Equation of a Sphere

The equation of a sphere defines all points equidistant from its center, given by (x – h)² + (y – k)² + (z – l)² = r², where (h, k, l) is the center and r is the radius. This article provides an in-depth exploration of the equation of a sphere, its properties, applications, and related concepts.

A sphere is a three-dimensional geometric shape that is perfectly symmetrical and has all points on its surface equidistant from a single point called the center. This distance from the center to any point on the surface is known as the radius of the sphere.

Key Characteristics of Sphere

Some of the key characteristics of sphere are:

• Shape: A sphere is round, similar to a ball, and does not have any edges or vertices.
• Surface: The surface of a sphere is smooth and continuous, with no flat areas.
• Symmetry: A sphere has infinite rotational symmetry; it looks the same from any direction

The equation of a sphere in three-dimensional space with center coordinates (0, 0, 0) and radius “r” is:

x² + y² + z² = r²

Here, (x, y, z) represents any point on the sphere’s surface. The equation signifies that the sum of the squares of the differences between the coordinates of any point on the sphere and the center point is equal to the square of the radius.

General Equation of Sphere

The general equation of a sphere in three-dimensional space is represented by the equation:

(x – h)² + (y – k)² + (z – l)² = r²

Where (h, k, l) denotes the coordinates of the sphere’s center and r is its radius.

Parametric Equations of a Sphere

The parametric equations of a sphere in can be expressed as:

x = h + r × sin(θ) × cos(φ),

y = k + r × sin(θ) × sin(φ), and

z = l + r × cos(θ)

Where (θ, φ) are spherical coordinates representing the inclination angle and azimuth angle, respectively.

Read More about Parametric Equation.

Geometrical Interpretation of the Equation of a Sphere

The equation “(x – h)² + (y – k)² + (z – l)² = r²” defines a sphere as the locus of points equidistant from its center (h, k, l). This implies that any point (x, y, z) on the sphere’s surface maintains a constant distance r from the center, illustrating the geometric property of a sphere.

To derive the equation of a sphere, let’s start with a sphere centered at (h, k, l) with radius r. Now, consider any point (x, y, z) on the sphere’s surface.

The distance between this point and the sphere’s center, according to the three-dimensional distance formula, is:

Distance = √((x – h)² + (y – k)² + (z – l)²)

For this point to be on the sphere’s surface, this distance must equal the radius r: √((x – h)² + (y – k)² + (z – l)²) = r

Squaring both sides to eliminate the square root gives us: (x – h)² + (y – k)² + (z – l)² = r²

This equation defines all points in space equidistant from the center, hence describing the surface of the sphere.

Example: For a sphere centered at (3, −2, 1) with a radius of 5.

Solution:

Equation of sphere centered at (h, k, l) with radius r: (x – h)² + (y – k)² + (z – l)² = r²

Thus, equation of sphere centered at (3, -2, 1) is:

(x − 3)² + (y + 2)² + (z − 1)² = 5²

This equation represents all points (x, y, z) that are 5 units away from the point (3, −2, 1).

Equation of a sphere in vector form: “(r – r₀)² = R²”, where “r” represents a general point on the sphere’s surface, “r₀” denotes the center of the sphere, and “R” is the radius.

Equation of a sphere in cylindrical coordinates: “r² + (z – z₀)² = R²”, where “r” is the radial distance from the z-axis, “z” is the vertical distance, and “z₀” is the vertical position of the center of the sphere.

Equation of a sphere in spherical coordinates: “R = R₀”, where “R” is the distance from the origin to any point on the sphere’s surface, and “R₀” is the radius of the sphere.

Surface Area Equation of Sphere

The surface area (A) of a sphere with radius (r) is given by the formula:

Area = 4πr²

This formula calculates the total area of the sphere’s surface, which is covered by points equidistant from its center. The surface area of a sphere is a fundamental quantity used in various mathematical and physical contexts, such as in geometry, physics, and engineering.

Volume Equation of Sphere

The volume (V) of a sphere with radius (r) is given by the formula:

Volume = (4/3)πr³

This formula represents the total amount of space enclosed by the sphere. It’s derived by multiplying the volume of a sphere (4/3)πr³, where r is the radius. This volume formula is fundamental in various mathematical and physical contexts, such as in geometry, physics, and engineering.

Read More,

• Sphere
• Surface Area of Sphere
• Volume of Area of Sphere
• Mensuration

Problem: Find the equation of a sphere with center (2, -3, 1) and radius 5.

Solution:

Using the equation of a sphere: (x – h)² + (y – k)² + (z – l)² = r², where (h, k, l) is the center and r is the radius.

Plugging in the values, we get: (x – 2)² + (y + 3)² + (z – 1)² = 5².

So, the equation of the sphere is (x – 2)² + (y + 3)² + (z – 1)² = 25.

Problem: Calculate the surface area of a sphere with radius 7 cm.

Solution:

Using the surface area formula for a sphere:A = 4πr².

Plugging in the radius, we get: A = 4π(7)² = 4π(49) = 196π cm².

Problem: Find the equation of a sphere with center (-1, 2, 3) and passing through the point (4, -1, 6).

Solution:

Since the sphere passes through the given point, the distance between the center and the point is equal to the radius.

Using the distance formula: √((x₂ – x₁)² + (y₂ – y₁)² + (z₂ – z₁)²) = r, where (x₁, y₁, z₁) is the center and (x₂, y₂, z₂) is the point on the sphere.

Plugging in the values, we get: √((-1 – 4)² + (2 + 1)² + (3 – 6)²) = r

Simplifying: √((-5)² + (3)² + (-3)²) = r

⇒√(25 + 9 + 9) = r

⇒√43 = r

So, the equation of the sphere is: (x + 1)² + (y – 2)² + (z – 3)² = 43.

Problem: A sphere has surface area 154π square units. Find its radius.

Solution:

Given surface area A = 154π.

We know the surface area formula for a sphere: A = 4πr².

Equating the two, we get: 4πr² = 154π.

Dividing both sides by 4π: r² = 154/4.

Solving: r² = 38.5.

Taking the square root: r ≈ √38.5 ≈ 6.22 units.

Problem: A sphere has a volume of 288π cubic units. Find its radius.

Solution:

Given volume V = 288π.

We know the volume formula for a sphere: V = (4/3)πr³.

Equating the two, we get: (4/3)πr³ = 288π.

Multiplying both sides by (3/4π): r³ = 288 × (3/4).

Solving: r³ = 216.

Taking the cube root: r ≈ ∛216 ≈ 6 units.

Problem 1: Find the equation of a sphere with center (0, 0, 0) and radius 10.

Problem 2: Calculate the volume of a sphere with radius 3 units.

Problem 3: Determine the surface area of a sphere with radius 12 meters.

Problem 4: Find the equation of a sphere with center (3, -2, 5) and radius 6.

Problem 5: Calculate the surface area of a sphere with radius 9 units.

Problem 6: Determine the volume of a sphere with surface area 324π square units.

In conclusion, the sphere, a perfectly symmetrical three-dimensional shape, plays a vital role in various fields, from mathematics to engineering and medicine. Its equation, parametric representations, and geometric interpretations provide insights into spatial relationships and enable precise calculations. Understanding the properties and applications of spheres enhances problem-solving skills and contributes to advancements across disciplines.

What is shape of sphere?

A sphere is a perfectly round three-dimensional shape.

What is equation of Sphere?

Equation of a sphere in three-dimensional space with center coordinates (0, 0, 0) and radius “r” is:

x² + y² + z² = r²

Write general equation of sphere.

The general equation of a sphere in three-dimensional space is represented by the equation:

(x – h)² + (y – k)² + (z – l)² = r²

Write parametric equation of sphere.

The parametric equations of a sphere in can be expressed as:

x = h + r × sin(θ) × cos(φ),

y = k + r × sin(θ) × sin(φ), and

z = l + r × cos(θ)

What is surface area of sphere?

Surface are of sphere with radius “r” is: Area = 4πr²