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{
\bf Solutions to the ``QUIZ'' for Oct. 15, 2009
}
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{\bf 1.} Calculate the iterated integral
$$
\int_{1}^{2} \int_{-1}^{1} (x+ y^2) \, dx \, dy
\quad .
$$

{\bf Sol.} We first do the {\bf inner integral}
$$
 \int_{-1}^{1} (x+ y^2) \, dx
= {{x^2} \over {2}} + y^2 x \Bigl \vert_{-1}^{1}=
({{1^2} \over {2}} + y^2 \cdot 1)-
({{(-1)^2} \over {2}} + y^2 \cdot (-1))=
2y^2
$$

Now we do the {\bf outer integral}
$$
\int_{1}^{2} \left [ \int_{-1}^{1} (x+ y^2) \, dx  \right ] \, dy
=
\int_{1}^{2} 2y^2 \, dy =
{{2y^3} \over {3}} \Bigl \vert_{1}^{2}=
{{2 \cdot 2^3} \over {3}} -{{2 \cdot 1^3} \over {3}} =
{{16} \over {3}} -{{2} \over {3}} =
{{16} \over {3}} -{{2} \over {3}} =
{{14} \over {3}} \quad.
$$

{\bf Ans.}: ${{14} \over {3}}$.

{\bf Comment}: About \%90 got it perfectly. As usual, a few people
messed up the (easy!) arithmetic, and a couple of people tried
``shortcuts'', but messed up in using them.

{\bf 2.} 
Calculate the double integral
$$
\int\int_{R} {{x^2y} \over {x^3+1}} \, dA \quad, \quad
$$
$$
R=\{(x,y)\, \Bigl \vert \, 
0 \leq x \leq 1 \, , \, -1 \leq y \leq 1 \, \} \quad.
$$

{\bf Sol.}: Making it into an iterated integral we have
$$
\int_{0}^{1} \int_{-1}^{1} {{x^2y} \over {x^3+1}} \, dy \, dx \quad
$$
{\bf This} integrand has the property that it is {\bf separable}
i.e. a product of a function of $x$-alone (namely $x^2/(x^3+1)$)
and a function of $y$-alone (namely $y$), so it is
{\bf legitimate} to use the shortcut:
$$
\left ( \int_{0}^{1} {{x^2} \over {x^3+1}} \, dx \right )
\left ( \int_{-1}^{1} y \,dy  \right ) \quad .
$$
Since the second integral is obviously $0$, we don't even have to bother
to compute the first integral, since {\bf everything} times zero, is
$0$ ($0$ kills everything, and if I know that you are going to die,
why bother getting to know you).
The answer is $0$.

{\bf Comment}: About \%80 got it perfectly. A few people made it
harder than it should be, and messed up.


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