Question
Let $R$ represent the first-quadrant region bounded by the $y$-axis and the curves $y=2^x$ and $y=8 \cos \frac{\pi x}{6}$, as shown in the graph.
(a) Find the area of region $R$.
(b) Set up, but do not evaluate, an integral expression for the volume of the solid formed when $R$ is rotated around the $x$-axis.
(c) Set up, but do not evaluate, an integral expression for the volume of the solid whose base is $R$ and all cross sections in planes perpendicular to the $x$-axis are squares.
▶️Answer/Explanation
(a) Draw a vertical element of area, as shown.
$
\begin{aligned}
\Delta A & =\left(y_{\text {top }}-y_{\text {bottom }}\right) \Delta x=\left(8 \cos \frac{\pi x}{6}-2^x\right) \Delta x \\
A & =\int_0^2\left(8 \cos \frac{\pi x}{6}-2^x\right) d x \\
& =\frac{6}{\pi} \cdot 8 \int_0^2 \cos \frac{\pi x}{6} d x-\int_0^2 2^x d x \\
& =\left.\frac{48}{\pi} \cdot \sin \frac{\pi x}{6}\right|_0 ^2-\left.\frac{2^x}{\ln 2}\right|_0 ^2 \\
& =\frac{48}{\pi}\left(\sin \frac{\pi}{3}-\sin 0\right)-\left(\frac{2^2}{\ln 2}-\frac{2^0}{\ln 2}\right) \\
& =\frac{24 \sqrt{3}}{\pi}-\frac{3}{\ln 2}
\end{aligned}
$
(b) Use washers; then
$
\begin{gathered}
\Delta V=\left(r_2^2-r_1^2\right) \Delta x=\pi\left(y_{\text {top }}^2-y_{\text {bottom }}^2\right) \Delta x \\
V=\pi \int_0^2\left[\left(8 \cos \frac{\pi x}{6}\right)^2-\left(2^x\right)^2\right] d x
\end{gathered}
$
(c)
See the figure above.
$
\begin{aligned}
\Delta V & =s^2 \Delta x=\left(y_{\text {top }}-y_{\text {bottom }}\right)^2 \Delta x \\
V & =\int_0^2\left(8 \cos \frac{\pi x}{6}-2^x\right)^2 d x
\end{aligned}
$