Frustum of a Cone

If a cone is cut by a plane parallel to its base, the portion of a solid between this plane and the base is known as the frustum of a cone.

The volume denoted by ABCD in the figure is the frustum of the cone ABE.


frustum-cone-01
 

Volume of the Frustum of a Cone

Since we know that cone is a limit of a pyramid, therefore the frustum of a cone will be the limit of the frustum of a pyramid. But the volume of a pyramid is

V = \frac{1}{3}h\left[ {{A_1} + {A_2} + \sqrt {{A_1}{A_2}} } \right]


Where              {A_1} = \pi {R^2}
{A_2} = \pi {r^2}
\therefore                         = \frac{h}{3}\left( {\pi {R^2} + \pi {r^2} + \pi \sqrt {{R^2}{r^2}} } \right)
 = \frac{{\pi h}}{3}\left( {{R^2} + {r^2} + Rr} \right)


frustum-cone-02
Example:

A cone 12cm high is cut 8cm from the vertex to form a frustum with a volume of 190cu.cm. Find the radius of the cone.

 

Solution:

Given that
Height of cone             = 12{\text{cm}}
Height of the frustum       h = 12 - 8 = 4{\text{cm}}
Volume of the frustum      = 190{\text{cu}}.{\text{cm}}

            Now the volume of the frustum cone
 = \frac{{\pi h}}{3}\left( {{R^2} + {r^2} + Rr} \right)
or         190 = \frac{{3.1415 \times 4}}{3}\left[ {\left( {\frac{2}{3}{r^2}} \right) + {r^2} + \frac{2}{3}{r^2}} \right] = 4.19 \times \frac{{19}}{9}{r^2}
or         {r^2} = \frac{{190 \times 9}}{{19 \times 4.19}} = \frac{{1710}}{{79.61}} = 21.48
r = \sqrt {21.40} = 4.63

            Hence the required radius of the cone  = 4.63

 

Curved Surface Area of the Frustum of a Cone

Since a cone is the limiting case of a pyramid, therefore the lateral surface of the frustum of a cone can be deduced from the slant surface of the frustum of a pyramid, i.e. the curved (lateral) surface of the frustum of the cone.


frustum-cone-03

             = \frac{1}{2}\left( {{\text{sum of circumferences of bases}}} \right) \times \left( {{\text{slant height}}} \right)
 = \frac{1}{2}\left( {2\pi R + 2\pi r} \right)l\,\,\, = \pi \left( {R + r} \right)l
l, being the slant height of the frustum, R and r being the two radii of bases.

Note:
(1)        Total surface area of the frustum of a cone

 = \pi {R^2} + \pi {r^2} + \pi \left( {R + r} \right)l


(2)        To find the slant height of the cone, use the Pythagorean theorem.

 

Example:

A bucket is in the shape of the frustum of a right circular cone, as shown in the figure below. Find the volume and the total surface area of the bucket.

 

Solution:


frustum-cone-04

            Slant height                 l = \sqrt {{h^2} + {r^2}} \,\,\, = \sqrt {{{\left( {17} \right)}^2} + {{\left( {2.5} \right)}^2}} \,\,\, = 17{\text{cm}}
Lateral surface area      = \pi \left( {R + r} \right)l
 = \pi \left( {7.5 + 5} \right)\left( {17} \right)\,\,\,\, = 670\,{\text{sq}}{\text{.cm}}
\therefore            Base areas                    = \pi {R^2} + \pi {r^2}\,\,\,\, = \pi \left( {{R^2} + {r^2}} \right)
 = \pi \left( {56.25 + 25} \right)\,\,\,\, = 177 + 79
\therefore            Total surface area         = \pi {R^2} + \pi {r^2} + \pi \left( {R + r} \right)l
 = 177 + 79 + 670
 = 926\,{\text{sq}}{\text{.cm}}

            Volume            = \frac{{\pi h}}{3}\left( {{R^2} + {r^2} + Rr} \right)
 = \frac{{17h}}{3}\left( {{{\left( {7.5} \right)}^2} + {{\left( 5 \right)}^2} + \left( {7.5} \right)\left( 5 \right)} \right)
 = 2114\,{\text{cu}}{\text{.cm}}