The portion of a sphere intercepted between two parallel planes is called a zone (i.e. frustum).

(i) The volume of the zone (or frustum) of a sphere may be found by taking the difference between segment and the segment (see figure) that is

Where is the altitude, and are respectively the radii of the small circle (bases).

(ii) The surface area of the zone or frustum is equal to the circumference of the great circle of the sphere times the altitude of the same.

i.e.

Where

(iii) Total surface area of a zone

__Example__**:**

The sphere of radius **8cm** is cut by two parallel planes, one passing **2cm** from the center and other **6cm** from the center. Find the area of the zone and the volume of the segment between two planes if both planes are on the same side of the center.

__Solution__**:**

Surface Area of the zone

Now

__Example__**:**

A stone was rolled into a hemispherical basin 8cm diameter having depth of water in it, when the water immediately rose to the tip of the basin. What was the cubic content of the stone?

__Solution__**:**

Let the figure represents a vertical mid-section of the basin and stone. Let and indicate the level of the water before and after the rolling of the stone. Then

If , then

Now, the cubical content of the stone

nearly.

__Segment of a Sphere__**:**

For a special segment of one base, the radius of the lower base is equal to zero. Therefore,

In this case, the total surface area of the segment

__Example__**:**

Find the volume of a segment of a sphere whose height is and the diameter of whose base is 8cm.

__Solution__**:**

Given that:

Volume of the segment