# Volume of a Frustum of a Cone – Formula and Derivation

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,

$$l^2$$ = $$h^2$$ + $$(r_1 – r_2)^2$$

### 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.

$$r_1$$ = $$10\over 2$$ = 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 $$cm^2$$