Question

To calculate the volume of the frustum of a pyramid.

Short answer

The volume of the frustum of a pyramid is \(\frac{1}{3}h\left( {{a^2} + ab + {b^2}} \right)\).

The volume of the frustum of a pyramid is \({b^2}h\) when \(a = b\).

The volume of the frustum of a pyramid is \(\frac{1}{3}{b^2}h\) when \(a = 0\).

Step-by-step solution

Given data

The frustum of a pyramid with square base of side \(b\), square top of side \(a\), and height \(h\).

Concept used of Frustum of pyramid

The portion of a solid (normally a cone or pyramid) that lies between one or two parallel planes which cuts it.

Solve to find the volume

Consider that the Equation of line \(x = \frac{{\Delta x}}{{\Delta y}}y + (x\)-intercept) ….(1)

Substitute \(\frac{a}{2} - \frac{b}{2}\) for \(\Delta x,h - 0\) for \(\Delta y\), and \(\frac{b}{2}\) for \(x\)-intercept in Equation (1).

\(\begin{aligned}x = &\frac{{\frac{a}{2} - \frac{b}{2}}}{{h - 0}}y + \frac{b}{2}\\ &= \frac{{a - b}}{{2h}}y + \frac{b}{2}\end{aligned}\)

The dimensions of the frustum of a pyramid as shown in Figure 1 .

Refer to Figure 1 .

Consider that the frustum of a pyramid is obtained by the line \(x = \frac{{a - b}}{{2h}}y + \frac{b}{2}\).

The region lies between \(a = 0\) and \(b = h\).

The expression to find the volume of the cap of a sphere as shown below.

\(V = \int_a^b A (y)dy\) ……(2)

Find the area of the cap of a sphere as shown below.

\(\begin{aligned}A(y) &= {(2x)^2}\\ &= {\left( {2\left( {\frac{{a - b}}{{2h}}y + \frac{b}{2}} \right)} \right)^2}\\ &= {\left( {\frac{{a - b}}{h}y + b} \right)^2}\\ &= \frac{{{{(a - b)}^2}}}{{{h^2}}}{y^2} + {b^2} + \frac{{2(a - b)}}{h}yb\end{aligned}\)

Which means, \(A(y) = \frac{{{{(a - b)}^2}}}{{{h^2}}}{y^2} + \frac{{2b(a - b)}}{h}y + {b^2}\)

Substitute 0 for a, h for \(b\), and \(\frac{{{{(a - b)}^2}}}{{{h^2}}}{y^2} + \frac{{2b(a - b)}}{h}y + {b^2}\) for \(A(y)\) in Equation (2).

\(\begin{aligned}V &= \int_0^h {\left( {\frac{{{{(a - b)}^2}}}{{{h^2}}}{y^2} + \frac{{2b(a - b)}}{h}y + {b^2}} \right)} dy\\ &= \left( {\frac{{{{(a - b)}^2}}}{{{h^2}}}\frac{{{y^3}}}{3} + \frac{{2b(a - b)}}{h}\frac{{{y^2}}}{2} + {b^2}y} \right)_0^h\\ &= \left( {\frac{{{{(a - b)}^2}}}{{{h^2}}}\frac{{{h^3}}}{3} + \frac{{2b(a - b)}}{h}\frac{{{h^2}}}{2} + {b^2}h - 0} \right)\\ &= \left( {\left( {{a^2} + {b^2} - 2ab} \right)\frac{h}{3} + bh(a - b) + {b^2}h} \right)\end{aligned}\)

On simplify further,

\(\begin{aligned}V &= \frac{1}{3}\left( {{a^2}h + {b^2}h - 2abh + 3abh - 3{b^2}h + 3{b^2}h} \right)\\ &= \frac{1}{3}\left( {{a^2}h + {b^2}h - 2abh} \right)\\ &= \frac{1}{3}\left( {{a^2} + {b^2} - 2ab} \right)\end{aligned}\) …..(3)

Therefore, the volume of the frustum of a pyramid is \(\frac{1}{3}\left( {{a^2} + {b^2} - 2ab} \right)\).

Calculate the volume of the frustum of a pyramid if \(a = b\)

Substitute \(b\) for \(a\) in Equation (3).

\(\begin{aligned}V &= \frac{1}{3}\left( {{b^2} + {b^2} + b \times b} \right)\\ &= \frac{1}{3}h \times 3{b^2}\\ &= {b^2}h\end{aligned}\)

Calculate the volume of the frustum of a pyramid if \(a = 0\).

Substitute 0 for \(a\) in Equation (3).

\(\begin{aligned}V &= \frac{1}{3}\left( {{0^2} + {b^2} + 0 \times b} \right)\\ &= \frac{1}{3}h \times {b^2}\\ &= \frac{1}{3}{b^2}h\end{aligned}\)

The volume of the frustum of a pyramid is \(\frac{1}{3}{b^2}h\) when \(a = 0\).