Can the volume of a sphere be negative?

No, the volume of a sphere is always positive because the radius is always positive.

Why do we use spherical coordinates to find the volume of a sphere?

Spherical coordinates simplify the integration process for spherical objects due to their natural alignment with the geometry of the sphere, reducing the complexity of the calculations.

How does the volume of a sphere change with respect to its radius?

The volume of a sphere increases exponentially as the radius increases, following a cubic relationship.

What are the challenges in computing the volume of irregular spherical objects?

For irregular objects, the volume cannot be easily calculated using simple geometric formulas, and advanced methods are required.

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Volume of a Sphere

Master Spherical Capacity in Minutes; Learn how to calculate the total three-dimensional space enclosed inside a perfectly round object. Understand the fundamental formula, see how it relates to liquid capacity, and master high-yield board exam problems.

Class: 10 Mathematics (CBSE)
Chapter: Surface Areas and Volumes
Estimated Learning Time: 15–20 Minutes

1.0Learning Outcomes

After completing this lesson, you will be able to:

State the mathematical formula for the volume of a complete sphere.
Convert between diameter and radius accurately before solving volume parameters.
Calculate the capacity of a sphere in cubic units (cm³, m³) and liquid liters.
Solve board exam problems on melting, recasting, and combining spherical shapes.

Volume of Sphere

A Sphere is a three-dimensional circle. Just like a circle, in a sphere, all the points on the surface are at equal distance from a fixed point, “the Centre”. The equal distance from the centre to its surface is known as the radius(r), and the line passing through the centre of the sphere is called the diameter of the sphere. In the geometrical plane, it occupies three axes, namely the x,y, and z-axis. Volume is an important physical property of a sphere, which we will discuss in this article.

2.0Definition of Volume of a Sphere

The Volume of a sphere can be defined as the amount of three-dimensional space enclosed within the surface of a sphere. In simple words, the Volume of a sphere is the capacity of a sphere, which can be calculated in cubic units, like m³, cm³, in³ etc. The volume of a sphere formula can be written as:

Volume of Sphere = $\frac{4}{3} π r^{3}$

Here,

$π = \frac{22}{7}$

r = Radius of the sphere

3.0Volume of Sphere and Key Concepts

Volume of Sphere with Diameter:

If in a given sphere, the diameter d of the sphere is given, the radius can be written as r = d2. (r = \frac{d}{2}) Hence, the volume of the sphere can be rewritten as:

Volume of sphere with Diameter:

$\frac{4}{3} π (\frac{d}{2})^{3}$

Or

Volume of sphere with Diameter:

$\frac{1}{6} π d^{3}$

The Volume of Sphere with Area:

If the total surface area A of a sphere is given, then the volume of the sphere can be found by finding the value of the radius “r” first, then putting it in the volume of the sphere formula. Mathematically, this can be represented as:

Surface Area of Sphere = $4 π r^{2}$

$r = \frac{A}{4 π}$

Volume of sphere = $\frac{4}{3} π \frac{A}{4 π}$

4.0Derivation of Volume of Sphere

The volume of a sphere can be derived using integral calculus by using two different types of integrations that are:

The volume of a sphere by integration
The Volume of a sphere by triple integration

Let’s derive the volume of a sphere formula using both of these methods.

Volume of a Sphere by Integration

Imagine a sphere as a bunch of numerous thin circular disks, which are piled one over another, as shown in the figure. By the diagram, it is clear that all these discs will have a different diameter with a collinear centre. Now, select any of these thin discs with radius r and thickness “dy” located at the y distance from the x-axis. Hence, according to the concept of integration, the volume of the sphere can be expressed as the product of the area of the circle and its thickness dy. Mathematically, this can be written as:

$d V = π r^{2} d y$

By using Pythagoras theorem, r can be expressed in the form of its thickness (y) and the Radius of the sphere, say R, as:

$d V = π (R^{2} - y^{2}) d y$

Now, integrating the total volume of the sphere,

$V = \int_{y = - R}^{y = + R} d v$

$V = \int_{y = - R}^{y = + R} π (R^{2} - y^{2}) d y$

$V = π [R^{2} y - \frac{y ^{3}}{3}]_{y = - R}^{y = + R}$

Substituting the limits,

$V = π [(R^{3} - \frac{R ^{3}}{3}) - (- R^{3} + \frac{R ^{3}}{3})]$

$V = π [2 R^{3} - \frac{2 R ^{3}}{3}]$

$V = \frac{π}{3} [6 R^{3} - 2 R^{3}]$

$V = \frac{π}{3} (4 R^{3})$

Hence, the Volume of the sphere will be:

$V = \frac{4}{3} π R^{3}$

The Volume of a Sphere by Triple Integration

To derive the volume of a sphere with the triple integration method, we will use spherical coordinates by using the standard equation of a sphere. The standard equation of a sphere of radius r with origin as the centre can be written as:

$x^{2} + y^{2} + z^{2} = r^{2}$

Here, x, y, and z are related to the spherical coordinates of the sphere r $\overset{,}{˚} θ, ϕ$ as:

$x = r sin θ cos ϕ$

$y = r sin θ sin ϕ$

$z = r cos θ$

Where,

r is represented as the radial distance from the origin (which will be constant for a sphere),
θ is the polar angle (or colatitude) measured from the positive z-axis, ranging from 0 to π,
ϕ is the azimuthal angle (or longitude) measured in the xy-plane from the positive x-axis, ranging from 0 to 2π.

The volume of spherical coordinates is given by:

$d V = r^{2} sin θ d r d θ d ϕ$

Now, using triple integration to find the volume of a sphere:

$V = \int_{0}^{2 π} \int_{0}^{π} \int_{0}^{r} r^{2} sin θ d r d θ d ϕ$ ……(1)

Step 1: Integrating with respect to r:

$\int_{0}^{r} r^{2} d r = \frac{r ^{3}}{3}$

Step 2: Integrate with respect to $θ$

$\int_{0}^{2 π} sin θ d θ = 2$

Step 3: Integrate with respect to $ϕ$

$\int_{0}^{π} d ϕ = 2 π$

Putting values of Steps 1, 2, & 3 in equation (1)

$V = (2 π) (2) (\frac{r ^{3}}{3})$

$V = \frac{4 π r ^{3}}{3}$

$V = \frac{4}{3} π r^{3}$

5.0Volume of Sphere Examples: Numericals

Problem 1: Find the volume of a sphere whose surface area is 154 cm².

Solution:

Surface area of sphere = 154

$4 π r^{2} = 154$

$r^{2} = \frac{154 \times 7}{4 \times 22} = \frac{49}{4}$

$r = \frac{49}{4} = \frac{7}{2}$

Now,

Volume of Sphere = $\frac{4}{3} π r^{3}$

Volume of Sphere = $\frac{4}{3} \times \frac{22}{7} \times (\frac{7}{2})^{3}$

Volume of Sphere = $\frac{22 \times 4 \times 7 \times 7 \times 7}{3 \times 7 \times 2 \times 2 \times 2}$

Volume of Sphere =36.67 $c m^{3}$

Problem 2: A spherical water tank has a diameter of 10 meters. How much water can it hold in terms of volume?

Solution: Given, Diameter = 10m

radius will be $\frac{10}{2}$ or 5 meters

Volume of Sphere = $\frac{4}{3} π r^{3}$

Volume of Sphere = $\frac{4}{3} \times \frac{22}{7} \times 5^{3}$

Volume of Sphere = $523.6 m^{3} (a pp ro x)$

Problem 3: A hollow sphere has an outer radius of 6 cm & an inner radius of 4 cm. What is the volume of the material making up the hollow sphere?

Solution: Let Outer radius (R) = 6cm and inner radius(r) = 4cm

The volume of material = Volume of outer surface – Volume of the inner surface

The volume of material = $\frac{4}{3} π R^{3} - \frac{4}{3} π r^{3} = \frac{4}{3} π (R^{3} - r^{3})$

The volume of material = $\frac{4}{3} π (R^{3} - r^{3}) = \frac{4}{3} π (6^{3} - 4^{3})$

$V = \frac{4}{3} π (6 - 4) (6^{2} + 6 \times 4 + 4^{2})$

$V = \frac{4}{3} \times \frac{22}{7} \times (2) (76) = 636.2 c m^{3}$

6.0EUREKA by ALLEN – Learn Better, Score Higher

With EUREKA from the ALLEN, you will achieve the highest level of academic achievement through an innovative approach designed specifically for Class 10 students. EUREKA provides world-class education using top instructors, cutting-edge artificial intelligence, and a proven method for preparing for board examinations; thus creating a complete learning environment for success. Through personalized assistance, compelling content, and ongoing assessment of progress toward maximum achievement, students receive everything they need to realise their full academic capabilities.

Key Features of EUREKA Class 10 Courses:
High-quality interactive lessons Engaging story-driven learning approach Comprehensive board exam preparation Structured answer writing practice Instant performance feedback Real-time learning analytics AI-powered doubt support anytime NCERT & CBSE curriculum alignment Complete flexibility in learning schedules

Explore Now

7.0Supporting Study Materials

This study material CBSE Notes and NCERT Solutions for the Chapter "Surface Areas and Volumes" on Volume of Sphere Topics is designed according to the latest CBSE Class 10 Mathematics syllabus and NCERT guidelines. It provides clear explanations of key concepts, definitions, formulas, and important questions to help students understand the volume of a sphere, volume of a hemisphere, word problems involving conversion of solids, and prepare effectively for examinations.

CBSE Class 10 Maths Notes Chapter 12 Surface Areas and Volumes	NCERT Solutions for Class 10 Maths Chapter 12: Surface Areas and Volumes

8.030-Second Quick Revision

A sphere is a perfectly round three-dimensional shape.
Volume measures the space occupied inside the sphere.
Radius (r) is the distance from the centre to any point on the surface.
The volume of a sphere depends on the cube of its radius.
Formula: Volume = (4/3)πr³
Doubling the radius increases the volume by 8 times.
The unit of volume is cubic units (cm³, m³, etc.).
Commonly used in problems involving balls, globes, and spherical containers.
Volume questions often require direct application of the formula.
Remember: Volume = Space Inside the Sphere.

9.0Recommended Next Topics

Volume of a Hemisphere (Open vs Closed structures)
Surface Area of a Sphere and Hemisphere
Volume of a Right Circular Cone
Complex combination solids (e.g., cylinders capped with hemispherical ends)

Join ALLEN!

(Session 2026 - 27)

Name

Mobile Number

Class

Choose class

Goal

Choose your goal

Preferred Programs

Preferred Mode

State

Choose State

I agree to

I authorise ALLEN Career Institute Pvt Ltd to send me regular updates via Phone calls, Whatsapp, SMS, Robocalls (Automated Calls), Emails, or on postal address.

Volume of a Sphere

Master Spherical Capacity in Minutes; Learn how to calculate the total three-dimensional space enclosed inside a perfectly round object. Understand the fundamental formula, see how it relates to liquid capacity, and master high-yield board exam problems.

Class: 10 Mathematics (CBSE)
Chapter: Surface Areas and Volumes
Estimated Learning Time: 15–20 Minutes

1.0Learning Outcomes

After completing this lesson, you will be able to:

State the mathematical formula for the volume of a complete sphere.
Convert between diameter and radius accurately before solving volume parameters.
Calculate the capacity of a sphere in cubic units (cm³, m³) and liquid liters.
Solve board exam problems on melting, recasting, and combining spherical shapes.

Volume of Sphere

A Sphere is a three-dimensional circle. Just like a circle, in a sphere, all the points on the surface are at equal distance from a fixed point, “the Centre”. The equal distance from the centre to its surface is known as the radius(r), and the line passing through the centre of the sphere is called the diameter of the sphere. In the geometrical plane, it occupies three axes, namely the x,y, and z-axis. Volume is an important physical property of a sphere, which we will discuss in this article.

2.0Definition of Volume of a Sphere

The Volume of a sphere can be defined as the amount of three-dimensional space enclosed within the surface of a sphere. In simple words, the Volume of a sphere is the capacity of a sphere, which can be calculated in cubic units, like m³, cm³, in³ etc. The volume of a sphere formula can be written as:

Volume of Sphere = $\frac{4}{3} π r^{3}$

Here,

$π = \frac{22}{7}$

r = Radius of the sphere

3.0Volume of Sphere and Key Concepts

Volume of Sphere with Diameter:

If in a given sphere, the diameter d of the sphere is given, the radius can be written as r = d2. (r = \frac{d}{2}) Hence, the volume of the sphere can be rewritten as:

Volume of sphere with Diameter:

$\frac{4}{3} π (\frac{d}{2})^{3}$

Or

Volume of sphere with Diameter:

$\frac{1}{6} π d^{3}$

The Volume of Sphere with Area:

If the total surface area A of a sphere is given, then the volume of the sphere can be found by finding the value of the radius “r” first, then putting it in the volume of the sphere formula. Mathematically, this can be represented as:

Surface Area of Sphere = $4 π r^{2}$

$r = \frac{A}{4 π}$

Volume of sphere = $\frac{4}{3} π \frac{A}{4 π}$

4.0Derivation of Volume of Sphere

The volume of a sphere can be derived using integral calculus by using two different types of integrations that are:

The volume of a sphere by integration
The Volume of a sphere by triple integration

Let’s derive the volume of a sphere formula using both of these methods.

Volume of a Sphere by Integration

Imagine a sphere as a bunch of numerous thin circular disks, which are piled one over another, as shown in the figure. By the diagram, it is clear that all these discs will have a different diameter with a collinear centre. Now, select any of these thin discs with radius r and thickness “dy” located at the y distance from the x-axis. Hence, according to the concept of integration, the volume of the sphere can be expressed as the product of the area of the circle and its thickness dy. Mathematically, this can be written as:

$d V = π r^{2} d y$

By using Pythagoras theorem, r can be expressed in the form of its thickness (y) and the Radius of the sphere, say R, as:

$d V = π (R^{2} - y^{2}) d y$

Now, integrating the total volume of the sphere,

$V = \int_{y = - R}^{y = + R} d v$

$V = \int_{y = - R}^{y = + R} π (R^{2} - y^{2}) d y$

$V = π [R^{2} y - \frac{y ^{3}}{3}]_{y = - R}^{y = + R}$

Substituting the limits,

$V = π [(R^{3} - \frac{R ^{3}}{3}) - (- R^{3} + \frac{R ^{3}}{3})]$

$V = π [2 R^{3} - \frac{2 R ^{3}}{3}]$

$V = \frac{π}{3} [6 R^{3} - 2 R^{3}]$

$V = \frac{π}{3} (4 R^{3})$

Hence, the Volume of the sphere will be:

$V = \frac{4}{3} π R^{3}$

The Volume of a Sphere by Triple Integration

To derive the volume of a sphere with the triple integration method, we will use spherical coordinates by using the standard equation of a sphere. The standard equation of a sphere of radius r with origin as the centre can be written as:

$x^{2} + y^{2} + z^{2} = r^{2}$

Here, x, y, and z are related to the spherical coordinates of the sphere r $\overset{,}{˚} θ, ϕ$ as:

$x = r sin θ cos ϕ$

$y = r sin θ sin ϕ$

$z = r cos θ$

Where,

r is represented as the radial distance from the origin (which will be constant for a sphere),
θ is the polar angle (or colatitude) measured from the positive z-axis, ranging from 0 to π,
ϕ is the azimuthal angle (or longitude) measured in the xy-plane from the positive x-axis, ranging from 0 to 2π.

The volume of spherical coordinates is given by:

$d V = r^{2} sin θ d r d θ d ϕ$

Now, using triple integration to find the volume of a sphere:

$V = \int_{0}^{2 π} \int_{0}^{π} \int_{0}^{r} r^{2} sin θ d r d θ d ϕ$ ……(1)

Step 1: Integrating with respect to r:

$\int_{0}^{r} r^{2} d r = \frac{r ^{3}}{3}$

Step 2: Integrate with respect to $θ$

$\int_{0}^{2 π} sin θ d θ = 2$

Step 3: Integrate with respect to $ϕ$

$\int_{0}^{π} d ϕ = 2 π$

Putting values of Steps 1, 2, & 3 in equation (1)

$V = (2 π) (2) (\frac{r ^{3}}{3})$

$V = \frac{4 π r ^{3}}{3}$

$V = \frac{4}{3} π r^{3}$

5.0Volume of Sphere Examples: Numericals

Problem 1: Find the volume of a sphere whose surface area is 154 cm².

Solution:

Surface area of sphere = 154

$4 π r^{2} = 154$

$r^{2} = \frac{154 \times 7}{4 \times 22} = \frac{49}{4}$

$r = \frac{49}{4} = \frac{7}{2}$

Now,

Volume of Sphere = $\frac{4}{3} π r^{3}$

Volume of Sphere = $\frac{4}{3} \times \frac{22}{7} \times (\frac{7}{2})^{3}$

Volume of Sphere = $\frac{22 \times 4 \times 7 \times 7 \times 7}{3 \times 7 \times 2 \times 2 \times 2}$

Volume of Sphere =36.67 $c m^{3}$

Problem 2: A spherical water tank has a diameter of 10 meters. How much water can it hold in terms of volume?

Solution: Given, Diameter = 10m

radius will be $\frac{10}{2}$ or 5 meters

Volume of Sphere = $\frac{4}{3} π r^{3}$

Volume of Sphere = $\frac{4}{3} \times \frac{22}{7} \times 5^{3}$

Volume of Sphere = $523.6 m^{3} (a pp ro x)$

Problem 3: A hollow sphere has an outer radius of 6 cm & an inner radius of 4 cm. What is the volume of the material making up the hollow sphere?

Solution: Let Outer radius (R) = 6cm and inner radius(r) = 4cm

The volume of material = Volume of outer surface – Volume of the inner surface

The volume of material = $\frac{4}{3} π R^{3} - \frac{4}{3} π r^{3} = \frac{4}{3} π (R^{3} - r^{3})$

The volume of material = $\frac{4}{3} π (R^{3} - r^{3}) = \frac{4}{3} π (6^{3} - 4^{3})$

$V = \frac{4}{3} π (6 - 4) (6^{2} + 6 \times 4 + 4^{2})$

$V = \frac{4}{3} \times \frac{22}{7} \times (2) (76) = 636.2 c m^{3}$

6.0EUREKA by ALLEN – Learn Better, Score Higher

With EUREKA from the ALLEN, you will achieve the highest level of academic achievement through an innovative approach designed specifically for Class 10 students. EUREKA provides world-class education using top instructors, cutting-edge artificial intelligence, and a proven method for preparing for board examinations; thus creating a complete learning environment for success. Through personalized assistance, compelling content, and ongoing assessment of progress toward maximum achievement, students receive everything they need to realise their full academic capabilities.

Key Features of EUREKA Class 10 Courses:
High-quality interactive lessons Engaging story-driven learning approach Comprehensive board exam preparation Structured answer writing practice Instant performance feedback Real-time learning analytics AI-powered doubt support anytime NCERT & CBSE curriculum alignment Complete flexibility in learning schedules

Explore Now

7.0Supporting Study Materials

This study material CBSE Notes and NCERT Solutions for the Chapter "Surface Areas and Volumes" on Volume of Sphere Topics is designed according to the latest CBSE Class 10 Mathematics syllabus and NCERT guidelines. It provides clear explanations of key concepts, definitions, formulas, and important questions to help students understand the volume of a sphere, volume of a hemisphere, word problems involving conversion of solids, and prepare effectively for examinations.

CBSE Class 10 Maths Notes Chapter 12 Surface Areas and Volumes	NCERT Solutions for Class 10 Maths Chapter 12: Surface Areas and Volumes

8.030-Second Quick Revision

A sphere is a perfectly round three-dimensional shape.
Volume measures the space occupied inside the sphere.
Radius (r) is the distance from the centre to any point on the surface.
The volume of a sphere depends on the cube of its radius.
Formula: Volume = (4/3)πr³
Doubling the radius increases the volume by 8 times.
The unit of volume is cubic units (cm³, m³, etc.).
Commonly used in problems involving balls, globes, and spherical containers.
Volume questions often require direct application of the formula.
Remember: Volume = Space Inside the Sphere.

9.0Recommended Next Topics

Volume of a Hemisphere (Open vs Closed structures)
Surface Area of a Sphere and Hemisphere
Volume of a Right Circular Cone
Complex combination solids (e.g., cylinders capped with hemispherical ends)

Frequently Asked Questions

Can the volume of a sphere be negative?

Why do we use spherical coordinates to find the volume of a sphere?

How does the volume of a sphere change with respect to its radius?

What are the challenges in computing the volume of irregular spherical objects?

Frequently Asked Questions

Can the volume of a sphere be negative?

Why do we use spherical coordinates to find the volume of a sphere?

How does the volume of a sphere change with respect to its radius?

What are the challenges in computing the volume of irregular spherical objects?

Join ALLEN!

Volume of a Sphere

1.0Learning Outcomes

Volume of Sphere

2.0Definition of Volume of a Sphere

3.0Volume of Sphere and Key Concepts

4.0Derivation of Volume of Sphere

Volume of a Sphere by Integration

The Volume of a Sphere by Triple Integration

5.0Volume of Sphere Examples: Numericals

6.0EUREKA by ALLEN – Learn Better, Score Higher

7.0Supporting Study Materials

8.030-Second Quick Revision

9.0Recommended Next Topics

On this page

Join ALLEN!

Volume of a Sphere

1.0Learning Outcomes

Volume of Sphere

2.0Definition of Volume of a Sphere

3.0Volume of Sphere and Key Concepts

4.0Derivation of Volume of Sphere

Volume of a Sphere by Integration

The Volume of a Sphere by Triple Integration

5.0Volume of Sphere Examples: Numericals

6.0EUREKA by ALLEN – Learn Better, Score Higher

7.0Supporting Study Materials

8.030-Second Quick Revision

9.0Recommended Next Topics

On this page