Table of Contents

Last modified on March 11th, 2022

A sphere is a three-dimensional symmetrical solid. Its shape is spherical which means completely round. It can be defined as the set of all the points equidistant from a fixed point, known as the center. The diagram below shows the shape of a sphere.

Unlike other solids such as a cube, cuboid, cone, cylinder, prism, or pyramid, it has a continuous surface. As a result, a sphere has no flat faces, vertices or edges.

Soccer balls, cricket balls, basketballs, yarns, globe, pendulum bobs, glass marbles, Christmas decorations, planets, and the human eyeball are common examples of sphere-shaped objects in real life.

If a sphere is cut into 2 equal halves, each half is known as a *hemisphere* or half-sphere or semi-sphere. (hemi in Greek and semi in Latin).

A sphere, within its shape, includes the center, radius, diameter, and the circumference. Given below is a diagram showing its parts.

The distance from the center to any point on the sphere’s surface is the **radius**, and it’s denoted by ‘r’. The radius is twice the diameter. **Diameter **is a straight line segment between 2 opposite points on the surface passing through the center. It is denoted by ‘d’. It is also referred to as the height of a sphere.

**Circumference**, a concept in circle, is also included in a sphere. It is simply the length of the largest possible circle inside a sphere. This circle is also referred to as the *great circle. *It is denoted by ‘C’.

Some important points about a sphere are:

- Its radius and circumference are constant. If radius is increased, the circumference increases, thereby increasing the overall size of the sphere. Thus, the circumference and size of the sphere decreases when the radius is decreased.
- The curvature of its surface is the same and uniform throughout (constant mean curvature).
- It has the largest volume over the smallest surface area as compared to any 3-dimensional object.

We have learned about the equation of a circle in the coordinate axes. Now let us learn about the equation of a sphere.

The equation of a sphere can be written both in standard and general forms.

The equation of a sphere in standard form is written as:

**( x – h)^{2} + (y – k)^{2} + (z – l)^{2} = r^{2}, **here (h, k, l) = center, r = radius

When the center is at the origin (0, 0, 0), the equation of a sphere in standard form becomes:

(*x* – 0)^{2} + (*y* – 0)^{2} + (*z *– 0)^{2} = *r*^{2}

**= > x^{2} + y^{2} + z^{2} = r^{2}**

The equation of a sphere in general (expanded) form is written as:

** x^{2} + y^{2} + z^{2} + 2ux + 2vy + 2wz + d = 0**, here (-u, -v, -w) = center, u, v, w, & d = constants

For finding the radius,

**${r=\sqrt{u^{2}+v^{2}+w^{2}-d}}$***, *here (-u, -v, -w) = center, u, v, w, & d = constants

Let us solve some examples to illustrate the above concept.

**Find the equation of a sphere in standard form whose center is at (0, 0, 0) and radius is 5 cm.**

Solution:

As we know, the equation of a sphere in standard form is:**(x – h) ^{2} + (y – k)^{2} + (z – l)^{2} = r^{2}, **here (0, 0, 0) = center, r = 5 cm

= > (x – 0)

= > x

**Find the equation of a sphere in general form whose center is at (2, 4, 6) and radius is 8 cm.**

Solution:

As we know, the equation of a sphere in general form is:**x ^{2} + y^{2} + z^{2} + 2ux + 2vy + 2wz + d = 0**, here (-u, -v, -w) = (2, 4, 6)

∴ x

∴ x

Solving for d:

d = u

= 4 + 16 + 36 – 64

= – 8

∴The equation is:

x

Like all other 3-d shapes, we can calculate the surface area and the volume of a sphere.

The volume of a sphere is the space it occupies in any 3-dimensional plane. It determines its density or the amount of space it occupies. It is expressed in cubic units such as m^{3}, cm^{3}, and mm^{3}. The formula is given below:

**Volume ( V) = (4/3)πr^{3}**, here π = 22/7 = 3.141, r = radius

Let us solve an example to understand the above concept better.

**Find the volume of a sphere whose radius is 3 cm.**

Solution:

As we know,**Volume ( V) = (4/3)πr^{3}**, here π = 22/7 = 3.141, r = 3 cm

∴

= 113.1 cm

The surface area of a sphere is the total region covered by its outer surface. It is expressed in square units such as m^{2}, cm^{2}, and mm^{2}. The formula is given below:

**Surface Area ( SA) = 4πr^{2}**, here π = 22/7 = 3.141, r = radius

Let us solve an example to understand the above concept better.

**Find the surface area** **of a sphere whose radius is 8 cm.**

Solution:

As we know,**Surface Area ( SA) = 4πr^{2}**, here π = 22/7 = 3.141, r = 8 cm

∴

= 804.25 cm

We have learned about the radius, diameter, and circumference of a sphere at the beginning of this article. Let us learn the formulas to find them.

It is half the diameter of a sphere. The basic formula to calculate the radius of a sphere is:

**Radius ( r) = (3V/4π)^{1/3}**, here V = volume, π = 22/7 = 3.141

Let us solve some examples to understand the concept better.

**Find the radius of a sphere with a volume is 3054 cm ^{3}.**

Solution:

As we know,**Radius ( r) = (3V/4π)^{1/3}**, here V = 3054 cm

∴

= 9 cm

Finding the radius of a sphere when the **DIAMETER **is known

**Find the radius of a sphere with a diameter of 14 cm.**

Solution:

Since radius is half the diameter,**Radius ( r) = d/2**, here d = 14 cm

∴

= 7 cm

The basic formula to calculate the diameter of a sphere is:

**Diameter ( d) = (6V/π)^{1/3}**, here V = volume, π = 22/7 = 3.141

Let us solve some examples to understand the concept better.

**Find the diameter of a sphere whose volume is 905 cm^{3}.**

Solution:

As we know,**Diameter ( d) = (6V/π)^{1/3}**, here V = 905 cm

∴

≈ 12 cm

The basic formula to calculate the circumference of a sphere is:

**Circumference ( C) = 2πr**, here π = 22/7 = 3.141,

Let us solve an example involving the above concept.

**Find the circumference of a sphere with a radius of 5 cm.**

Solution:

As we know,**Circumference ( C) = 2πr**, here π = 22/7 = 3.141, r = 5 cm

∴

= 31.41 cm

Last modified on March 11th, 2022