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\centerline{MATH 134 -- Calculus with Fundamentals 2}
\centerline{Discussion Day 2 -- More on volumes with ``washer'' cross-sections}
\centerline{February 22, 2018}
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\noindent
{\it Background}
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Recall that when a solid of revolution has ``washer'' cross sections 
with outer radius $r_{out}$ and inner radius $r_{in}$ (both possibly depending on $x$), 
the volume is 
computed by
$${\rm Volume} = \int_a^b \pi(r_{out})^2 - \pi(r_{in})^2\ dx$$
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\noindent
{\it Questions}
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\begin{enumerate}
\item[(1)] Find the volume of the solid obtained by rotating the region bounded
by $y = 4$, $y = x/2$ for $x = 0$ to $x = 1$ about the $x$-axis.  (It's a solid cylinder
with a cone ``drilled out.'')  First draw the 
region, then identify $r_{out}, r_{in}$ for the rotation about the $x$-axis, set up 
the volume integral, then carry out the integration. 
\item[(2)]  Same region as in (1), but rotated about the vertical line $x = -1$.
\item[(3)]  Now consider the solid obtained by rotating the region inside the 
circle $x^2 + (y - 3)^2 = 1$ about the $x$-axis.  Since that region does
not extend all the way down to the $x$-axis, this solid will also have 
washer cross-sections.
\begin{enumerate}
\item[(a)] Solve the equation of the circle for $y$ to get 
$$y = 3 \pm \sqrt{1 - x^2}$$
One of the choices of the sign will give you the outer radius and one
will give you the inner radius.  Which is which?
\item[(b)] Set up the volume integral. Note that the limits of integration correspond
to the $x$-values farthest to the left and to the right on the given circle.  What
are they?
\item[(c)] Square out the two terms in $\pi (r_{out})^2 - \pi (r_{in})^2$, and 
simplify.  Explain why what you get is 
$${\rm Volume} = 12\pi \int_{-1}^1 \sqrt{1 - x^2}\ dx$$
\item[(d)]  The integral in part (c) is one you can do just by recognizing it as
an \emph{area}.  What is the value?  ({\it DON'T}\/ try to find an antiderivative; 
we don't know how to do that yet!)
\item[(e)]  A famous theorem of the ancient Greek mathematician Pappus of Alexandria (ca. 290 - ca. 350 CE) says (in a particular case) 
that the volume of the solid obtained by rotating a circular disk about an axis that
does not intersect the circle is
equal to {\it the area of the circle, times the distance the center of the circle travels
as it rotates around the line}.  Check that your answer from part (d) agrees with this(!) 
\end{enumerate}
\end{enumerate}

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