Find the volume $V$ of the solid whose base is the region bounded by the graphs of $y=\sqrt{x}$ and $y=\dfrac{1}{8}x^{2}$, if slices perpendicular to the base along the $x$-axis have cross sections that are squares.

Figure 43

Solution We begin by graphing the region bounded by the graphs of $y=\sqrt{x}$ and $y=\dfrac{1}{8}x^{2}$, as shown in Figure 43. The points of intersection of the two graphs are found by solving the equation $$\begin{eqnarray*} \sqrt{x} &=&\dfrac{1}{8}x^{2} \\[2pt] x &=&\dfrac{1}{64}x^{4} \\[2pt] x^{4}-64x &=&0 \\[2pt] x( x^{3}-64) &=&0 \\[2pt] x &=&0 \quad\hbox{or}\quad x=4 \end{eqnarray*}$$

The two graphs intersect at the points $\left( 0,0\right)$ and (4,2).

The solid with slices perpendicular to the base along the $x$-axis that are squares is shown in Figure 44(a). See Figure 44(b). A slice perpendicular to the $x$-axis at $x_{i}$ is a square with side $s_{i}=\displaystyle\sqrt{x_{i}}-\dfrac{1}{8}x_{i}^{2}.$ The area $A_{i}$ of the cross section at $x_{i}$ is $$A_{i}=s_{i}^{2}=\left( \sqrt{x_{i}}-\dfrac{1}{8} x_{i}^{2}\right) ^{2}$$

Figure 44

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See Figure 44(c). The volume $V_{i}$ of a typical slice is $$\begin{eqnarray*} V_{i}&=& \hbox{(Area of the cross section)(Thickness of the slice)}\\[3pt] &=&A (x_{i}) \Delta x=\left( \sqrt{x_{i}}-\dfrac{1}{8}x_{i}^{2}\right) ^{2}\Delta x \end{eqnarray*}$$

The volume $V$ of the solid is $$\begin{eqnarray*} V &=&\int_{0}^{4}A(x) {\it dx}=\int_{0}^{4}\left( \sqrt{x}-\dfrac{1}{8} x^{2}\right) ^{2}{\it dx}=\int_{0}^{4}\left( x-\dfrac{1}{4}x^{5/2}+\dfrac{1}{64} x^{4}\right) {\it dx} \notag \\[5pt] &=&\left[ \dfrac{x^{2}}{2}-\dfrac{1}{14}x^{7/2}+\dfrac{1}{320}x^{5}\right] _{0}^{4}=8-\dfrac{128}{14}+\dfrac{1024}{320}=\dfrac{72}{35}\hbox{ cubic units } \end{eqnarray*}$$