\section{Change of Variables}
\begin{enumerate}[1.]

\item {\it Evaluate the double integrals. }

\medskip

{\it (a) $\displaystyle \iint_D y\, dA$, where $D=\{(x,y):\ y\geq 0,\ 1\leq x^2+y^2\leq 9\}$.}



\begin{align*}
&\int_0^{\pi} \int_1^3  r\sin\theta\, r dr\, d\theta
=\int_0^{\pi} \int_1^3  r^2\sin\theta\,  dr\, d\theta
\\
&=\int_0^{\pi}   \frac{1}{3}r^3\sin\theta\, \Big|_{r=1}^3  d\theta
=\int_0^{\pi}   (9-\frac{1}{3})\sin\theta\,   d\theta
\\
&=\int_0^{\pi}   \frac{26}{3} (-\cos \theta)\Big|_0^\pi
= \frac{26}{3}\cdot 2 =\frac{52}{3}
\end{align*}





(b) $\displaystyle $ 



\begin{align*}
\iint_{\mathbb{R}^2} \frac{dA}{(4+x^2+y^2)^5}
&= \int_0^{2\pi} \int_0^\infty  \frac{r\, dr\, d\theta}{(4+r^2)^5}
\\
&= \int_0^{2\pi} \int_0^\infty   \frac{1}{2}(4+r^2)^{-5}  (2r)  \, dr\, d\theta
\\
&\ \ =\int_0^{2\pi}   \frac{-1}{8}(4+r^2)^{-4} \Big|_0^\infty  \, d\theta
=\int_0^{2\pi}   \frac{1}{8}4^{-4}  \, d\theta =  \frac{2\pi}{8}4^{-4} =     \frac{\pi}{4^5}.
\\
\end{align*}

\bigskip




\item {\it Evaluate the integral $\iint_D (\frac{x}{y}-1)dA$, if $D$ is the region in the first
quadrant bounded by the hyperbolas $x^2-y^2=1$, $x^2-y^2=4$, and the
lines $x=2y$, $x=3y$.  }

\bigskip

Let $u=x^2-y^2$, $v=x/y$, then the region of integration is given by
$1\leq u\leq 4$, $2\leq v\leq 3$.  To find the Jacobian, we first compute
$$\frac{\partial(u,v)}{\partial(x,y)}=
\left|\begin{matrix} 2x&-2y\\ 1/y&-x/y^2\\ \end{matrix}\right|
=-2\frac{x^2}{y^2}+2=2-2v^2,$$
so that
$$\frac{\partial(x,y)}{\partial(u,v)}=\frac{1}{2-2v^2}.$$
Hence
$$\int\int_D (\frac{x}{y}-1)dA
=\int_1^4\int_2^3(v-1)\Big|\frac{1}{2-2v^2}\Big|dvdu$$
$$=\int_1^4\int_2^3\frac{v-1}{2(v^2-1)}dvdu
=\int_1^4\int_2^3\frac{1}{2(v+1)}dvdu$$
$$=\int_1^4\frac{\ln(v+1)}{2}\Big|_2^3 du
=\int_1^4\frac{\ln 4-\ln 3}{2} du$$
$$=\frac{3}{2}(\ln 4-\ln 3).$$
We used that $|2-2v^2|=2v^2-2$ for $v\in[2,3]$.

\bigskip

\item {\it Use a change of variables to evaluate the integral ${\iint_D (x-y)dxdy}$,
where $D$ is the region in the first quadrant bounded by the hyperbolas $x^2-y^2=1$, 
$x^2-y^2=4$, and the lines $x+y=3$, $x+y=6$.}

\medskip

Let $u=x^2-y^2$, $v=x+y$, then the region of integration is $1\leq u\leq 4$, $3\leq v\leq 6$.
(We can solve this for $x,y$: $x-y=u/v$, $x=\frac{1}{2}(v+\frac{u}{v})$, 
$y=\frac{1}{2}(v-\frac{u}{v})$.  This is, however, optional.)  We compute the Jacobian:
$$
\frac{\partial(u,v)}{\partial(x,y)}=\left|\begin{matrix}
2x & -2y \\
1& 1\\
           \end{matrix}\right|=2x+2y=2v,\ 
\frac{\partial(x,y)}{\partial(u,v)}=\frac{1}{2v}.
$$
(Alternatively, one can compute $\frac{\partial(x,y)}{\partial(u,v)}$ directly.  This leads to
a slightly longer computation, but is equally correct.)

\smallskip

Thus our integral is
\begin{align*}
&\int_1^4\int_3^6 \frac{u}{v}\cdot\frac{1}{2v}dvdu
=\int_1^4\int_3^6 \frac{u}{2v^2}dvdu
=\int_1^4\frac{-u}{2v}\Big|_{v=3}^6du=\int_1^4\frac{u}{2}\Big(\frac{1}{3}-\frac{1}{6}\Big)du\\
&=\int_1^4\frac{u}{12}du=\frac{u^2}{24}\Big |_1^4=\frac{15}{24}=\frac{5}{8}.
\end{align*}

\bigskip

\item {\it Evaluate the area of the region in the first
quadrant bounded by the hyperbolas $xy=1$, $xy=5$, and the
parabolas $y=x^2$, $y=4x^2$.  (Hint: introduce new coordinates in which the region
of integration is a rectangle.)}

\bigskip
Let $u=xy$ and $v=y/x^2$, so that
\begin{equation*}
R=\{(u,v): 1\le u\le 5,\;1\le v\le 4\}.
\end{equation*}
We also need the Jacobian $\frac{\partial(x,y)}{\partial(u,v)}$. It is easiest to compute the Jacobian of the inverse transformation $\frac{\partial(u,v)}{\partial(x,y)}$ and then take its reciprocal. We find that
\begin{align*}
\frac{\partial(u,v)}{\partial(x,y)} &= \left|
\begin{array}{cc}
y & x\\
-2y/x^3 & 1/x^2
\end{array} \right|\\
&= \frac{y}{x^2} + 2\frac{y}{x^2}= 3\frac{y}{x^2}= 3v,
\end{align*}
so that
\begin{equation*}
\frac{\partial(x,y)}{\partial(u,v)} = \left(\frac{\partial(u,v)}{\partial(x,y)}\right)^{-1} = \frac{1}{3v}.
\end{equation*}
Then the area of our region is
\begin{align*}
\operatorname{Area}(R) &= \int_R dA=\int_1^5 \int_1^4 \left|\frac{\partial(x,y)}{\partial(u,v)}\right|\,dv\,du\\
&=\int_1^5 \int_1^4 \frac{1}{3v}\;dv\,du\\
&=\int_1^5 \frac{1}{3}\ln v\Bigr|_1^4\;du\\
&=\int_1^5 \frac{1}{3}\ln 4\;du=\frac{4}{3}\ln 4.
\end{align*}
We used $\ln v$ and not $\ln |v|$ for the antiderivative of $1/v$ because $v>0$ on $R$.

\bigskip

\item {\it
Determine whether the integral
$\displaystyle \iint_{\rr^2} \frac{|y| }{(1+ x^2+y^2)^3 }\, dA$ is convergent.}


\medskip

We write
\begin{align*}
\iint_{\rr^2} \frac{|y| }{(1+ x^2+y^2)^3 }\, dA &=  \iint_{x^2+y^2\leq 1} \frac{|y| }{(1+ x^2+y^2)^3 }\, dA
\\
&+ \iint_{x^2+y^2\geq 1} \frac{|y| }{(1+ x^2+y^2)^3 }\, dA,
\end{align*}
and estimate each part separately:
$$ \iint_{x^2+y^2\leq 1} \frac{|y| }{(1+ x^2+y^2)^3 }\, dA \leq  \iint_{x^2+y^2\leq 1} {|y| }\, dA
\leq  \iint_{x^2+y^2\leq 1} 1\, dA =\pi,
 $$

\begin{align*}
 \iint_{x^2+y^2\geq 1} & \frac{|y| }{(1+ x^2+y^2)^3 }\, dA  
 \leq  \iint_{x^2+y^2\geq 1} \frac{\sqrt{x^2+y^2} }{(1+ x^2+y^2)^3 }\, dA \\
 &= \int_1^\infty \int_0^{2\pi} \frac{r}{(1+r^2)^3} r d\theta dr 
 = 2\pi \int_1^\infty  \frac{r}{(1+r^2)^3} r  dr  
 \\
 & = 2\pi \int_1^\infty  \frac{r^2}{(1+r^2)^3}dr \leq  2\pi \int_1^\infty  \frac{r^2}{(r^2)^3} dr 
  \\
 &= 2\pi \int_1^\infty r^{-4} dr =- \frac{2\pi}{3} r^{-3}\Big|_{r=1}^\infty 
 = \frac{2\pi}{3}.
 \end{align*}
 Hence the integral is convergent. 
\end{enumerate}

