Chapter 6.2, Problem 34E

To determine

To calculate: The volume of the frustum of a pyramid.

Answer to Problem 34E

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 $a=0$.

Explanation of Solution

Given:

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

Calculation:

Consider that the Equation of line $x=\frac{\Delta x}{\Delta y}y+\left(x\text{-intercept}\right)$ (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).

$\begin{array}{c}x=\frac{\frac{a}{2}-\frac{b}{2}}{h-0}y+\frac{b}{2}\\ =\frac{a-b}{2h}y+\frac{b}{2}\end{array}$

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.

$V={\displaystyle {\int }_a^{b}A\left(y\right)dy}$ (2)

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

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

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

$\begin{array}{c}V={\displaystyle {\int }_0^h\left(\frac{{\left(a-b\right)}^2}{h^2}{y}^2+\frac{2b\left(a-b\right)}{h}y+b^2\right)}dy\\ ={\left(\frac{{\left(a-b\right)}^2}{h^2}\frac{y^3}{3}+\frac{2b\left(a-b\right)}{h}\frac{y^2}{2}+b^{2}y\right)}_0^h\\ =\left(\frac{{\left(a-b\right)}^2}{h^2}\frac{h^3}{3}+\frac{2b\left(a-b\right)}{h}\frac{h^2}{2}+b^{2}h-0\right)\\ =\left(\left(a^2+b^2-2ab\right)\frac{h}{3}+bh\left(a-b\right)+b^{2}h\right)\end{array}$

On simplify further,

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

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

Calculate the volume of the frustum of a pyramid if $a=0$.

Substitute 0 for a in Equation (3).

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

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