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\centerline
{
\bf Solutions to the ``QUIZ'' for Lecture 13
}
\bigskip

{\bf 1.} 
Change the order of integration in
$$
\int_{1}^{4} \int_{0}^{\ln y} f(x,y) \, dx \, dy \quad .
$$

{\bf Sol.}: The {\bf region} of integration is
$$
D=\{ (x,y) \vert 1 \leq y \leq 4, 0 \leq x \leq \ln y \}
$$
This is a type II (Horizontally Simple) region.
The {\bf main road} is along the $y$-axis from $(0,1)$ to $(0,4)$,
and the ``side-streets'' extend from the $y$-axis (where $x=0$)
to the curve $x=\ln y$. Drawing it, we get that the
projection on the $x$-axis of that region is between
$x=0$ and $x= \ln 4$, and making this the main road,
the vertical side-streets extend from the curve $x=\ln y$,
that we should rewrite, from the new perspective as $y=e^x$, to
the horizontal line $y=4$. So a type I description is
$$
D=\{ (x,y) \vert 0 \leq x \leq \ln 4, e^x \leq y \leq 4 \} \quad 
$$

{\bf Finally} the integral is
$$
\int_{0}^{\ln 4} \int_{e^x}^{4} f(x,y) \, dx \, dy \quad .
$$
This is the {\bf ans.} (type: double integral of a general function
on a specific domain).

{\bf 2.} 
Evaluate
$$
\int_{0}^{2} \int_{y/2}^{1} {{1} \over {(x^2+1)^2} }  \, dx \, dy \quad ,
$$
by inverting the order of integration and evaluating the new iterated
integral.

{\bf Sol.} The region implied by the given iterated integral is
$$
D=\{ (x,y) \vert 0 \leq y \leq 2, y/2 \leq x \leq 1 \} \quad 
$$
This is a type II description.
If you draw it, the {\bf main road} is along the $y$-axis
from $y=0$ to $y=2$ and the horizontal side streets are
from the line $x=y/2$ to the vertical line $x=1$.
This turns out to be a triangle with vertics $(0,0),(1,0),(1,2)$.

Let's find a type I description of $D$. The projection on the
$x$-axis is from $x=0$ to $x=1$ and 
the horizontal side streets starts at $y=0$ and end at the line
$x=y/2$ that we now must write at $y=2x$.
So the type I description of $D$ is:
$$
D=\{ (x,y) \vert 0 \leq x \leq 1, 0 \leq y \leq 2x \} \quad  .
$$
Now we can {\bf interchange} the order of integration and get
$$
\int_{0}^{1} \int_{0}^{2x} {{1} \over {(x^2+1)^2} }  \, dy \, dx \quad .
$$
We first do the {\bf inner} integral
$$
\int_{0}^{2x} {{1} \over {(x^2+1)^2} }  \, dy
={{1} \over {(x^2+1)^2} } \left ( \int_{0}^{2x} \, dy \right ) \quad
$$
$$
={{1} \over {(x^2+1)^2} } \left ( y \Bigl \vert_{0}^{2x} \right ) \quad
={{1} \over {(x^2+1)^2} } \left ( 2x-0 \right ) \quad
={{2x} \over {(x^2+1)^2} }  \quad .
$$
Now we are ready to do the {\bf outer} integral
$$
\int_{0}^{1} {{2x} \over {(x^2+1)^2} }  \, dx \quad .
$$
This calls for a {\bf u-substitution}.
$u=x^2+1$, $2x \, dx=du$, and when $x=0$, $u=1$, and when $x=1$, $u=2$.
$$
\int_{1}^{2} {{1} \over {u^2} }  \, du=
\int_{1}^{2} u^{-2}  \, du= {{u^{-1}} \over {-1}} \Bigl \vert_{1}^{2}
$$
$$
= {{-1} \over {u}} \Bigl \vert_{1}^{2}=
 {{-1} \over {2}} - {{-1} \over {1}} = {{1} \over {2}} \quad .
$$
{\bf Ans.}: ${{1} \over {2}}$ (type: number).


{\bf Commnets}: I did not allow enough time for these problems, but please study the solutions carefully.

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