Question

To calculate the volume of frustrum of the right circular cone.

Short answer

The volume of a frustum of the circular cone is \(\frac{1}{3}\pi h\left( {{R^2} + Rr + {r^2}} \right)\).

Step-by-step solution

Given data

The frustum of a right circular cone with height \(h\), lower base radius \(R\), and top radius \(r\).

Concept used of Right Circular Cone 

The volume of a right circular cone can be calculated by the use of the base radius and height of the cone.

Solve to find the volume

The dimension of the frustum of a right circular cone is as shown in Figure 1 .

From Figure 1, consider that the frustum of a right circular cone is obtained by rotation of the line about \(y\)-axis.

The expression to find the volume of the frustum of a right circular cone as shown below.

\(v = \int_a^b A (y)dy\) ….(1)

Find the area of the frustum of a right circular cone as shown below.

\(\begin{aligned}dA(y) &= \pi {r^2}\\ &= \pi {\left( {R - \frac{{R - r}}{h}y} \right)^2}\\ &= \pi \left( {{R^2} + \frac{{{{(R - r)}^2}}}{{{h^2}}}{y^2} - 2R\left( {\frac{{R - r}}{h}y} \right)} \right)\\ &= \pi \left( {{R^2} - 2R\left( {\frac{{R - r}}{h}y} \right) + \frac{{{{(R - r)}^2}}}{{{h^2}}}{y^2}} \right)\end{aligned}\)

Substitute 0 for a, h for \(b\), and \(\pi \left( {{R^2} - 2R\left( {\frac{{R - r}}{h}y} \right) + \frac{{{{(R - r)}^2}}}{{{h^2}}}{y^2}} \right)\) for \(A(y)\) in Equation (1).

\(\begin{aligned}V = &\int_0^h \pi \left( {{R^2} - \frac{{2R(R - r)}}{h}y + {{\left( {\frac{{R - r}}{h}} \right)}^2}{y^2}} \right)dy\\ &= \pi \left( {{R^2}y - \frac{{2R(R - r)}}{h}\frac{{{y^2}}}{2} + {{\left( {\frac{{R - r}}{h}} \right)}^2}\frac{{{y^3}}}{3}} \right)_0^h\\ &= \pi \left( {{R^2}h - \frac{{2R(R - r)}}{h}\frac{{{h^2}}}{2} + {{\left( {\frac{{R - r}}{h}} \right)}^2}\frac{{{h^3}}}{3} - 0} \right)\\ &= \pi \left( {{R^2}h - R(R - r)h + {{(R - r)}^2}\frac{h}{3}} \right)\end{aligned}\)

Simply further,

\(\begin{aligned}V &= \pi \left( {{R^2}h - {R^2}h + Rrh + \left( {{R^2} - {r^2} - 2Rr} \right)\frac{h}{3}} \right)\\ &= \pi \left( {Rrh + \left( {{R^2} - {r^2} - 2Rr} \right)\frac{h}{3}} \right)\\ &= \pi \left( {\frac{{Rrh + {R^2}h - {r^2}h - 2Rrh}}{3}} \right)\\ &= \pi \left( {\frac{{Rrh + {R^2}h - {r^2}h - 2Rrh}}{3}} \right)\end{aligned}\)

Which means, \(V = \frac{1}{3}\pi h\left( {{R^2} + Rr + {r^2}} \right)\)

Therefore, the volume of the frustum of a right circular cone is \(\frac{1}{3}\pi h\left( {{R^2} + Rr + {r^2}} \right)\).