frac{308 \times 7}{22 \times 8} \$ \text{m}^2$"
Simplify the fraction. Divide 308 by 22: $308 \div 22 = 14$.
"$r^2 = \frac{\cancel{308}^{14} \times 7}{\cancel{22}_1 \times 8} \$ \text{m}^2$"
"$r^2 = \frac{14 \times 7}{8} \$ \text{m}^2$"
Simplify further by dividing 14 and 8 by 2:
"$r^2 = \frac{\cancel{14}^7 \times 7}{\cancel{8}_4} \$ \text{m}^2$"
"$r^2 = \frac{7 \times 7}{4} \$ \text{m}^2 = \frac{49}{4} \$ \text{m}^2$"
Now, take the square root of both sides to find $r$. Since $r$ is a length, it must be positive:
"$r = \sqrt{\frac{49}{4}} \$ \text{m}$"
"$r = \frac{\sqrt{49}}{\sqrt{4}} \$ \text{m} = \frac{7}{2} \$ \text{m}$"
"$\mathbf{r = 3.5 \$\$ m}$"
The radius of the base of the cylinder is 3.5 metres (m).
Volume Formulas for Cones
A cone is a three-dimensional solid that tapers smoothly from a flat base (usually circular) to a single point called the apex or vertex. We typically deal with a Right Circular Cone, where the base is a circle, and the apex is directly above the center of the base.
To calculate the volume of a right circular cone, we need two key dimensions:
- Radius ($r$): The radius of the circular base.
- Perpendicular Height ($h$): The distance from the apex to the center of the base, measured along the axis of the cone. (Note: The slant height ($l$) is involved in surface area formulas but not directly in the volume formula).
Relationship between Volume of Cone and Cylinder
There is a definite relationship between the volume of a cone and the volume of a cylinder that has the same base radius and the same height. It is a well-established geometric fact, often demonstrated through experiments (like filling one from the other) or derived using calculus, that the volume of a cone is exactly one-third the volume of a cylinder with identical base radius and height.
Formula Derivation:
Let's use the relationship between the volume of a cone and a cylinder. The volume of a cylinder with radius $r$ and height $h$ is $V_{\text{cylinder}} = \pi r^2 h$.
The volume of a cone with the same radius $r$ and height $h$ is one-third of the cylinder's volume:
$\text{Volume}_{\text{cone}} = \frac{1}{3} \times \text{Volume}_{\text{cylinder \$\$ with \$\$ same \$\$ base \$\$ and \$\$ height}}$
... (1)
Substitute the formula for the volume of the cylinder:
$\mathbf{Volume_{\text{cone}} = \frac{1}{3} \pi r^2 h}$
... (2)
Alternatively, this formula is a specific instance of the general volume formula for any pyramid (a solid tapering from a polygonal base to an apex): Volume $= \frac{1}{3} \times (\text{Area of Base}) \times (\text{Height})$. For a cone, the base is a circle with area $\pi r^2$.
Volume $= \frac{1}{3} \times (\pi r^2) \times h = \frac{1}{3} \pi r^2 h$
[Using general pyramid volume formula]
Formula:
The formula for the Volume of a right circular cone is:
$\textbf{Volume of Cone} = \mathbf{\frac{1}{3} \pi \times (Radius)^2 \times Height}$
Or simply, $\mathbf{V = \frac{1}{3} \pi r^2 h}$. The unit is cubic units (e.g., $\text{cm}^3$, $\text{m}^3$).
Note that the slant height ($l$) is not needed for calculating the volume. If $r$ and $h$ are given, you can calculate the volume directly. If $r$ and $l$ are given, you would first need to find $h$ using $h = \sqrt{l^2 - r^2}$ before calculating the volume.
Examples
Example 1. Find the volume of a cone whose base radius is $6$ cm and height is $14$ cm. (Use $\pi = \frac{22}{7}$)
Answer:
Given:
Cone.
Base radius, $r = 6$ cm.
Height, $h = 14$ cm.
Value of $\pi = \frac{22}{7}$.
To Find:
Volume of the cone, $V$.
Solution:
Using the formula for the volume of a cone, $V = \frac{1}{3} \pi r^2 h$. Using formula (2) derived above:
"$V = \frac{1}{3} \pi r^2 h$"
Substitute the given values for $\pi$, $r$, and $h$:
"$V = \frac{1}{3} \times \frac{22}{7} \times (6 \$ \text{cm})^2 \times 14 \$ \text{cm}$"
[Substituting values]
"$V = \frac{1}{3} \times \frac{22}{7} \times 36 \$ \text{cm}^2 \times 14 \$ \text{cm}$"
Simplify by cancelling common factors (7 with 14, and 3 with 36):
"$V = \frac{1}{\cancel{3}_1} \times \frac{22}{\cancel{7}_1} \times \cancel{36}^{12} \times \cancel{14}^2 \$ \text{cm}^3$"
"$V = 1 \times 22 \times 12 \times 2 \$ \text{cm}^3$"
Perform the multiplication:
"$V = 22 \times (12 \times 2) \$ \text{cm}^3 = 22 \times 24 \$ \text{cm}^3$"
"$\mathbf{V = 528 \$\$ cm^3}$"
The volume of the cone is 528 cubic centimetres ($\text{cm}^3$).
Example 2. The volume of a cone is $1232$ $\text{cm}^3$. If its base radius is $7$ cm, find its height. (Use $\pi = \frac{22}{7}$)
Answer:
Given:
Cone.
Volume, $V = 1232 \$ \text{cm}^3$.
Base radius, $r = 7$ cm.
Value of $\pi = \frac{22}{7}$.
To Find:
Height of the cone, $h$.
Solution:
Using the formula for the volume of a cone, $V = \frac{1}{3} \pi r^2 h$. Using formula (2):
"$V = \frac{1}{3} \pi r^2 h$"
We are given $V$ and $r$, and need to find $h$. Substitute the given values:
"$1232 \$ \text{cm}^3 = \frac{1}{3} \times \frac{22}{7} \times (7 \$ \text{cm})^2 \times h$"
[Substituting values]
"$1232 = \frac{1}{3} \times \frac{22}{7} \times 49 \times h$"
Simplify the terms multiplying $h$. Cancel 7 with 49:
"$1232 = \frac{1}{3} \times 22 \times \cancel{49}^7 \times h$"
"$1232 = \frac{1}{3} \times (22 \times 7) \times h$"
"$1232 = \frac{1}{3} \times 154 \times h$"
"$1232 = \frac{154}{3} h$"
Rearrange the equation to solve for $h$. Multiply both sides by $\frac{3}{154}$:
"$h = 1232 \times \frac{3}{154} \$ \text{cm}$"
Simplify the fraction. Divide 1232 by 154. Let's try dividing by common factors. Both are even. $1232 \div 2 = 616$, $154 \div 2 = 77$. 616 is divisible by 77 ($616 = 8 \times 77$).
"$h = \cancel{1232}^{616} \times \frac{3}{\cancel{154}_{77}} \$ \text{cm}$"
[Divide by 2]
"$h = \cancel{616}^8 \times \frac{3}{\cancel{77}_1} \$ \text{cm}$"
[Divide by 77]
"$h = 8 \times 3 \$ \text{cm}$"
"$\mathbf{h = 24 \$\$ cm}$"
The height of the cone is 24 centimetres (cm).
Volume Formulas for Spheres and Hemispheres
Spheres and hemispheres are three-dimensional shapes with curved surfaces. Their volume formulas are distinct from polyhedrons and cylinders/cones, though they are related through higher mathematical concepts or historical derivations.
Sphere
A sphere is a perfectly symmetrical solid where all points on its surface are equidistant from a central point. This distance is called the radius ($r$).
Formula for Volume of a Sphere:
The formula for the volume of a sphere is one of the famous results from ancient Greek mathematics, attributed to Archimedes. Unlike simple solids like cuboids or cylinders, its derivation requires methods that were advanced for his time (and now typically involve calculus). However, the formula itself is relatively simple and fundamental.
The formula for the volume ($V$) of a sphere with radius $r$ is:
$\mathbf{Volume_{\text{sphere}} = \frac{4}{3} \pi r^3}$
... (1)
This formula shows that the volume depends on the cube of the radius, which is expected for a three-dimensional measure.
Archimedes' Discovery (Relationship with Cylinder):
Archimedes also discovered a remarkable relationship between the volume of a sphere and the volume of the smallest right circular cylinder that can completely contain it. This circumscribing cylinder would have a radius equal to the sphere's radius ($r$) and a height equal to the sphere's diameter ($2r$).
The volume of this circumscribing cylinder is:
Volume of Cylinder $= \pi \times (\text{cylinder radius})^2 \times (\text{cylinder height})$
$= \pi \times (r)^2 \times (2r)$
$= 2 \pi r^3$
Archimedes proved that the volume of the sphere is exactly two-thirds of the volume of this circumscribing cylinder:
Volume of Sphere $= \frac{2}{3} \times (\text{Volume of Circumscribing Cylinder})$
$= \frac{2}{3} \times (2 \pi r^3)$
$\mathbf{= \frac{4}{3} \pi r^3}$
This relationship confirms the formula for the volume of a sphere.
Formula:
The formula for the Volume of a sphere with radius $r$ is:
$\textbf{Volume of Sphere} = \mathbf{\frac{4}{3} \pi \times (Radius)^3}$
Or simply, $\mathbf{V = \frac{4}{3} \pi r^3}$. The unit is cubic units (e.g., $\text{cm}^3$, $\text{m}^3$).
Hemisphere
A hemisphere is exactly half of a sphere, obtained by cutting the sphere through its center with a plane. It consists of a curved surface and a flat circular base. Let the radius be $r$. The radius of the flat circular base is the same as the radius of the original sphere.
Formula for Volume of a Hemisphere:
The volume of a hemisphere is simply half the volume of the corresponding full sphere with the same radius.
$\text{Volume}_{\text{hemisphere}} = \frac{1}{2} \times (\text{Volume}_{\text{sphere \$\$ with \$\$ same \$\$ radius}})$
... (2)
Substitute the formula for the volume of a sphere from (1):
$\text{Volume}_{\text{hemisphere}} = \frac{1}{2} \times \left( \frac{4}{3} \pi r^3 \right)$
[Substituting from (1) into (2)]
Simplify the expression:
"$= \frac{1}{\cancel{2}_1} \times \frac{\cancel{4}^2}{3} \pi r^3$"
$\mathbf{Volume_{\text{hemisphere}} = \frac{2}{3} \pi r^3}$
... (3)
Formula:
The formula for the Volume of a hemisphere with radius $r$ is:
$\textbf{Volume of Hemisphere} = \mathbf{\frac{2}{3} \pi \times (Radius)^3}$
Or simply, $\mathbf{V = \frac{2}{3} \pi r^3}$. The unit is cubic units (e.g., $\text{cm}^3$, $\text{m}^3$).
Examples
Example 1. Find the volume of a sphere whose radius is $21$ cm. (Use $\pi = \frac{22}{7}$)
Answer:
Given:
Sphere.
Radius, $r = 21$ cm.
Value of $\pi = \frac{22}{7}$.
To Find:
Volume of the sphere.
Solution:
Using the formula for the volume of a sphere, $V = \frac{4}{3} \pi r^3$. Using formula (1) derived above:
"$V = \frac{4}{3} \pi r^3$"
Substitute the given values for $\pi$ and $r$:
"$V = \frac{4}{3} \times \frac{22}{7} \times (21 \$ \text{cm})^3$"
[Substituting values]
"$V = \frac{4}{3} \times \frac{22}{7} \times 21 \$ \text{cm} \times 21 \$ \text{cm} \times 21 \$ \text{cm}$"
Simplify by cancelling common factors (3 and 7 with 21):
"$V = \frac{4}{\cancel{3}_1} \times \frac{22}{\cancel{7}_1} \times \cancel{21}^7 \times \cancel{21}^3 \times 21 \$ \text{cm}^3$"
[Cancel 3 with 21, Cancel 7 with 21]
"$V = 4 \times 22 \times 7 \times 3 \times 21 \$ \text{cm}^3$"
"$V = (4 \times 22) \times (7 \times 3) \times 21 \$ \text{cm}^3$"
"$V = 88 \times 21 \times 21 \$ \text{cm}^3$"
"$V = 88 \times 441 \$ \text{cm}^3$"
Perform the multiplication: $88 \times 441 = 88 \times (400 + 40 + 1) = 35200 + 3520 + 88 = 38808$.
$\begin{array}{cc}& & 4 & 4 & 1 \\ \times & & & 8 & 8 \\ \hline & 35 & 2 & 8 \\ 352 & 8 & \times \\ \hline 388 & 0 & 8 \\ \hline \end{array}$"$\mathbf{V = 38808 \$\$ cm^3}$"
The volume of the sphere is 38808 cubic centimetres ($\text{cm}^3$).
Example 2. Find the volume of a solid hemisphere of radius $3.5$ cm. (Use $\pi = \frac{22}{7}$)
Answer:
Given:
Solid hemisphere.
Radius, $r = 3.5$ cm.
Value of $\pi = \frac{22}{7}$.
To Find:
Volume of the hemisphere.
Solution:
Using the formula for the volume of a hemisphere, $V = \frac{2}{3} \pi r^3$. Using formula (3) derived above:
"$V = \frac{2}{3} \pi r^3$"
Substitute the given values for $\pi$ and $r$. It's helpful to write $3.5$ as a fraction: $3.5 = 3 \frac{1}{2} = \frac{7}{2}$.
"$V = \frac{2}{3} \times \frac{22}{7} \times (3.5 \$ \text{cm})^3$"
[Substituting values]
"$V = \frac{2}{3} \times \frac{22}{7} \times \left(\frac{7}{2} \$ \text{cm}\right)^3$"
"$V = \frac{2}{3} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times \frac{7}{2} \$ \text{cm}^3$"
Simplify by cancelling common factors (7 with 7, and 2 with one of the 2s):
"$V = \frac{\cancel{2}_1}{3} \times \frac{22}{\cancel{7}_1} \times \frac{\cancel{7}^1}{\cancel{2}_1} \times \frac{7}{2} \times \frac{7}{2} \$ \text{cm}^3$"
[Cancel 2 with 2, Cancel 7 with 7]
"$V = \frac{1}{3} \times 22 \times 1 \times \frac{7}{2} \times \frac{7}{2} \$ \text{cm}^3$"
"$V = \frac{22 \times 49}{3 \times 4} \$ \text{cm}^3$"
Simplify further by dividing 22 and 4 by 2:
"$V = \frac{\cancel{22}^{11} \times 49}{3 \times \cancel{4}_2} \$ \text{cm}^3$"
"$V = \frac{11 \times 49}{3 \times 2} \$ \text{cm}^3 = \frac{539}{6} \$ \text{cm}^3$"
Perform the division to get a decimal value:
"$539 \div 6 \approx 89.8333...$"
"$\mathbf{V \approx 89.83 \$\$ cm^3}$"
[Rounding to two decimal places]
The volume of the solid hemisphere is approximately 89.83 cubic centimetres ($\text{cm}^3$).
Volume of Solid Figures (Consolidated)
This table provides a consolidated summary of the formulas for calculating the volume of the standard solid shapes discussed in the previous sections. These formulas are derived from the concept of stacking base areas (for prisms and cylinders) or are based on established geometric relationships (for pyramids, cones, and spheres).
Summary Table of Volume Formulas
| Solid Shape | Dimensions | Volume Formula ($V$) | General Principle |
|---|---|---|---|
| Cuboid | Length $l$, Breadth $b$, Height $h$ | $\mathbf{V = lbh}$ | Area of Base × Height |
| Cube | Side $a$ | $\mathbf{V = a^3}$ | Area of Base × Height |
| Right Circular Cylinder | Radius $r$, Height $h$ | $\mathbf{V = \pi r^2 h}$ | Area of Circular Base × Height |
| Right Circular Cone | Radius $r$, Height $h$ | $\mathbf{V = \frac{1}{3} \pi r^2 h}$ | $\frac{1}{3}$ × Area of Circular Base × Height |
| Sphere | Radius $r$ | $\mathbf{V = \frac{4}{3} \pi r^3}$ | (Related to circumscribing cylinder volume) |
| Hemisphere | Radius $r$ | $\mathbf{V = \frac{2}{3} \pi r^3}$ | $\frac{1}{2}$ × Volume of Sphere |
General Principles for Volume:
Based on the structures of these solids, we can observe general patterns for volume calculation:
- For solids with a constant cross-sectional area from base to top (like Prisms and Cylinders):
$\mathbf{Volume = Area \$\$ of \$\$ Base \times Height}$
- For solids that taper uniformly from a base to a single apex (like Pyramids and Cones):
$\mathbf{Volume = \frac{1}{3} \times Area \$\$ of \$\$ Base \times Height}$
These general principles help in understanding and remembering the volume formulas for these categories of solids.
Always ensure that all dimensions used in the formula are in the same units. The unit of volume will be the cube of the linear unit.