# Zone or Frustum of a Sphere

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

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

Where $$h$$ is the altitude, and $${r_1}$$ and $${r_2}$$ 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. $$S = 2\pi rh$$

Where $$S = {\text{area of a zone}}$$

$$h = {\text{altitude}}$$

$$r = {\text{radius of the sphere}}$$

(iii) The total surface area of a zone $$ = 2\pi rh + \pi {r_1}^2 + \pi {r_2}^2$$

__Example__**:**

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

__Solution__**:**

The surface area of the zone $$S = 2\pi rh\,\,\,\,\, = 2 \times 3.1415 \times 8 \times 4$$

$$ = 201.06\,{\text{sq}}{\text{.cm}}$$

Now $${r_1} = \sqrt {{{\left( {OB} \right)}^2} – {{\left( {OE} \right)}^2}} $$

$$ = \sqrt {{8^2} – {2^2}} \,\,\,\, = \sqrt {60} $$

$${r_2} = \sqrt {{{\left( {OD} \right)}^2} – {{\left( {OF} \right)}^2}} $$

$$ = \sqrt {{8^2} – {6^2}} \,\,\,\, = \sqrt {28} $$

$$\therefore $$ $$V = \frac{{\pi h}}{6}\left( {{h^2} + 3{r_1}{h^2} + 3{r_2}{h^2}} \right)$$

$$ = 586.43\,{\text{cu}}{\text{.cm}}$$

__Example__**:**

A stone was rolled into a hemispherical basin 8cm in diameter having a $$3\frac{1}{2}{\text{m}}$$ depth of water in it, when the water immediately rose to the top of the basin. What was the cubic content of the stone?

__Solution__**:**

Let the figure represent a vertical mid-section of the basin and stone. Let $$DE$$ and $$AB$$ indicate the level of the water before and after the stone rolled into the basin. Then

$$GC = 3\frac{1}{2}{\text{m}},\,\,\,FG = \frac{1}{2}{\text{m}},\,\,\,FB = 4{\text{m}}$$

If $$GE = x{\text{m}}$$, then

$${x^2} = 3\frac{1}{2} \times 4\frac{1}{2}\,\,\,\, = \sqrt {\frac{{63}}{4}} {\text{m}}$$

$$ \Rightarrow $$ $$x = \frac{{\sqrt {63} }}{2}{\text{m}}$$

Now, the cubical content of the stone:

$$ = \,{\text{the cubical content of the water displaced}}$$

$$ = \,{\text{the cubical content of the same ABED}}$$

$$\frac{{\pi \left( {\frac{1}{2}} \right)}}{6}\left[ {3\left( {\frac{{63}}{4} + 16} \right) + {{\left( {\frac{1}{2}} \right)}^2}} \right]{\text{cu}}{\text{. m}}$$

$$ = \frac{\pi }{{12}} \times \frac{{121}}{2} = 25\,{\text{cu}}{\text{.m}}$$ nearly.

__Segment of a Sphere__

For a special segment of one base, the radius of the lower base $${r_1}$$ is equal to zero. Therefore,

$$V = \frac{{\pi h}}{6}\left( {{h^2} + 3{r_2}^2} \right)$$

$$\therefore $$ In this case, the total surface area of the segment $$ = 2\pi rh + \pi {r^2}$$

__Example__**:**

Find the volume of a segment of a sphere whose height is $$4\frac{1}{2}{\text{cm}}$$ and the diameter of whose base is 8cm.

__Solution__**:**

Given that:

$$h = 4\frac{1}{2}{\text{cm}},\,\,\,\,\,{r_1} = 4{\text{cm}}$$

$$\therefore $$ The volume of the segment $$ = \frac{{\pi h}}{6}\left( {{h^2} + 3{r_2}^2} \right)$$

$$ = \frac{{\pi \times 9}}{{2 \times 6}}\left[ {{{\left( {\frac{9}{2}} \right)}^2} + 3 \times {4^2}} \right]$$

$$ = \frac{{22 \times 9 \times 27 \times 3}}{{7 \times 2 \times 6 \times 4}}$$

$$ = 160.8\,{\text{cu}}{\text{.cm}}$$