The volume of a Sphere: It is the amount of space contained by a sphere. The volume is calculated by the integration method and measured in cubic units. The formula to calculate the volume of a sphere is,

where r is the radius of the sphere.

The size of the sphere determines the volume of the sphere and is based on the radius. The more the radius, the more the volume would be. A sphere has three axes- the X-axis, Y-axis and Z-axis. Some examples for sphere which we all see in our daily lives include basketball, football, globe, etc. Let us learn more about the volume of a sphere, its formula, derivation and its types. 

What is a Sphere?

A sphere is a three-dimensional geometrical shape. A ball, an orange are common examples of sphere. The three coordinate axes, (x)-axis, (y)-axis and (z)-axis, define the shape of a sphere. Like a circle, the sphere also has a centre point.

Every point on the surface of a sphere is equidistant from its centre. This fixed distance from the centre to any point on the sphere’s circumference is called the sphere’s radius.

Consider a football. Can we say the shape of the football is a circle? No! But there is a connection of circle in a football. How? Try rotating a circle along any of its diameters. Can you see that the football takes the shape of a fear?

Is it a similar shape to a football? Yes, it is. The sphere is a three-dimensional shape obtained while rotating a circle along any of its diameters.

Types of Spheres

There are two types of spheres:

1. Solid Sphere

A solid sphere is a three-dimensional object. It is a sphere which is filled up with the material it is made up of. For example, marbles are solid spheres.

2. Hollow Sphere

A hollow sphere is a sphere that has only the outer spherical boundary and nothing filled inside.

For easy understanding, consider an ice cream ball like shown below.

An ice cream ball with ice cream inside is an example of a solid sphere and after the ice cream is eaten , the empty ball is an example of the hollow sphere. In the upcoming section of this article, we will learn how to calculate the volume of solid sphere and hollow sphere, with examples.

What is Volume?

The volume of a three-dimensional solid is the amount of space it occupies or space enclosed by a boundary or the capacity to hold something. Read more about Definition of volume for different geometrical figures.

For example, the volume of a cuboidal box indicates the amount of water or any substance contained in it. If we consider the case of a ball ice cream, the amount of ice cream that can be filled inside the ball without any gap is the volume of the ice cream ball.

What is Volume of Sphere

The sphere volume is the amount of space contained by it. Let us understand this. Try poking a football and making a hole. Fill it with water to the brim. Now, pour the water into a beaker and measure how much water is contained in it. That is the volume of that football. In this section we shall learn more about volume of solid sphere and volume of hollow sphere through various examples.

Let us check another example:
Fill a mug with water up to its brim and keep it inside a bucket. Now slowly drop a tennis ball into that. Some water will overflow from the mug to the bucket. The amount of water displaced indicates the volume of the tennis ball dropped.

The volume of a sphere is measured in cubic units like \({{\rm{m}}^3},\,{\rm{c}}{{\rm{m}}^3}\)etc.

 

The Value of \(\pi \)

A constant term \(\pi \) is used in the formula for calculating the area of a circle. \(\pi \) is a constant term, also known as Archimedes’ Constant. It can be defined as the ratio of the circumference of a circle to its diameter. It is an irrational number whose value is \(3.141592653589793238\). For calculation purpose, the value of \(\pi \) is approximately taken as \(3.14\) when used as a decimal number and is taken as \(\frac{{22}}{7}\) when used as a fraction.

Definition of the Volume of a Sphere?

The volume of the sphere is calculated by the integration method. Assuming the sphere is made up of numerous thin circular disks arranged one over the other. These circular disks will have continuously varying radius, but their centres will be collinear.

The volume can be written as the product of the area of the circle and its thickness. Integrating the product of radius and thickness, we will obtain the volume of the sphere.

How to Calculate the Volume of Sphere?

The volume of a sphere is equal to four-thirds of the product of \(\pi \) and the cube of the radius.

(a) Volume of a Sphere When the Radius of the Sphere Is Given

The volume of a sphere can be calculated using the formula,
Volume of a sphere is \(V = \frac{4}{3}\pi {r^3}\) cubic units
Where \(r\) is the radius of the sphere.

(b) Volume of a Sphere When the Diameter of the Sphere Is Given

Volume of a sphere is \(V = \frac{4}{3}\pi {\left( {\frac{d}{2}} \right)^3}\) cubic units
Where \(d\) is the diameter of the sphere.

Factors Affecting the Volume of a Sphere

The volume of the sphere is determined by the size of the sphere. The size is based on the radius of the sphere. The more the radius, the more will be the volume of a sphere.

 

The volume of a sphere is given by \(V = \frac{4}{3}\pi {r^3},\) where \(r\) is the radius of the sphere.

 

In the formula for volume of a sphere, \(\frac{4}{3}\) and \(\pi\) are constants. The volume of a sphere directly depends on radius of the sphere:

 

\(V \propto {r^3}\)

 

If we compare two spheres of different size, the sphere with greater radius will be having larger volume. This could be easily understood with the help of figures as given below.

Here, the pink sphere is having a radius of \(1\,{\rm{m}}\) and blue sphere is having a radius of \(2\,{\rm{m}}{\rm{.}}\) As the radius of the blue sphere is more than that of pink, volume of blue sphere will be larger than the the pink sphere.

Volume of a Sphere Unit

The unit of the radius of a sphere can be any units of length like \({\rm{mm,}}\,{\rm{cm,}}\,{\rm{m,}}\) etc. Since the radius is cubed in the formula of volume of a sphere, the unit should also be cubed. Hence, the unit of volume of a sphere will be the cube of any unit of length like \({\rm{m}}{{\rm{m}}^3}{\rm{,}}\,{\rm{c}}{{\rm{m}}^3}{\rm{,}}\,{{\rm{m}}^3}{\rm{,}}\) etc.

Applications of the Volume of a Sphere

The knowledge of the volume of a sphere helps us in many ways.

There are a lot of games played with football, basketball, tennis ball, cricket ball, snooker balls, golf ball, etc. All these balls are spherical but vary in size and texture. The size of these balls is decided by the radius of the sphere.

All our planets, including Earth, are considered to be nearly spherical in shape. The Sun, the Moon and the stars are also nearly spherical. Steel or metal balls used in rolling mechanisms such as wheels, tooling, bearings, are spherical.

The knowledge of volume is essential in calculating the mass of objects. Suppose, if we need to find the mass of a metal ball, we can find out if we know the density of the material as:
\({\rm{mass}} = {\rm{volume}} \times {\rm{density}}\)

Hemisphere and its Volume

Let us learn the definition of hemisphere and the formula to calculate the volume of a hemisphere, through a general example.

What is Hemisphere?

Take a lemon that is spherical and cut it into two halves. The two pieces of lemon that we have now are nothing but two hemispheres. A hemisphere is an exact half of a sphere. In essence, two identical hemispheres make a sphere.

Volume of a Hemisphere

As a hemisphere is a three-dimensional shape and exactly the half of a sphere, the volume of the hemisphere will also be exactly half of the volume of a sphere.

 

Volume of a hemisphere is \(V = \frac{2}{3}\pi {r^3}\) cubic units
Where \(r\) is the radius of the hemisphere.

Related Topics to Study

Check out a few interesting articles related to the topic.

1. Sphere 2. Volume of a Cuboid 3. Volume of a Cube 4. Volume of Solids 5. Surface Area of a Sphere 6. Volume of a Hemisphere

Solved Examples

Below are a few solved examples that can help in getting a better idea.

Q.1. Find the volume of a sphere whose radius is \(6\,{\rm{cm}}\) considering \(\pi = \frac{{22}}{7}.\)
Ans: Given the radius of a sphere, \(r = 6\,{\rm{cm}}\)
We know that the volume of a sphere is calculated as
\(V = \frac{4}{3}\pi {r^3}.\)
So, the volume of a sphere of radius \(6\,{\rm{cm}} = \frac{4}{3} \times {6^3}\,{\rm{c}}{{\rm{m}}^3}\)
\( = \frac{4}{3} \times \frac{{22}}{7} \times 6 \times 6 \times 6\,{\rm{c}}{{\rm{m}}^3}\)
\( = \frac{{4 \times 22 \times 6 \times 6 \times 2}}{{3 \times 7}}{\rm{c}}{{\rm{m}}^3}\)
\( = 905.14\,{\rm{c}}{{\rm{m}}^3}.\)

Q.2. Find the volume of sphere whose diameter is \(6\,{\rm{cm}}\) considering \(\pi = \frac{{22}}{7}.\)
Ans: Given, the diameter of a sphere \(6\,{\rm{cm}}\)
So, the radius of the sphere \( = \frac{{{\rm{diameter}}}}{2} = \frac{{6\,{\rm{cm}}}}{2} = 3\,{\rm{cm}}\)
We know that the volume of a sphere is calculated as
\(V = \frac{4}{3}\pi {r^3}.\)
So, the volume of a sphere of radius \(3\,{\rm{cm}} = \frac{4}{3} \times \frac{{22}}{7} \times 3 \times 3 \times 3\,{\rm{c}}{{\rm{m}}^3}\) \( = \frac{{4 \times 22 \times 3 \times 3}}{7}{\rm{c}}{{\rm{m}}^{\rm{3}}}\)
\( = 113.14\,{\rm{c}}{{\rm{m}}^3}\)

Q.3. The volume of a sphere is \(523\,{\rm{cubic}}\,{\rm{meters}}.\) Find the radius of the sphere considering \(\pi = 3.14.\)
Ans: Given the volume of a sphere \( = 523\,{\rm{cubic}}\,{\rm{meters}}.\)
We know that the volume of a sphere is calculated as:
\(V = \frac{4}{3}\pi {r^3}\)
We can find the radius of the sphere from the volume of a sphere as
\(r = \sqrt[3]{{\frac{{V \times 3}}{{4\pi }}}}\)
So, the radius of the given sphere
\(r = \sqrt[3]{{\left( {\frac{{523 \times 3}}{{4 \times 3.14}}} \right)}}{\rm{m}}\)
\( = \sqrt[3]{{\left( {124.9} \right)}}{\rm{m}}\)
\( \approx 5\,{\rm{m}}\)

Q.4. What is the amount of water that can be held by a spherical ball of radius \(21\,{\rm{cm}}?\)
Ans: To find the amount of water that can be held by a spherical ball, we need to find the volume of the ball. Since the ball is a sphere, the volume of the ball can be found using the volume of a sphere.
We know that the volume of a sphere is calculated as
\(V = \frac{4}{3}\pi {r^3}\)
Given, the radius of the ball \( = 21\,{\rm{cm}}\)
So, the volume of the ball \( = \frac{4}{3} \times \frac{{22}}{7} \times 21 \times 21 \times 21\,{\rm{c}}{{\rm{m}}^3}\)
\( = \frac{{4 \times 22 \times 21 \times 21}}{1}\,{\rm{c}}{{\rm{m}}^3}\)
\( = 38808\,{\rm{c}}{{\rm{m}}^3}\)
Hence, the amount of water that can be held by the spherical ball of radius \(21\,{\rm{cm}}\) is \(38,808\;\) cubic centimetres.

Q.5. Rachana has three wax marbles of radii \(6\,{\rm{cm}},\,8\,{\rm{cm}}\) and \(10\,{\rm{cm}}.\) She melted all three marbles and recast them into a single solid marble. Find the radius of the new single solid marble?
Ans: The marbles are nothing but spheres. When two or more spheres are melted and recast into another big sphere, their volume will remain the same.
That is,
Volume of the new single solid marble \( = \) Volume of marble of \(6\,{\rm{cm}} + \) Volume of marble of \(8\,{\rm{cm}} + \) Volume of marble of \(10\,{\rm{cm}}\)
We know that the volume of a sphere is calculated as \(V = \frac{4}{3}\pi {r^3}\)
Considering the radius of the single solid marble as \(R,\) we get the volume as
\(\frac{4}{3}\pi {R^3} = \frac{4}{3}\pi \times {6^3} + \frac{4}{3}\pi \times {8^3} + \frac{4}{3}\pi \times {10^3}\)
\({R^3} = {6^3} + {8^3} + {10^3}\)
\({R^3} = 1728\)
\(R = 12\)
Hence, the radius of the single solid marble is \(12\,{\rm{cm}}{\rm{.}}\)

Q.6. A Christmas tree was decorated with \(10\) Christmas balls each of radius \(2\,{\rm{cm}}{\rm{.}}\) Find the amount of glitter that can be filled in it.
Ans: Radius of Christmas ball \( = 2\,{\rm{cm}}\)
Number of Christmas balls \(= 10\)
To find the amount of glitter that can be filled in it, we should find the volume of the Christmas ball.
Volume of Christmas ball \( = \) Volume of a sphere of \(2\,{\rm{cm}}\)
We know that the volume of a sphere is calculated as
\(V = \frac{4}{3}\pi {r^3}\)
So, the volume of one Christmas ball \( = \frac{4}{3}\pi \times {2^3}\,{\rm{c}}{{\rm{m}}^3}\)
\( = \frac{4}{3} \times \frac{{22}}{7} \times 2 \times 2 \times 2\,{\rm{c}}{{\rm{m}}^3}\)
\( = 33.52\,{\rm{c}}{{\rm{m}}^3}\)
Hence, the volume of \(10\) christmas balls \( = 335.2\,{\rm{c}}{{\rm{m}}^3}.\)

Q.7. A hemispherical bowl has a radius of \(21\,{\rm{cm}}.\) How much amount of water it can contain?
Ans: To find the amount of water that can be held by a hemispherical bowl, we need to find the volume of the bowl. Since the bowl is a hemisphere, the volume of the bowl can be found using the volume of a hemisphere.
We know that the volume of a hemisphere is calculated as
\(V = \frac{2}{3}\pi {r^3}\)
Given, the radius of the ball \( = 21\,{\rm{cm}}\)
So, the volume of the ball \( = \frac{2}{3} \times \frac{{22}}{7} \times 21 \times 21 \times 21\,{\rm{c}}{{\rm{m}}^3}\)
\( = \frac{{2 \times 22 \times 21 \times 21}}{1}{\rm{c}}{{\rm{m}}^3}\)
\( = 19404\,{\rm{c}}{{\rm{m}}^3}\)
Hence, the amount of water that can be held by the hemispherical bowl of radius \(21\,{\rm{cm}}\) is \(19,404\;\) cubic centimetres.

Q.8. Find the amount of water displaced by a solid spherical ball of diameter \(42\,{\rm{cm}}{\rm{.}}\)
Ans: The amount of water displaced by a sphere is its volume. Radius of the sphere \( = \frac{{42}}{2}\,{\rm{cm = 21}}\,{\rm{cm}}{\rm{.}}\)
We know that the volume of a sphere is calculated as
\(V = \frac{4}{3}\pi {r^3}\)
So, the amount of water displaced \( = \) Volume of the sphere of \(21\,{\rm{cm}}\)
\( = \frac{4}{3}\pi \times {21^3}\,{\rm{c}}{{\rm{m}}^3}\)
\( = \frac{4}{3} \times \frac{{22}}{7} \times 21 \times 21 \times 21\,{\rm{c}}{{\rm{m}}^3}\)
\( = \frac{{4 \times 22 \times 21 \times 21}}{1}\,{\rm{c}}{{\rm{m}}^3}\)
\(=38808\,{\rm{c}}{{\rm{m}}^3}\)
Hence, \(38808\,{\rm{c}}{{\rm{m}}^3}\) of water is displaced by a solid spherical ball of diameter \(42\,{\rm{cm}}{\rm{.}}\)

Q.9. The diameter of a metallic ball is \(14\,{\rm{cm}}{\rm{.}}\) What is the mass of the ball, if the density of the metal is \(10\,{\rm{g}}\,{\rm{per}}\,{\rm{c}}{{\rm{m}}^{\rm{3}}}.\)
Ans: The mass of the ball can be found as
\({\rm{mass}} = {\rm{volume}} \times {\rm{density}}\)
Radius of the sphere \( = \frac{{14}}{2}{\rm{cm}} = 7\,{\rm{cm}}\)
The metallic ball is a sphere and the volume of a sphere is calculated as
\(V = \frac{4}{3}\pi {r^3}\)
\( = \frac{4}{3} \times \frac{{22}}{7} \times 7 \times 7 \times 7\,{\rm{c}}{{\rm{m}}^3}\)
\( = \frac{{4 \times 22 \times 7 \times 7}}{3}{\rm{c}}{{\rm{m}}^3}\)
\( = 1437.33\,{\rm{c}}{{\rm{m}}^3}\)
Hence, the mass of the ball \( = 1437.33\,{\rm{c}}{{\rm{m}}^3} \times 10\,{\rm{g}}\,{\rm{per}}\,{\rm{c}}{{\rm{m}}^{\rm{3}}}\)
\( = 14373.3\,{\rm{g}}\)

Q.10. Madhurima was preparing mango ice cream. She was filling the ice cream in ice cream balls. Each ice cream ball is of radius \(3\,{\rm{cm}}.\) How much amount of ice cream can be filled in \(20\) such balls?
Ans: Radius of ice cream ball \( = 3\,{\rm{cm}}\)
Number of ice cream balls \(=20\)
To find the amount of ice cream that can be filled in it, we should find the volume of the ice cream ball.
Volume of ice cream ball \( = \) Volume of a sphere of \(3\,{\rm{cm}}\)
We know that the volume of a sphere is calculated as
\(V = \frac{4}{3}\pi {r^3}\)
So, the volume of one ice cream ball \( = \frac{4}{3}\pi {3^3}\,{\rm{c}}{{\rm{m}}^3}\)
\( = \frac{4}{3} \times \frac{{22}}{7} \times 3 \times 3 \times 3\,{\rm{c}}{{\rm{m}}^3}\)
\( = 113.14\,{\rm{c}}{{\rm{m}}^3}\)
therefore the volume of \(20\) ice cream balls \( = 2262.8\,{\rm{c}}{{\rm{m}}^3}\)
Hence, \(2262.8\,{\rm{c}}{{\rm{m}}^3}\) of ice cream can be filled in \(20\) ice cream balls.

FAQs on Volume of a Sphere

Below are some Frequently Asked Questions on the Volume of a Sphere:

Q1. Why is \(\frac{4}{3}\) used in the volume of a sphere?
Ans: The sphere is made up of infinite circles stacked over each other, with their centres in a straight line. So the volume of the sphere is the sum of the product of the area of those circles and its thickness. Integrating the product of radius and thickness, we will obtain the volume of the sphere as \(\frac{4}{3}\pi {r^3}.\)
Hence, \(\frac{4}{3}\) is used in the volume of a sphere.

Q2. What is the TSA of a sphere?
Ans: The total surface area (TSA) of a sphere is found by \(4\pi {r^2}.\)

Q3. What is the formula for calculating the volume of a sphere?
Ans: The formula for calculating the volume of a sphere is \(V = \frac{4}{3}\pi {r^3}\) where \(r\) is the radius of the sphere.

Q4. What is the unit of volume of a sphere?
Ans: The unit of volume of a sphere is given in cubic units like \({{\rm{m}}^{\rm{3}}}{\rm{,}}\,{\rm{c}}{{\rm{m}}^{\rm{3}}}\) etc.

Q5. How to find the radius of the sphere when the volume of the sphere is given?
Ans: When the volume of a sphere is given, the radius of the sphere can be found by using the formula for the volume of a sphere.
Volume of a sphere \(V = \frac{4}{3}\pi {r^3}\)
So, radius \(r = \sqrt[3]{{\frac{{V \times 3}}{{4\pi }}}}\)

Q6. How to find the diameter of the sphere when the volume of the sphere is given?
Ans: When the volume of a sphere is given, the radius of the sphere can be found by using the formula for the volume of a sphere.
Volume of a sphere \(V = \frac{4}{3}\pi {r^3}\)
So, radius \(r = \sqrt[3]{{\frac{{V \times 3}}{{4\pi }}}}\)
\({\rm{Diameter}} = 2\,r = 2\sqrt[3]{{\frac{{V \times 3}}{{4\pi }}}}\)

Q7. How many parameters are required to find the volume of a sphere?
Ans: Since \(\frac{4}{3}\) and \(\pi\) are constants, the only required parameter to find the volume of a sphere is either the radius or the diameter of the sphere.

Q8. What is the relation between a sphere and a hemisphere?
Ans: A hemisphere is an exact half of a sphere.

Q9. What is the volume of a hemisphere?
Ans: As a hemisphere is a three-dimensional shape and exactly the half of a sphere, the volume of the hemisphere will also be exactly the half of the volume of a sphere.
Volume of a hemisphere is \(V = \frac{2}{3}\pi {r^3}\) cubic units, where \(r\) is the radius of the hemisphere.

Q10. Why is a sphere \(\frac{2}{3}\) of a cylinder?

Ans: Let us consider a cylinder and sphere of the same radius and height. That is, consider a sphere of radius \(r\) and \(a\) cylinder of base radius \(r\) and a height \(\left( h \right)\) of \(2r.\)
Taking the ratio of their volumes, we get
Volume of a sphere : Volume of a cylinder \( = \frac{4}{3}\pi {r^3}:\pi {r^2}h\)
\( = \frac{4}{3}\pi {r^3}:\pi {r^2}2r\)
\( = \frac{4}{3}\pi {r^3}:2\pi {r^3}\)
\( = \frac{4}{3}:2\)
\( = \frac{2}{3}:1\)
Hence, a sphere is \(\frac{2}{3}\) of a cylinder.

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