\def\whichexam{Exam II}
\def\examdate{March 16, 2004}

\long\def\newp #1.{
\par\noindent\vtop\bgroup\noindent{\bf #1.}
}
\long\def\problembreak{\par\egroup\vfill\eject\vtop\bgroup}
\def\endp{\par\noindent\vrule width\hsize depth 0pt height .1pt\egroup\par\bigskip}

\let\ds=\displaystyle
\font\sm=cmr7
\magnification1200
\centerline{\whichexam}
\centerline{\examdate}
\def\Int{\displaystyle\int}
\def\Lim{\displaystyle\lim}
\def\frac#1#2{{{#1}\over{#2}}}
\def\smfrac#1#2{{{\scriptstyle#1}\over{\scriptstyle#2}}}
\bigskip
\def\frac#1#2{{{\textstyle#1}\over{\textstyle#2}}}
\def\mbox#1{\hbox{\rm#1}}

%:12
\newp 12.
Evaluate:
$$
\int \frac{x^2}{\sqrt{1-4 x^2}}\ dx.
$$
Try a trig. substitution with $\sin\theta = 2x$.
Then $dx = \frac{1}{2} \cos\theta\> d \theta$, 
$x = \frac{\sin\theta}{2}$ and $\sqrt{1-4 x^2} = \cos\theta$, so
$$\eqalign{
\int \frac{x^2}{\sqrt{1-4 x^2}}\ dx = &\int \frac{\sin^2\theta}{4\cos\theta} \frac{1}{2}\cos\theta\> d \theta =
\frac{1}{8}\int \sin^2\theta\> d \theta = \frac{1}{8}\int \frac{1-\cos2\theta}{2}\> d \theta
\cr = &
\frac{1}{16}\biggl( \theta - \frac{\sin(2\theta)}{2}\biggr) + C =
\frac{1}{16}\big(\theta - (\sin\theta)(\cos\theta)\bigr) +C
\cr}$$
Now $\theta = \arcsin(2x)$, $\sin \theta = 2x$ and $\cos\theta = \sqrt{1-4x^2}$, so
$$\int \frac{x^2}{\sqrt{1-4 x^2}}\ dx  = \frac{1}{16}\biggl(
\arcsin(2x) - 2x\sqrt{1-4x^2}\biggr) + C$$
\endp

%:13
\newp 13.
Consider the integral
$$
\int_0^2 (2 x + 3)\ dx.
$$
\item{(a)} (5 pts.) Evaluate this integral exactly.
\item{(b)} (8 pts.) Using the Trapezoidal Rule with $n=4$ find an approximation
  to the integral.
\item{(c)} (2 pts.) Explain your answer in part (b).{\bf Hint:}Consider the error.
\bigskip
\item{(a)} $\int_0^2 (2x+3)\ dx = x^2+3x\big\vert^2_0 = (2^2+3\cdot 2) - ( 0^2 + 3\cdot 0)=10$.
\item{(b)} $$\eqalign{
T_4  = & \frac{h}{2} \left[ f(0) + 2 f(1/2) + 2 f(1) + 2 f(3/2) 
    + f(2) \right] \cr
   = & \frac{1/2}{2} \left[ 3 + 2 (4) + 2 (5) + 2 (6) + 7 \right] 
   =  \frac{40}{4} = 10.
}$$
\item{(c)}
The error bound for the Trapezoidal Rule involves the {\sl second\/}
derivative of the integrand, $f(x)$.  Notice that for this problem
$f^{\prime\prime}(x) = 0$ so that we may take $K=0$ and
$$
\left| E_T \right| \leq \frac{K (b-a)^3}{12 n^2} = 0.
$$
The error is guaranteed to be zero.

\endp
%: 14
\newp 14.
Find the area of the surface obtained by rotating the curve about
the $y${--}axis:
$$
x = \sqrt{a^2 - y^2} \qquad 0 \leq y \leq a.
$$
Specifically,
\par\item{(a)}(9 pts.) Write an integral which gives the surface area using the general formula
for surface area.
\par\item{(b)} (6 pts.) Evaluate your integral.

{\bf Solution 1:} For revolution about the $y$-axis
$$A \; = \; \int 2 \pi x ds \; = \; \int_0^a 2 \pi f(y) \sqrt{1 +
\bigl(f^\prime(y)\bigr)^2} \; dy . $$
We have
$$ f'(y) = \frac{1}{2} (a^2 - y^2)^{-1/2} (-2y) = \frac{-y}{\sqrt{a^2 -y^2}}
.$$ Thus
$$\eqalign{
A =&  \int_0^a 2 \pi \sqrt{a^2 -y^2} \sqrt{ 1+ \left(
\frac{-y}{\sqrt{a^2 -y^2}} \right)^2} \; dy \cr
=& \int_0^a 2 \pi \sqrt{a^2 -y^2} \sqrt{ 1+
\frac{y^2}{a^2 -y^2} } \; dy \; = \; \int_0^a 2 \pi  \sqrt{ a^2 - y^ 2 +
y^2} \; dy \cr
=& \int_0^a 2 \pi  \sqrt{ a^2} \; dy \; = \;  \int_0^a 2 \pi a dy
\; = \; \Bigl. 2\pi a y \Bigr|_0^a \; = \; 2\pi a (a - 0) \cr
=& 2 \pi a^2.
}$$

{\bf Solution 2:}
$$A \; = \; \int 2 \pi x ds \; = \; \int_0^a 2 \pi x \sqrt{1 +
\left(\frac{dx}{dy}\right)^2} \; dx.$$
We have $x = \sqrt{a^2 - y^2}$ for $0 \leq y \leq a$ implies $y =
\sqrt{a^2 - x^2}$ for $0 \leq x \leq a$, and
$$\frac{dx}{dy} = \frac{1}{2} (a^2 - x^2)^{-1/2} (-2x) =
\frac{-x}{\sqrt{a^2 -x^2}} .$$ Thus
$$\eqalign{
A =&  \int_0^a 2 \pi x \sqrt{ 1+ \left(
\frac{-x}{\sqrt{a^2 -x^2}} \right)^2} \; dx \; = \; \int_0^a 2 \pi x \sqrt{
1+  \frac{x^2}{a^2 -x^2} } \; dx \cr
=&  \int_0^a 2 \pi x \sqrt{ \frac{a^2 - x^ 2 +  x^2}{a^2 - x^2}} \; dx
\; = \; \int_0^a 2 \pi x \frac{a}{\sqrt{a^2 - x^2}} \; dx \; = \;
-\int_{a^2}^0 \pi a  \frac{1}{\sqrt{u}} \; du
}$$
for $u = a^2 - x^2$ and thus $du = -2xdx$.  Thus
$$\eqalign{
A =& \Bigl. - \pi a 2\sqrt{u} \Bigr|_{a^2}^0 \; =\; - 2\pi a (0 - a) \cr
=& 2 \pi a^2.
}$$
\endp

\end

