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\centerline{MATH 136 -- Calculus 2}
\centerline{Volumes by Slicing}
\centerline{March 10, 2020}
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\noindent
{\it Background}
If a solid extends along the $x$-axis between $a$ and $b$ and the
area of the cross-section by a plane $x = {\rm const}$ is given by some
function $A(x)$ for all $x$, then {\it Cavalieri's Principle} (named after
the Italian Renaissance mathematician Bonaventura Cavalieri, 1598-1647)
says that 
$${\rm Volume} = \int_a^b A(x)\ dx$$
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\noindent
{\it Questions}
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\begin{enumerate}
\item A circular cone extends from $x = 0$ to $x = 4$ along the $x$-axis.
The cross-section in each plane $x = {\rm constant}$ is a circle 
whose radius increases linearly from $r = 0$ 
at $x = 0$ to $r = 5$ by the time $x$ reaches $4$.  
\begin{enumerate}
\item Write the radius $r$ as a function of $x$ on the interval $0$ to $4$.
\item[] {\it Solution:}  The radius is 
$$r(x) = 5x/4$$
\item Write the area of the circular cross-section as a function of $x$.
\item[] {\it Solution:}  $A(x) = \pi (5x/4)^2 = \frac{25\pi x^2}{16}$.
\item Find the volume by applying the Cavalieri Principle equation above.
\item[] {\it Solution:}  The volume is 
$$\int_0^4 \frac{25\pi x^2}{16}\ dx = \left.\frac{25\pi x^3}{48}\right|_0^4 = 
\frac{100\pi}{3}.$$ 
\item Check your result with the formula for the volume of a cone from 
high school geometry.
\item[] {\it Solution:}  The formula is $V_{\rm cone} = \frac{1}{3} \pi R^2 h$
where $R$ is the radius of the base, and $h$ is the height.  Here $R = 5$ and $h = 4$, so 
$V_{\rm cone} = \frac{100\pi}{3},$
which agrees with the value computed above(!)
\end{enumerate}
\item A solid extends along the $x$-axis from $x = -3$ to $x = 3$.
The cross-section in each plane $x = {\rm constant}$ is a semicircle
with radius $e^x$.  Find the volume of the solid using Cavalieri's Principle.
\item[] {\rm Solution:}  The volume is 
$$V = \int_{-3}^3 \frac{\pi}{2} (e^x)^2\ dx = \int_{-3}^3 \frac{\pi}{2} e^{2x}\ dx 
= \left.\frac{\pi}{4}e^{2x}\right|_{-3}^3 = \frac{\pi}{4}(e^6 - e^{-6}).$$
\item A solid extends from $x = 0$ to $x = 10$ along the $x$-axis.  
The cross-section at $x$ is a square with side $\sqrt[4]{25 + x^2}$.
Set up the integral to compute the volume using Cavalieri's Principle.
Evaluate the integral using a trigonometric substitution or a table of
integrals.  
\item[] {\it Solution:}  The volume is 
$$\int_0^{10} (\sqrt[4]{x^2 + 25})^2\ dx = \int_0^{10} \sqrt{x^2 + 25}\ dx.$$
Using the trig subsitution $x = 5\tan\theta$ or a table, 
$$= \left. \frac{x\sqrt{x^2 + 25}}{2} + \frac{25}{2}\ln|x + \sqrt{x^2 + 25}|\right|_0^{10} \doteq 73.95.$$
\item In the previous problem, would it make an difference if the 
squares all had center point along the $x$-axis or if all the squares
had one corner at a point on the $x$-axis?
\item[] {\it Solution:}  No, the volumes would be equal(!)
\end{enumerate}
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