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\begin{document}

\begin{center}
{\Large Warm-Up Problems and Lecture Problems

Solutions

March 17, 2003}
\end{center}

\begin{enumerate}

\item  (Review)  Rotate the region in the plane bounded by $y=x^2$ and $y=4x$ 
	around the axis $y=-5$.
	Set up an integral representing the volume of this region.
	\\\sol
	\[
	V = \int_0^4 \pi (r_o^2-r_i^2) \; dx 
		= \pi \int_0^4 (5+4x)^2 - (5+x^2)^2 \; dx = \frac{1216\pi}{5}
	\]

\item   Sketch the parametric curve and find the length (at least set up the integral!).
	\begin{enumerate}
	\item  $x=t^3/3$, $y=t^2/2$, $0\leq t \leq 1$.
		\\
	(Remember that the formula for arc length is $L=\int_a^b \sqrt{(dx/dt)^2 + (dy/dt)^2}\;dt$.)
		\\\sol
		\[
		L = \int_0^1 \sqrt{ (t^2)^2 + (t)^2} \; dt = \int_0^1 t \sqrt{t^2+1} \; dt = \frac{1}{3}(2^{3/2} - 1)
		\]

	\item  $x=4\sin t$, $y=4 \cos t -5$, $0 \leq t \leq \pi$.
		\\\sol
		\[
		L = \int_0^\pi \sqrt{(4\cos t)^2 + (-4\sin t)^2}\; dt = \int_0^\pi 4 \; dt = 4\pi
		\]
	\end{enumerate}


\newpage

\lecture

\item  Set up the integral representing the length of the curve (try to compute the integral if you have time -- keep in mind that you might have to use numerics).
	\begin{enumerate}
	\item  $y=e^x$, from $(0,1)$ to $(1,e)$.
		\\\sol
		\[
		L=\int_0^1 \sqrt{1+e^{2x}} \; dx \cong 2.0035
		\]

	\item  $y=\frac{x^3}{6} + \frac{1}{2x}$, $\frac{1}{2} \leq x \leq 1$.
		\\\sol
		\[
		L = \int_{1/2}^1 \sqrt{1+ \left(\frac{x^2}{2}-\frac{1}{2x^2}\right)^2} \; dx \cong 0.6458
		\]

	\item  $y=\sin x$, $0\leq x \leq 2\pi$.
		\\\sol
		\[
		L = \int_0^{2\pi} \sqrt{1+\cos^2 x} \; dx \cong 3.8202
		\]

	\item  $x=y^4$, $-1\leq y \leq 2$.
		\\\sol
		\[
		L = \int_{-1}^2 \sqrt{1+(4y^3)^2}\; dy \cong 18.2471
		\]
	\end{enumerate}

\end{enumerate}

\end{document}