Triangles as Cross Sections (College Board AP® Calculus AB): Revision Note

Let $R$ be the region enclosed by the graph of $f\left(x\right)=\sqrt{4-x}$ and the $x$- and $y$-axes, as shown in the figure below.

Region $R$ is the base of a solid. For the solid, at each $x$ the cross section perpendicular to the $x$-axis is an equilateral triangle. Find the volume of the solid.

Answer:

Use $\mathrm{Volume}={\int }_a^{b}A\left(x\right)dx$

Use Pythagoras' theorem to work out the relevant lengths in the cross section

$\mathrm{height}=\sqrt{{\left(f\left(x\right)\right)}^2-{\left(\frac{f\left(x\right)}{2}\right)}^2}=\frac{\sqrt{3}}{2}f\left(x\right)$

At each $x$ the cross-sectional area is $\frac{1}{2}\times \mathrm{base}\times \mathrm{height}$

$A\left(x\right)=\frac{1}{2}·f\left(x\right)·\frac{\sqrt{3}}{2}f\left(x\right)=\frac{\sqrt{3}}{4}{\left(f\left(x\right)\right)}^2$

Now the volume integral can be used

$\begin{array}{rcl}\mathrm{Volume}& =& {\int }_0^4\frac{\sqrt{3}}{4}{\left(\sqrt{4-x}\right)}^{2}dx\\ & =& \frac{\sqrt{3}}{4}{\int }_0^4\left(4-x\right)dx\\ & =& \frac{\sqrt{3}}{4}{\left[4x-\frac{1}{2}{x}^2\right]}_0^4\\ & =& \frac{\sqrt{3}}{4}\left(\left(4\left(4\right)-\frac{1}{2}{\left(4\right)}^2\right)-\left(0\right)\right)\\ & =& 2\sqrt{3}\\ & =& 3.464101...\end{array}$

3.464 units cubed