# Sphere

## Definition

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).

## Parts and Properties

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:

1. 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.
2. The curvature of its surface is the same and uniform throughout (constant mean curvature).
3. 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.

## Equation

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

### In Standard Form

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

(xh)2 + (yk)2 + (z l)2 = r2, 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 = r2

= > x2 + y2 + z2 = r2

### In General Form

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

x2 + y2 + z2 + 2ux + 2vy + 2wz + d = 0, here (-u, -v, -w) = center, u, v, w, & d = constants

${r=\sqrt{u^{2}+v^{2}+w^{2}-d}}$, 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 = r2, here (0, 0, 0) = center, r = 5 cm
= > (x – 0)2 + (y – 0)2 + (z – 0)2 = r2
= > x2 + y2 + z2 = 25

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:
x2 + y2 + z2 + 2ux + 2vy + 2wz + d = 0, here (-u, -v, -w) = (2, 4, 6)
∴ x2 + y2 + z2 + 2.(-2). x + 2.(-4).y + 2.(-6).z + d = 0
∴ x2 + y2 + z2 -4x – 8y – 12z + d = 0
${r=\sqrt{u^{2}+v^{2}+w^{2}-d}}$
Solving for d:
d = u2 + v2 + w2 – r2, here (-u, -v, -w) = (-2, -4, -6), r = 8 cm
= 4 + 16 + 36 – 64
= – 8
∴The equation is:
x2 + y2 + z2 – 4x – 8y – 12z – 8 = 0

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

## Formulas

### Volume

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 m3, cm3, and mm3. The formula is given below:

Volume (V) = (4/3)πr3, 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)πr3, here π = 22/7 = 3.141, r = 3 cm
V = (4/3) × 3.141 × 33
= 113.1 cm3

### Surface Area

The surface area of a sphere is the total region covered by its outer surface. It is expressed in square units such as m2, cm2, and mm2. The formula is given below:

Surface Area (SA) = 4πr2, 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) = 4πr2, here π = 22/7 = 3.141, r = 8 cm
SA = 4 × 3.141 × 82
= 804.25 cm2

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 cm3.

Solution:

As we know,
Radius (r) = (3V/4π)1/3, here V = 3054 cm3, π = 22/7 = 3.141
r = (3 × 3054/4 × 3.141)1/3
= 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
r = 14/2
= 7 cm

### Diameter

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 cm3.

Solution:

As we know,
Diameter (d) = (6V/π)1/3, here V = 905 cm3, π = 22/7 = 3.141
d = (6 × 905/3.141)1/3
≈ 12 cm

### Circumference

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

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

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
C = 2× 3.141 × 5
= 31.41 cm

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