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\begin{center}
Math 132: Calculus for Physical and Life Sciences 2 \\ 
Solutions for Problem Set 6 \\ 
Due Friday, March 14, 2008, at the beginning of class.
\end{center}
\vspace{.1in}

Note: To save space, only a few of the graphs are shown.

\begin{enumerate}
\item Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line.  Sketch the region, and a typical disk or washer.
\begin{enumerate}

\item $y=x^2+2$, $x=1$, $x=2$, $y=0$; about the $x$-axis 

\vspace{.1in}
\sol
In the following plot we show the region being rotated and a
typical slice through the region by a plane perpendicular
to the $x$-axis (shown in red).  The red curve generates
the cross-section, which is a disk in this case.
 
\begin{center}
\includegraphics[width=2.8truein,height=2.8truein]{132PS6S1a.ps}
\end{center}

The cross-section by the plane $x = c$ perpendicular to the $x$-axis is a disk
with radius $x^2 + 2$ and area $\pi (x^2 + 2)^2$.  The volume is 
\begin{eqnarray*}
V &=& \int_1^2 \pi (x^2 + 2)^2\ dx\\
&=& \pi \int_1^2 x^4 + 4x^2 + 4\ dx\\
&=& \pi \left( \left.\frac{x^5}{5} + \frac{4x^3}{3} + 4x\right|_1^2\right)\\
&=& \pi \left[\left(\frac{32}{5} + \frac{32}{3} + 8\right) - \left(\frac{1}{5} + \frac{4}{3} + 4\right)\right]\\
&=& \frac{293\pi}{15}.
\end{eqnarray*}
\item $y=\frac{x}{2}+3$, $y=x^2$; about the $x$-axis

\vspace{.1in}
\sol
The line $y = \frac{x}{2} + 3$ crosses the parabola $y = x^2$ at
$x = -\frac{3}{2},2$. The cross-section by a plane $x = c$ is
a washer with inner radius $x^2$ and outer radius $\frac{x}{2} + 3$ (except
for one disk at $c = 0$ where the hole ``closes up'' for a single $x$-value).

\begin{center}
\includegraphics[width=2.8truein,height=2.8truein]{132PS6S1b.ps}
\end{center}

The volume is 
\begin{eqnarray*}
V &=& \int_{-3/2}^2 \pi \left(\frac{x}{2} + 3\right)^2 - \pi (x^2)^2\ dx\\
&=& \pi \int_{-3/2}^2 \frac{x^2}{4} + 3x + 9 - x^4\ dx\\
&=& \pi \left( \left.\frac{x^3}{12} + \frac{3x^2}{2} + 9x - \frac{x^5}{5} \right|_{-3/2}^2\right)\\
&=& \pi \left(\frac{274}{15} - \frac{-711}{80}\right)\\
&=& \frac{6517\pi}{240}.
\end{eqnarray*}
\item $y=\sin x$, $x=0$, $x=\pi$, $y=0$; about the line $y = 2$. 

\vspace{.1in}
\sol
The cross-section by a plane $x = c$ is a washer with inner radius
$2 - \sin(x)$ and outer radius $2$.  The volume is 
\begin{eqnarray*}
V &=& \int_0^\pi 4\pi - \pi (2 - \sin(x))^2 \ dx\\
&=& \pi \int_0^\pi 4 \sin(x) - \sin^2(x)\ dx\\
&=& \pi \left(\left. -4\cos(x) - \frac{x}{2} + \frac{1}{4}\sin(2x) \right|_0^\pi\right)\\
&=& \pi \left(4 - \frac{\pi}{2} + 4\right)\\
&=& 8\pi - \frac{\pi^2}{2}.
\end{eqnarray*}

\item $y=x^3$, $y=\sqrt{x}$, $x=0$, $x=1$; about the $x$-axis

\vspace{.1in}
\sol
The cross-section by a plane $x=c$ is a washer with outer radius
$\sqrt{x}$ and inner radius $x^3$.  
The volume is 
\begin{eqnarray*}
V &=& \int_0^1 \pi(\sqrt{x})^2 - \pi (x^3)^2 \ dx\\
&=& \pi \int_0^1 x - x^6\ dx\\
&=& \pi \left(\left. \frac{x^2}{2} - \frac{x^7}{7} \right|_0^1\right)\\
&=& \frac{5\pi}{14}.
\end{eqnarray*}

\item $y=\frac{1}{x}$, $y=x$, $y=2x$ for $x>0$; about the $x$-axis\vspace{.1in}

\vspace{.1in}
\sol
Between $x = 0$ and $x = \sqrt{2}/2$, the top edge of the region
is the line $y = 2x$ and the bottom is the line $y = x$.  From $x = \sqrt{2}/2$ 
to $x = 1$, the top edge is $y = \frac{1}{x}$ and the bottom is the line $y = x$. 
\begin{center}
\includegraphics[width=2.8truein,height=2.8truein]{132PS6S1e.ps}
\end{center} 
The solid generated rotating around the $x$-axis is the union of two solids meeting
only along the plane $x = \sqrt{2}/2$.  The cross-sections of the first are
the washers with inner radius $x$ and outer radius $2x$ given by rotating the 
slice to the left of the blue dashed line in the figure.  The cross-sections
of the second are also washers with inner radius $x$ and outer radius
$1/x$ obtained by rotating the slices to the 
right of the blue dashed line.  Hence the 
total volume is
\begin{eqnarray*}
V &=& \int_0^{\sqrt{2}/2} \pi(2x)^2 - \pi x^2 \ dx + \int_{\sqrt{2}/2}^1 \pi (1/x)^2 - \pi x^2\ dx\\
&=& \pi \left(\int_0^{\sqrt{2}/2} 3x^2 \ dx +  \int_{\sqrt{2}/2}^1 (1/x)^2 - x^2\ dx\right)\\
&=& \pi \left(\left( \left.x^3\right|_0^{\sqrt{2}/2}\right) + \left(\left.\frac{-1}{x} - \frac{x^3}{3} \right|_{\sqrt{2}/2}^1\right)\right)\\
&=& \frac{4\pi}{3}(\sqrt{2}-1).
\end{eqnarray*} 

\item $x=4-y^2$, $x=y^2$; about the $y$-axis

\vspace{.1in}
\sol
The solid will be generated by rotating the region between 
these two parabolas with $x \ge 0$:

\begin{center}
\includegraphics[width=2.8truein,height=2.8truein]{132PS6S1f.ps}
\end{center}

Since we rotate around the $y$-axis, the slices of interest
are the cross-sections
by planes $y = d$.  These are washers (except for the cross-section
at $y = 0$ where the hole in the center ``closes up'').  The
inner radius is $y^2$ and the outer radius is $4 - y^2$.  The 
parabolas cross at $y = \pm \sqrt{2}$.  The volume is
\begin{eqnarray*}
V &=& \int_{-\sqrt{2}}^{\sqrt{2}} \pi (4 - y^2)^2 - \pi(y^2)^2\ dy\\
&=& \pi \int_{-\sqrt{2}}^{\sqrt{2}} 16 - 8y^2\ dy\\
&=& \pi \left(\left. 16y - \frac{8y^3}{3}\right|_{-\sqrt{2}}^{\sqrt{2}}\right)\\
&=& \frac{64\pi\sqrt{2}}{3}.
\end{eqnarray*} 

\item $x=\frac{1}{y}$, $y=4$, $x=2$; about the $y$-axis

\vspace{.1in}
\sol
The cross-section by a plane $y = d$ is a washer with inner radius
$\frac{1}{y}$ and outer radius 2.  The lower limit of integration
comes from where $x = 2$ crosses $x = 1/y$.  So the volume is
\begin{eqnarray*}
V &=& \int_{1/2}^{4} 4\pi - \pi(1/y)^2\ dy\\
&=& \pi \int_{1/2}^4 4 - \frac{1}{y^2}\ dy\\
&=& \pi \left(\left.4y + \frac{1}{y}\right|_{1/2}^4\right)\\
&=& \frac{49\pi}{4}.
\end{eqnarray*} 

\end{enumerate}

\item The base of a certain solid is the circle $x^2+y^2=9$ and each cross section perpendicular to the $x$-axis is an equilateral triangle with one side across the base.  Find the volume of the solid.

\vspace{.1in}
\sol  The area of an equilateral triangle of side $s$ is 
\[
A = \frac{1}{2}(\text{base})(\text{height}) = \frac{1}{2}(s)\left(\frac{s\sqrt{3}}{2}\right) = \frac{s^2\sqrt{3}}{4}.
\]
For this solid, in the cross-section at $x = c$ will have $s = 2\sqrt{9 - x^2}$, so the volume
is 
\begin{eqnarray*}
V &=& \int_{-3}^3 A(x)\ dx\\
&=& \int_{-3}^3 \frac{4(9-x^2)\sqrt{3}}{4}\ dx\\
&=& \sqrt{3} \int_{-3}^3 9 - x^2\ dx\\
&=& \sqrt{3} \left(\left. 9x - \frac{x^3}{3}\right|_{-3}^3\right)\\
&=& 36\sqrt{3}.
\end{eqnarray*}  

\item A nose cone for a space reentry vehicle is designed so that a cross section, taken $x$ feet from the the tip and perpendicular to the axis of symmetry, is a circle of radius $\frac{1}{4}x^2$ ft. Find the volume of the nose cone given that its length is $20$ ft.

\vspace{.1in}
Taking the tip as $x = 0$ and the base $x = 20$, we have
\begin{eqnarray*}
V &=& \int_0^{20} A(x)\ dx\\
&=& \int_0^{20} \pi\left(\frac{1}{4}x^2\right)^2\ dx\\
&=& \frac{\pi}{16}\int_0^{20} x^4\ dx\\
&=& \frac{\pi}{16} \left(\left. \frac{x^5}{5}\right|_0^{20}\right)\\
&=& 40000\pi.
\end{eqnarray*}
(units are cubic feet).  
  
\item Graph the curve and find its exact length.

\begin{enumerate}
\item $x=4t+3$, $y=3t-2$,  $0\le t\le 2$ 

\vspace{.1in}
\sol
The graph is a straight line from the point $(3,-2)$ to the point $(11,4)$.
The arclength is 
\begin{eqnarray*}
L &=& \int_0^2 \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\ dt\\
&=& \int_0^2 \sqrt{16 + 9}\ dt\\
&=& 10.
\end{eqnarray*}
 
\item $x=\frac{1}{3}t^3$, $y=\frac{1}{2}t^2$, $0\le t\le 1$

\vspace{.1in}
\sol
The curve here is the portion of a ``cuspidal cubic'' in the plane shown in black here:
\begin{center}
\includegraphics[width=2.8truein,height=2.8truein]{132PS6S4b.ps}
\end{center}
The arclength is 
\begin{eqnarray*}
L &=& \int_0^1 \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\ dt\\
&=& \int_0^1 \sqrt{(t^2)^2 + t^2}\ dt\\
&=& \int_0^1 t\sqrt{t^2 + 1}\ dt  \text{ let $u = t^2 + 1$, $du = 2t\, dt$}\\
&=& \int_{u=1}^{u=2} \frac{1}{2}\sqrt{u}\ du\\
&=& \left(\left.\frac{1}{3} u^{3/2}\right|_1^2\right)\\
&=& \frac{1}{3}(2\sqrt{2} - 1).
\end{eqnarray*}

\item $x=3\sin (2t)$ and $y=3 \cos(2t)$, $0\le t \le \frac{\pi}{2}$ 

\vspace{.1in}
\sol
The curve is the arc of a circle with radius 3 and center $(0,0)$ starting
at $(0,3)$ and proceeding clockwise to $(0,-3)$ (one half of the full circle).
The arclength is
\begin{eqnarray*}
L &=& \int_0^{\pi/2} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\ dt\\
&=& \int_0^{\pi/2} \sqrt{36\cos^2(2t) + 36\sin^2(2t)}\ dt\\
&=& \int_0^{\pi/2} 6\ dt\\
&=& 3\pi.
\end{eqnarray*}
where we used the trigonometric identity $\cos^2(\theta) + \sin^2(\theta) = 1$
to simplify before integrating.

\item $y=3x^2+2$ and $2\le x \le 4$ 

\vspace{.1in}
The curve is a segment of a parabola opening up.  The parameter is $x$ so
the arclength is
\begin{eqnarray*}
L &=& \int_2^4 \sqrt{1 + \left(\frac{dy}{dx}\right)^2}\ dx\\
&=& \int_2^4 \sqrt{1 + 36x^2}\ dx \qquad \text{ so let $u = 6x$, $du = 6\,dx$ }\\
&=& \frac{1}{6} \int_{u=12}^{u=24} \sqrt{1+u^2}\ du \text{ which is the form \#21 in the table }\\
&=& \frac{1}{6}\left(\left.\frac{u}{2}\sqrt{1 + u^2} + \frac{1}{2}\ln|u + \sqrt{1+u^2}|\right|_{12}^{24}\right)\\
&=& 2 \sqrt{577} - \sqrt{145} + \frac{1}{12}\ln|24 + \sqrt{577}| - \frac{1}{12}\ln|12+\sqrt{145}|.
\end{eqnarray*}
\end{enumerate}

\item Use Simpson's Rule with $n=10$ to estimate the arc length of the curve $y=\ln x$ for $1\le x\le 4$.

\vspace{.1in}
\sol
The length is given by the integral 
\[
L = \int_1^4 \sqrt{1 + \frac{1}{x^2}}\ dx.
\]
Let $f(x) = \sqrt{1 + \frac{1}{x^2}}$.
For Simpson with $n = 10$, $\Delta x = \frac{3}{10}$ and we obtain the approximation
\begin{eqnarray*}
L &\doteq & \frac{(3/10)}{3}\left( 
f(1) + 
4 f\left(\frac{13}{10}\right) + 
2 f\left(\frac{16}{10}\right) + 
4 f\left(\frac{19}{10}\right) + 
2 f\left(\frac{22}{10}\right) + 
4 f\left(\frac{25}{10}\right) +  \right.\\
& &\left. 2 f\left(\frac{28}{10}\right) + 
4 f\left(\frac{31}{10}\right) +
2 f\left(\frac{34}{10}\right) + 
4 f\left(\frac{37}{10}\right) +
f(4)\right)\\
&\doteq & 3.343018.
\end{eqnarray*}
(This approximate value is correct to 3 decimal places if we round -- the 
exact value is about $3.342799189$.)
\item Find the average value of the function on the given interval. 

\begin{enumerate}
\item $y=x\cos(3x)$, $[0,{\pi}]$ 

\vspace{.1in}
\sol
By the definition, the average value is 
\[
f_{ave} = \frac{1}{\pi} \int_0^\pi x\cos(3x)\ dx.
\]
To evaluate this, we use parts with $u = x$, $dv = \cos(3x)\ dx$.
\begin{eqnarray*}
f_{ave} &=& \frac{1}{\pi} \int_0^\pi x\cos(3x)\ dx\\
&=& \frac{1}{\pi} \left( \left.\frac{x}{3}\sin(3x)\right|_0^\pi - \frac{1}{3}\int_0^\pi\sin(3x)\ dx \right)\\
&=& \frac{1}{\pi} \left( 0  + \left. \frac{1}{9}\cos(3x)\right|_0^\pi\right)\\
&=& \frac{-2}{9\pi}.
\end{eqnarray*}

\item $y=x\sqrt{1+3x^2}$, $[1,3]$

\vspace{.1in}
\sol
By the definition, the average value is 
\[
f_{ave} = \frac{1}{2} \int_1^3 x\sqrt{1+3x^2}\ dx.
\]
To evaluate this, we use the $u$-substitution $u = 1 + 3x^2$, so $du = 6x\, dx$.
\begin{eqnarray*}
f_{ave} &=& \frac{1}{2} \int_1^3 x\sqrt{1 + 3x^2}\ dx\\
&=& \frac{1}{2} \left( \int_{u=4}^{u=28} \frac{1}{6} u^{1/2}\ du \right)\\
&=& \frac{1}{18} \left( \left. u^{3/2} \right|_4^{28}\right)\\
&=& \frac{1}{9}(28\sqrt{7} - 4).
\end{eqnarray*}

\end{enumerate}

\item 
Find the number $b$ such that the average value of $f(x)=x^3-1$ on the interval $[0,b]$ is equal to $6$.

\vspace{.1in}
\sol
We want $b$ such that 
\begin{eqnarray*}
6 &=& \frac{1}{b} \int_0^b x^3 - 1\, dx\\
  &=& \frac{1}{b} \left(\left. \frac{x^4}{4} - x\right|_0^b \right)\\
  &=& \frac{b^3}{4} - 1.
\end{eqnarray*}
This is true for $b = {\root 3\of {28}} \doteq 3.04$.

\item The amount of a certain drug present in a patient's body for the first $4$ days after the drug has been administered is $$C(t)=5e^{-0.2t}$$ units. Determine the average amount of the drug present in the patient's body for the first $4$ days after the drug has been administered.

\vspace{.1in}
\sol
The average amount is 
\begin{eqnarray*}
C_{ave} &=& \frac{1}{4}\int_0^4 5e^{(-0.2)t}\, dt\\
        &=& \frac{1}{4}\left( \left. -25 e^{(-0.2)t}\right|_0^4\right)\\
        &=& \frac{25}{4}\left(1 - e^{-0.8}\right)\\
        &\doteq& 3.44\, \text{units}.
\end{eqnarray*}

\end{enumerate}



\end{document}
