Annulus is a Latin word meaning a ring. In mathematics annulus refer to the region between two concentric circles. Real Life examples of annulus are a circular path round a circular region, a circular running track etc. In a pipe or a tube with circular cross sections, the material is contained in the annular Cylinder formed.

## Area of Annulus

Area of annulus = Area of outer circle - Area of inner circle.
= πR2 - πr2 = π(R2 - r2) sq.units

### Solved Example

Question: A jogging path of width 4ft goes around a circular play area for children. The diameter of the play area is 90 ft. The administration is considering relaying the tiles in jogging path. if the cost of laying is quoted as 15 dollars per sq.ft, find the total cost for relaying tiles on the jogging path.

Solution:

The area of the path can be calculated using annulus area formula
where R = radius of the outer circle = 45 + 4 = 49 ft   and r = radius of the play area = 45 ft
Area of the jogging path = π(R2 - r2)
= π(492 - 452) = 1181.24 sq ft.
Cost of relaying per sq ft = 15 dollars
Total cost of relaying tiles on jogging path = 1181.24 x 15 = 17,718.60 dollars.

## Volume of Annulus

The walls of pipes or tubes form annular cylinders. Hence while finding the material content in a pipe or its weight, the volume of the annular cylinders are found.
Volume of annulus = Volume of the outer cylinder - volume of the inner cylinder
= πR2l - πr2l  = πl(R2 - r2) cubic units
Where R and r are correspondingly the outer and inner radii and l the length of the pipe.

### Solved Example

Question: A cylindrical tank has a wall width of 3 inches and inner diameter of 4 ft. If the height of the tank is 10ft, find the volume occupied by the walls of the tank ignoring the top and bottom bases. Round the answer to hundredth of a cubic ft.

Solution:

It is required to compute the annular volume here.
Inner diameter = 4ft.  Hence inner radius r = 2ft
Outer radius = r + wall width = 2 + 0.25 = 2.25 ft                       (3 inches = 0.25 ft)
Height of the cylinder h = 10 ft
Annular volume of the walls = πh(R2 - r2)
= π.10((2.25)2 -22) = 33.38 cubic feet.