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\centerline{MATH 136 -- Calculus 2}
\centerline{Practice Day on Volumes of Solids of Revolution}
\centerline{March 11, 2020}
\vskip 10pt
\noindent
{\it Background}
\vskip 10pt
As we have seen in some examples, let the region $R$ be defined by 
$0 \le y \le f(x)$ for all $x$ in $[a,b]$.
Then the volume of the solid of revolution obtained by rotating $R$ about
the $x$-axis is given by
\begin{equation}
\label{volbasic}
\text{V (circle cross-sections)} = \int_a^b \pi (f(x))^2\ dx
\end{equation}
since the cross-section at $x$ is a circle with radius $r = f(x)$.  
Note that this region ``extends all the way to the $x$-axis'' (the axis of rotation).
If there is a ``gap'' between the region and the axis of rotation, then the cross-sections
will be ``washers'' and we need to set up the volume integral differently:
\begin{equation}
\label{volwasher}
\text{V (washer cross-sections)} = \int_a^b \pi \left((\text{outer radius})^2 - (\text{inner radius})^2\right)\ dx
\end{equation}
Today we want to practice on using \eqref{volbasic} and \eqref{volwasher} to set up and compute
integrals for volumes of some regions of this type.
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\noindent
{\it Questions}
\vskip 10pt
For each problem, 
\begin{enumerate}
\item[(i)] Set up the volume integral for all of the following regions first. 
\item[(ii)] The go back and evaluate each integral using the FTC part I.  You 
may use a table of integrals anywhere here, as needed.  
\end{enumerate}
\begin{enumerate}
\item  The solid generated by rotating the region between $y = x^4$
and the $x$-axis, $[a,b] = [0,1]$.
%\item[] {\it Solution:}  (Circle cross-sections) The volume is computed by 
%$$V = \int_0^1 \pi (x^4)^2\ dx = \left. \pi\frac{x^9}{9}\right|_0^1 = \frac{\pi/9}$$
\item  The solid generated by rotating the region between $y = x^4$ and $y = x$, $[a,b] = [0,1]$.
%\item[] {\it Solution:}  (Washer cross-sections) Here the outer radius is $x$ and the inner radius is $x^4$ since
%$x > x^4$ for all $x$ with $0 < x < 1$.  The volume is 
%$$V = \int_0^1 \pi(x^2 - x^8)\ dx = \left. \pi\left(\frac{x^3}{3} - \frac{x^9}{9}\right)\right|_0^1 = 
%\frac{2\pi}{9}.$$ 
\item  The solid generated by rotating the region between $y = \sec(x)$, 
and $y = \cos(x)$,  $[a,b] = [0,\pi/4]$.
%\item[] {\it Solution:} (Washer cross-sections) The outer radius is $\sec(x)$ and the 
%inner radius is $\cos(x)$.  The volume is 
%\begin{align*}
%V &= \int_0^{\pi/4} \pi(\sec^2(x) - \cos^2(x))\ dx\\
%&= \left. \pi\left(\tan(x) - \frac{x}{2} - \frac{\cos(x)\sin(x)}{2}\right)\right|_0^{\pi/4}\\
%&= \pi\left(1 - \frac{\pi}{8} - \frac{1}{4}\right)
%&= \frac{6\pi - \pi^2}{8}.
%\end{align*} 
\item  The solid generated by rotating the region between 
$y = (1 - x^2)^{1/4}$ and the $x$-axis, $[a,b] = [0,1]$.
%\item[] {\it Solution:}  (Circle cross-sections) The volume is 
%$$\int_0^1 \pi(1 - x^2)^{1/2}\ dx = \pi\left.\left(\frac{x}{\sqrt{1 - x^2}}{2} + \frac{1}{2}\sin^{-1}(x)\right)
%\right|_0^1 = \frac{\pi^2}{4}.$$
\item  $f(x) = \sin(x)$, $[a,b] = [0,\pi]$.  
%\item[] {\it Solution:} (Circle cross-sections) The volume is 
%$$V = \int_0^\pi \pi\sin^2(x)\ dx = \pi\left.\left( \frac{x}{2} - \frac{\sin(x)\cos(x)}{2}\right)\right|_0^\pi
%= \frac{\pi^2}{2}.$$  
\end{enumerate}
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