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

Solutions

March 12, 2003}
\end{center}

\begin{enumerate}

\item   Consider the function $y=x^2$.
	\begin{enumerate}
	\item  For a given $x$-value, consider the vertical line segment from the $x$-axis to the graph of $y=x^2$.  Let $r_1(x)$ denote the length of this line segment.   Find a nice expression for $r_1(x)$.
	\\\sol  The length of this line segment is the $y$-value, so $r_1(x)=x^2$.

	\item  For a given $y$-value, consider the horizontal line segment from the $y$-axis to the graph of $y=x^2$.  Let $r_2(y)$ denote the length of this line segment.   Find a nice expression for $r_2(y)$.
	\\\sol  The length of this line segment is the $x$-value, so $r_2(y)=\sqrt{y}$.
	\end{enumerate}


\item  Find the area of an equilateral triangle with side length $a$.
	\\\sol
	Draw the picture.  We know the area of a triangle is $\frac{1}{2}bh$ where $b$ is the length of a base and $h$ is the length of the height (perpendicular to the base).  So, you have to use a little geometry and the Pythagorean Theorem to see if the base has length $a$, then the height has length $\frac{\sqrt{3}}{2}a$.
Therefore the area is $\frac{\sqrt{3}}{4}a^2$.


\newpage

\lecture

\item  Take the region bounded by $y=e^x$, the $x$-axis, $y=0$ and $y=1$.
	Rotate the region around the $x$-axis and compute the volume.
	\\Here are the steps:
	\begin{enumerate}
	\item  Draw the region.
	\item  For a given $x$-value, find the height of the vertical line segment from the $x$-axis to the curve $y=e^x$.  (This will be the radius of the circular cross section).
	\\\sol Height is equal to $y$, or $e^x$.

	\item  Find the area of a cross section ($A=\pi r^2$).
	\\\sol $A(x)=\pi(e^x)^2 = \pi e^{2x}$.

	\item  Set up the integral representing the volume.
	\\\sol
	\[
	\mbox{Vol} = \int_0^1 \pi e^{2x}\; dx
	\]

	\item  Compute the integral.
	\[
	\mbox{Vol} = \int_0^1 \pi e^{2x}\; dx = \frac{\pi}{2}(e^2-1) \cong 10.0359
	\]

	\end{enumerate}





\end{enumerate}

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