Find the volume $V$ of a pyramid of height $h$ with a square base, each side of length $b$.

Solution We position the pyramid with its vertex at the origin and its axis along the positive $x$-axis. Then the area $A$ of a typical cross section at $x$ is a square. Let $s$ denote the length of the side of the square at $x$. See Figure 42(a). We form two triangles: one with height $x$ and side $s$, the other with height $h$ and side $b.$ These triangles are similar (AAA), as shown in Figure 42(b). Then we have $$\begin{equation*} \dfrac{x}{s}=\dfrac{h}{b}\qquad\hbox{ or }\qquad s =\dfrac{b}{h}x \end{equation*}$$

Figure 42

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So, $s$ is a function of $x$, and the area $A$ of a typical cross section is $A=s^{2}=\dfrac{b^{2}}{h^{2}}x^{2}$. See Figure 42(c). The volume $V$ of a typical cross section is $V=\dfrac{ b^{2}}{h^{2}}x^{2}\Delta x$. See Figure 42(d).

Since $A$ is a continuous function of $x$, where the slice occurred, we have $$\begin{equation*} V=\int_{0}^{h}A(x) \,{\it dx}=\displaystyle \int_{0}^{h}\dfrac{b^{2}}{h^{2}} x^{2}{\it dx}= \frac{b^{2}}{h^{2}}\int_{0}^{h}x^{2}{\it dx}=\frac{b^{2}}{h^{2}} \left[ \frac{x^{3}}{3} \right] _{0}^{h}=\frac{1}{3}b^{2}h\hbox{ cubic units} \end{equation*}$$