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\centerline{MATH 134 -- Calculus with Fundamentals 2}
\centerline{Practice Day on computing volumes of solids via integrals}
\centerline{February 20, 2018}
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\noindent
{\it Background}
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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}
{\rm Volume} = \int_a^b \pi (f(x))^2\ dx.
\end{equation}
Today we want to practice on using \eqref{volbasic} to set up and compute
integrals for volumes of some regions of this type.
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\noindent
{\it Questions}
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For each problem, 
\begin{enumerate}
\item[(i)] Set up the volume integral
\item[(ii)]  Evaluate it using the FTC part I.
\end{enumerate}
\begin{enumerate}
\item  $f(x) = x^4$, $[a,b] = [0,1]$.
\item  $f(x) = \sec(x)$, $[a,b] = [0,\pi/4]$.
\item  $f(x) = \sqrt{x}\cdot (1 - x^2)^{1/4}$, $[a,b] = [0,1]$.  
\item  $f(x) = \sin(x)$, $[a,b] = [0,\pi]$.  For this one you will want to 
use a trigonometric identity to rewrite your function before integrating:
$$\sin^2(x) = \frac{1}{2}(1 - \cos(2x))$$
(The reason this works is because of two other trig identities:
\begin{align*}
\cos^2(x) + \sin^2(x) &= 1\\
\cos^2(x) - \sin^2(x) &= \cos(2x).
\end{align*}
Subtract the second from the first and solve for $\sin^2(x)$ to see the identity above.) 
\end{enumerate}
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\caption{The region for problem 4.}
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