Semicircles as Cross Sections (College Board AP® Calculus AB): Study Guide

Let $R$ be the triangular region with vertices $\left(0,0\right)$, $\left(0,2\right)$ and $\left(4,0\right)$, 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 a semicircle. Find the volume of the solid.

Answer:

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

To define $A\left(x\right)$, first find the equation of the line through points $\left(0,2\right)$ and $\left(4,0\right)$

$\mathrm{gradient}=\frac{0-2}{4-0}=-\frac{1}{2}$

$\begin{array}{rcl}y-0& =& -\frac{1}{2}\left(x-4\right)\\ y& =& 2-\frac{1}{2}x\end{array}$

That's the diameter of each semicircle; to find the radius divide by two

$\mathrm{radius}=\frac{\left(2-{\displaystyle \frac{1}{2}}x\right)}{2}=1-\frac{1}{4}x$

At each $x$ the cross-sectional area is $\frac{1}{2}\pi r^2$

$\begin{array}{rcl}A\left(x\right)& =& \frac{1}{2}\pi {\left(1-\frac{1}{4}x\right)}^2\\ & =& \frac{1}{2}\pi \left(1-\frac{1}{2}x+\frac{1}{16}{x}^2\right)\\ & =& \frac{\pi }{32}\left(16-8x+x^2\right)\end{array}$

Now the volume integral can be used

$\begin{array}{rcl}\mathrm{Volume}& =& {\int }_0^4\frac{\pi }{32}\left(16-8x+x^2\right)dx\\ & =& \frac{\pi }{32}{\int }_0^4\left(16-8x+x^2\right)dx\\ & =& \frac{\pi }{32}{\left[16x-4x^2+\frac{1}{3}{x}^3\right]}_0^4\\ & =& \frac{\pi }{32}\left(\left(16\left(4\right)-4{\left(4\right)}^2+\frac{1}{3}{\left(4\right)}^3\right)-\left(0\right)\right)\\ & =& \frac{\pi }{32}\left(64-64+\frac{64}{3}\right)\\ & =& \frac{2\pi }{3}\\ & =& 2.094395......\end{array}$

2.094 units cubed (to 3 decimal places)