Surface Area of Part of a Sphere

Introduction to Sphere:

Sphere is the one of the type of the geometrical figure.  Sphere is the three dimensional figure of the circle.  Surface area is the area covered by the sphere.  Radius of the sphere is the distance between centers to the circumference of the sphere.  Here we have to discuss about the surface area of sphere with example problems.

Brief Explanation Surface Area of Part of a Sphere

Sphere:

Sphere is a solid 3D shape figure.  It is look like a ball.  The picture of the sphere is look like in the following diagram

Here A is the area of the sphere and r is the radius of the sphere. And o is the center of the sphere.

Surface Area:

The surface area of the part of the sphere is mentioned by the following formulas,

Total surface area of the sphere =4 p r2 Square units
Curved surface area of the sphere=3 p r2 Square units
In the above formulas we have to substitute the radius of the sphere and the value of p is always equal to 3.14 or `(22)/(7)` . We can get the surface area of the sphere.  Curved surface area of the sphere is also called as lateral area of the sphere.


Example and Practice Problems

Find the Surface area of part of a sphere if the radius of the sphere is 105 cm.

Solution:

Given radius of the sphere r=105 cm.

Curved Surface Area:

A= 3 p r2 Square units

Here we have to substitute the values of r and p then we can get,

A= 3 x `(22)/(7)` x 105 Square centimeter

Divided by 7 we can get,

A=3 x 22 x15 Square centimeter

Multiplying this we can get,

A=990 Square centimeter

Total Surface Area:

A= 4 p r2 Square units

Here we have to substitute the values of r and p then we can get,

A= 4 x`(22)/(7)` x 105 Square centimeter

Divided by 7 we can get,

A=4 x 22 x15 Square centimeter

Multiplying this we can get,

A=1320 Square centimeter

Practice problems:

Find the surface area of the  sphere if the radius is 0.25 meter,        ans = 0.785 squrae meter
Find the surface area of the  sphere if the radius is 82 inches           ans = 84453.44 square inches.

Rectangular Pyramid Vertices

Introduction to rectangular pyramid vertices

The pyramid is the solid form objects through a polygon for a bottom.  Each faces joint on one point. The bottom of a rectangular pyramid is for all time a rectangle.  The rectangular pyramid includes the five vertices. It normally contains five sides.  Let us see about the rectangular pyramid vertices.

Rectangular Pyramid Vertices

A rectangular pyramid contains five vertices.

Rectangular pyramid is geometrical form within math. Usually pyramids are objects which contain a pyramid like formation through a triangular or else rectangular or else square or else pentagonal base etc. The bottom is which categorized the kind of pyramids. Pyramid through a rectangular bottom is identified as rectangular pyramid.

The rectangular pyramid contains the five vertices and five sides, eight edges.

In the above rectangular pyramid contains five vertices.

The volume of right rectangular pyramid = `1/3`  x base x height

Example

Given the length = 12 cm, width = 15 cm, height = 18 cm.

The volume of right rectangular pyramid = `1/3`  x base x height

= `1/3`  x (length x width) x height.

= `1/3`  x (12 x 15) x 18

= `1/3`  x 180 x 18

= `1/3`  x 3240

= 1080

Therefore the volume of right rectangular pyramid is 1080 cm3

Between, if you have problem on these topics geometric probability formula, please browse expert math related websites for more help on math word problem solver online.

Examples for Rectangular Pyramid

Example 1

Compute the volume of rectangular pyramid if length = 8 cm, width = 10 cm, height = 14 cm.                             

Solution

Given the length = 8 cm, width = 10 cm, height = 14 cm.

The volume of right rectangular pyramid = `1/3`  x base x height

= `1/3`  x (length x width) x height.

= `1/3`  x (8 x 10) x 14

= `1/3` * 80 x 14

= `1/3` * 1120

= 373.3

Therefore the volume of right rectangular pyramid is 373.3 cm3

Example 2

Compute the volume of rectangular pyramid if length = 6 cm, width = 12 cm, height = 20 cm.                             

Solution

Given the length = 6 cm, width = 12 cm, height = 20 cm.

The volume of right rectangular pyramid = `1/3`  x base x height

= `1/3`  x (length x width) x height.

= `1/3`  x (6 x 12) x 20

= `1/3`  x 72 x 20

= `1/3`  x 1440

= 480

Therefore the volume of right rectangular pyramid is 480 cm3

Polar Equation Cartesian

Introduction on polar equation Cartesian :

Polar Equations:

The polar equation system is a two-dimensional equation system in which each point on a plane is determined by a distance from a fixed point and an angle from a fixed direction.

Cartesian Coordinates:

A Cartesian coordinate system specifies each point uniquely in a plane by a pair of numerical coordinates, which are the signed distances from the point to two fixed perpendicular directed lines, measured in the same unit of length.

Complex Form of Polar Equation Cartesian:
A complex number is of the form x + iy, where x, y belongs to R and 'i' is called imaginary unit (i = `sqrt(-1)` )

Let z = x+ iy,

Real part of z = Re(z) = x, and imaginary part of z = Im(z) = y.

The point (2, 8) written as 2 + 8i. Cartesian form are used to solve non linear shallow -water equations on the sphere.

Let, z1 = a + ib; z2 = c + id

z1 = z2; a = c; b = d.

Sinz = Sin (a+ib)

= Sina Cos hb + iCosa Sin hb

Cosz = Cos (a+ib)

= Cosa Cos hb  - iSina Sin hb

Sinz = Cosz

Sina Cos hb = Cosa Cos hb

Cosa Sin hb = Sina Sin hb

I am planning to write more post on geometric probability formula, combination probability formula. Keep checking my blog.

Steps for Solving Polar Equation Cartesian:

The following steps are used to solve from Cartesian form to polar form,

Find the value of mod(z), | z | = `sqrt(x^2 + y^2)`

Find the `theta` value by using `tan theta = (y)/(x)`

Find q in radians.

By writing the equation in polar form, `z = r(cos theta + i sin theta)`.

Example for Solving Polar Equation Cartesian:

Question: Solve z = 1 + i in polar form.

Solution:  1) r = | z | = `sqrt(1^2 + 1^2) = sqrt(1 + 1) = sqrt2`

2) `tan theta = ` `1 / 1`  => `theta = ` 45°

3) `theta = (pi)/(4)`

4) The polar form is, z = r (`cos theta + i sin theta` )

=> z = `sqrt(2)(cos (pi /4) + sin(pi/4))`

The Polar form of z = 1 + i is, z = `sqrt(2)(cos (pi /4) + sin (pi/4))`