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{
\bf Solutions to the ``QUIZ'' of Oct. 26, 2009
}

\bigskip


{\bf 1.} 
Use polar coordinates to compute the double
integral
$$
\int \, \int_D xy \, dA \quad ,
$$
where 
$$
D= \{ (x,y) \, \vert \, x^2+y^2 \leq 1 \, , \,
x \geq 0 \, , \, y \geq 0 \, \} \quad .
$$

{\bf Sol.}: The {\bf region} is the quarter-circle, center origin,
radius $1$, in the first quadrant. Its {\bf polar coordinates}
description is:
$$
\{(r, \theta) \, \, \vert \, \, 0 \leq \theta \leq \pi/2 \quad , \quad 0 \leq r \leq 1 \}
$$
doing the {\bf translation} we get
$$
\int_{0}^{\pi/2} \int_{0}^{1} 
(r \cos \theta) (r \sin \theta ) r \, dr \, d \theta
$$
This is the {\bf most important}  part. From here, Maple can do it.
Doing it by hand, we have
$$
\int_{0}^{\pi/2} \int_{0}^{1} 
r^3 \cos \theta  \sin \theta \, dr \, d \theta=
\left ( \int_{0}^{\pi/2} \cos \theta  \sin \theta \, d \theta  \right )
\left ( \int_{0}^{1} r^3 \, dr  \right )
$$
$$
=
\left ( \int_{0}^{\pi/2}   {{\sin 2\theta} \over {2}} \, d \theta  \right )
\left ( \int_{0}^{1} r^3 \, dr  \right )
=
\left ( {{-\cos 2 \theta} \over {4}} \Bigl \vert_{0}^{\pi/2} \right )
\left (  {{r^4} \over {4}} \Bigl \vert_{0}^{1} \right )
$$
$$
={{-\cos \pi -- \cos 0} \over {4}} \cdot {{1} \over {4}}
={{2} \over {4}} \cdot {{1} \over {4}}
={{1} \over {8}} \quad .
$$

{\bf Ans.}: ${{1} \over {8}}$ (type=number).

{\bf Comment}: About \%60 of the students were able to  set it up correctly. 
Only about \%20 finished it up completely.
Many people just left it unevaluated (as I told you to do if you were in a rush).

{\bf 2.} Evaluate the iterated integral by converting it to polar coordinates
$$
\int_0^1 \, \int_0^{\sqrt{1-y^2}} e^{x^2+y^2} \, dx \, dy  \quad .
$$

{\bf Sol.} This is an iterated integral over a {\bf type-II} region (since the $dy$ integration is outside).
Looking at the limits of integration, the region in question is
$$
\{ (x,y) \, \vert \, 0 \leq y \leq 1 \, , \, 0 \leq x \leq \sqrt{1-y^2} \} \quad .
$$
This is the quarter-circle, center origin, radius $1$, that is located at the first-quadrant.
A polar-coordinates description is
$$
\{ (r, \theta ) \, \vert \, 0 \leq \theta  \leq \pi/2 \, , \,   0 \leq r \leq 1 \} \quad .
$$
Since $x^2+y^2=r^2$ and $dydx=dA=r dr d \theta$, the integral equals:
$$
\int_0^{\pi/2} \, \int_0^{1} e^{r^2} \, r dr \, d\theta  \quad .
$$
This is the set-up. In real life you would go to Maple. But here we have to do it by hand.

The inner integral is:
$$
\int_0^{1} r e^{r^2} dr \quad .
$$
Doing the substitution $u=r^2$ we get $du=2r dr$, and when $r=0, u=0$ and when $r=1,u=1$.
So this equalus
$$
\int_0^{1}  e^{u}/2= (1/2)e^u \Bigl \vert_{0}^{1}= (1/2)(e^1-e^0)= {{e-1} \over {2}} \quad .
$$
Now we do the {\bf outer integral}.
$$
\int_0^{\pi/2} \, 
{{e-1} \over {2}}
\, d\theta=  {{e-1} \over {2}} \theta \Bigl \vert_{0}^{\pi/2}= {{e-1} \over {2}} \cdot {{\pi} \over {2}}=
{{(e-1)\pi} \over {4}}  \quad .
$$

{\bf Ans.} ${{(e-1)\pi} \over {4}}$ (type number).


{\bf Comment}: About \%50 of the students were able to correctly set it up. About \%20 finished it up
completely. Many people just left it unevaluated (as I told you to do if you were in a rush).





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