Zone or Frustum of a Sphere

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 EBD and the segment ABC (see figure) that is

Where h is the altitude, {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) Total surface area of a zone  = 2\pi rh + \pi  {r_1}^2 + \pi {r_2}^2


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.

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}}

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


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


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.


Given that:
            h  = 4\frac{1}{2}{\text{cm}},\,\,\,\,\,{r_1} = 4{\text{cm}}
\therefore            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}}