Chapter 6.2, Problem 36E

To determine

To calculate: The volume of the described solid (pyramid).

Answer to Problem 36E

The volume of the pyramid is $\underset{\_}{\frac{\sqrt{3}}{12}{a}^{2}h}$.

Explanation of Solution

Given information:

The pyramid with height h and base b an equilateral triangle with side a.

Consider the side length at any height is $\alpha$.

The dimension of the pyramid as shown in Figure 1.

Refer to Figure 1.

The side length is 0 at the height $h$ and a at $h=0$.

Apply proportion for the cross section.

$\begin{array}{l}\frac{\alpha }{a}=\frac{h-y}{h}\\ \alpha =a\left(1-\frac{y}{h}\right)\end{array}$

The region lies between $a=0$ and $b=h$.

Calculation:

The expression to find the volume of the pyramid as shown below.

$V={\displaystyle {\int }_a^{b}A\left(y\right)dy}$ (1)

Find the area of the equilateral triangle as shown below.

$\begin{array}{c}A\left(y\right)=\frac{\sqrt{3}}{4}{\alpha }^2\\ =\frac{\sqrt{3}}{4}{\left(a\left(1-\frac{y}{h}\right)\right)}^2\\ =\frac{\sqrt{3}{a}^2}{4}\left(1+\frac{y^2}{h^2}-2\frac{y}{h}\right)\end{array}$

Substitute 0 for a, h for b, and $\frac{\sqrt{3}{a}^2}{4}\left(1+\frac{y^2}{h^2}-2\frac{y}{h}\right)$ for $A\left(y\right)$ in Equation (1).

$\begin{array}{c}V={\displaystyle {\int }_0^h\frac{\sqrt{3}{a}^2}{4}\left(1+\frac{y^2}{h^2}-2\frac{y}{h}\right)dy}\\ =\frac{\sqrt{3}{a}^2}{4}{\left(y+\frac{1}{h^2}\frac{y^3}{3}-\frac{2}{h}\frac{y^2}{2}\right)}_0^h\\ =\frac{\sqrt{3}{a}^2}{4}\left(h+\frac{1}{h^2}\frac{h^3}{3}-\frac{h^2}{h}-0\right)\\ =\frac{\sqrt{3}{a}^2}{4}\left(h+\frac{h}{3}-h\right)\end{array}$

$=\frac{\sqrt{3}}{12}{a}^{2}h$

Therefore, the volume of the pyramid is $\underset{\_}{\frac{\sqrt{3}}{12}{a}^{2}h}$.