Area of Part of Sphere

A 3-dimensional item shaped similar to a ball.

Each point on the surface is the similar distance from the middle.

Sphere facts:

It is completely symmetrical

It have no edges or vertices

It is not a polyhedron

Every point on the surface is the similar distance from the middle.


Area and Volume of Sphere:

Area of Sphere = 4 × π × r2

Volume of Sphere = (`4/3` ) × π × r3

Example for Area of part of sphere:

Locate the area of part of sphere x2 + y2 + z2 = 81 that lies over the cone z = sqrt(x2 + y2)

The cone and sphere cross when equally equations are satisfied. Substitute z from the cone into the sphere equation to obtain:

x2 + y2 + (sqrt(x2 + y2))2 = 81

2x2 + 2y2 = 81

x2 + y2 = 81/2

This is a sphere by means of radius sqrt (81/2). State you have a spherical method for the area of a sphere. So integrate through spherical coordinates. If θ is the angle concerning z counterclockwise as of +x (0 ≤ θ ≤ 2π), and φ is the angle as of z (0 ≤ φ ≤ π), and r is distance as of the origin. The spherical to Cartesian change are:

x = r sin φ cos θ

y = r sin φ sin θ

z = r cos φ

When y = 0 and x is positive, then x = sqrt (81/2) and θ = 0. Also r = 9 all over the place on the circle of junction, since that is the radius of the sphere. This means that:

sqrt(81/2) = 9×sin φ

sin φ = sqrt(81/2)/9 = sqrt(1/2) = sqrt(2)/2

φ = sin-1(sqrt(2)/2) = π/4

This is the utmost value of φ. The differential of area is:

dS = (r×dφ)×(r×sin φ×dθ) = r2×sinφ×dφ×dθ

integrated for φ = 0 to π/4 and θ = 0 to 2π

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Another Example for Area of part of sphere:

Locate the area of part of sphere x2 + y2 + z2 = 1:

Locate the area of part of sphere  x2 + y2 + z2 = 1 that lies above the cone z = sqrt(x2 + y2)

The cone will cut the sphere in a plane at right angles to the z axis at

z2 = 1-x2 - y2 and z2 = x2 + y2

z = sqrt (1/2)

Therefore you have a spherical limit with a height of 1-sqrt(1/2 ) as of a sphere of radius 1

along with the surface area of sphere is 2 pi r h = 2 * pi * 1 * [1-sqrt(1/2)]