Here you will learn formula for the volume of a frustum of a cone with derivation and examples based on it.

Let’s begin –

## What is Frustum of Cone ?

**Frustum of a cone** is the solid obtained after removing the upper portion of the cone by a plane parallel to its base. The lower portion is the frustum of a cone.

**Height** : The perpendicular distance between the aforesaid plane and the base is called the height of the frustum.

## Volume of a Frustum of a Cone

The formula for volume of a frustum of a cone is

V = \(h\over 3\)\([A_1 + \sqrt{A_1A_2} + A_2]\)

[\(A_1\), \(A_2\) are the areas of bottom and top of the frustum]

V = \(\pi h\over 3\)\([{r_1}^2 + \sqrt{r_1r_2} + {r_2}^2]\)

where h is the height of the frustum, \(r_1\), \(r_2\) are the radii of the base and the top of frustum of a cone.

**Note** : Height h of the frustum is given by the relation,

\(l^2\) = \(h^2\) + \((r_1 – r_2)^2\)

**Also Read** : Area of a Frustum of a Cone – Formula and Derivation

### Derivation :

Let \(r_1\) and \(r_2\) be the radii of the bottom and top of the frustum respectively and h be the height of the frustum.

The frustum of a cone is the solid obtained after removing the upper portion (small cone) of it by a plane parallel to the base of the big cone.

Volume of frustum = Volume of bigger cone – Volume of smaller cone

= \(1\over 3\)\(\pi {r_1}^2 h_1\) – \(1\over 3\)\(\pi {r_2}^2 h_2\)

Since, [ \(h_1\over r_1\) = \(h_2\over r_2\) = k ]

= \(\pi k\over 3\)\([{r_1}^3 – {r_2}^3]\)

= \(\pi k\over 3\) (\(r_1 – r_2\)) \(({r_1}^2 + {r_1r_2} + {r_2}^2)\)

Volume = \(1\over 3\) (\(kr_1 – kr_2\)) \((\pi{r_1}^2 + \pi{r_1r_2} + \pi{r_2}^2)\)

V = \(1\over 3\) (\(h_1 – h_2\)) \((\pi{r_1}^2 + \sqrt{(\pi{r_1}^2)(\pi{r_2}^2)} + \pi{r_2}^2)\)

Volume = \(h\over 3\)\([A_1 + \sqrt{A_1A_2} + A_2]\)

Where \(A_1\), \(A_2\) are the areas of bottom and top of the frustum respectively and h = \(h_1 – h_2\).

Volume of frustum of Cone = \(h\over 3\)\([A_1 + \sqrt{A_1A_2} + A_2]\)

= \(h\over 3\) \((\pi{r_1}^2 + \sqrt{(\pi{r_1}^2)(\pi{r_2}^2)} + \pi{r_2}^2)\)

Volume = \(\pi h\over 3\)\([{r_1}^2 + \sqrt{r_1r_2} + {r_2}^2]\)

**Example** : A friction clutch is in the form of the frustum of a cone, the diameters of the ends being 8 cm, and 10 cm and length 8 cm. Find its Volume.

**Solution** : Here, radius,

\(r_1\) = \(10\over 2\) = 5cm,

\(r_2\) = \(8\over 2\) = 4cm

Slant height, l = 8 cm

Height h of the frustum is given by the relation,

\(l^2\) = \(h^2\) + \((r_1 – r_2)^2\)

\(\implies\) 64 = \(h^2\) + \((5 – 4)^2\) or \(h^2\) = 63 \(\implies\) h = 7.937

\(\therefore\) Volume = \(\pi h\over 3\)\([{r_1}^2 + \sqrt{r_1r_2} + {r_2}^2]\)

Volume = \(3.14 \times 7.937\over 3\) \(\times\) [25 + 20 + 16] = 506.75 \(cm^2\)