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\centerline
{
\bf Solutions to the ``QUIZ'' for Lecture 16
}
\bigskip

{\bf 1.}  Compute the Jacobian of the transfomation
$$
\Phi(r,s)=(rs,r+s)
$$
\bigskip
{\bf Sol.:} Here $x=rs,y=r+s$ and
$$
J=(x_r)(y_s)-(x_s)(y_r)=(s)(1)-(r)(1)=s-r \quad .
$$

{\bf Ans.} $s-r$ (type: function of $r$ and $s$).

{\bf Comment}: Almost everyone got it right.
\bigskip
{\bf 2.} Let $\D=\Phi(\R)$ where $\Phi(u,v)=(u+v,v^2)$ and
$R=[0,6] \times [1,2]$. Calculate
$$
\int\int_\D y \, dA \quad .
$$
(Note: it is not necessary to compute $D$).
\bigskip
{\bf Sol.}  Here the transformation is $x=u+v,y=v^2$.
The Jacobian is $J=(x_u)(y_v)-(x_v)(y_u)=(1)(2v)-(1)(0)=2v$.

By the change of variable formula we have
$$
\int\int_\D y \, dA =
\int\int_\R y J \, dA =
\int\int_\R (v^2) (2v) \, dA =
\int\int_\R 2v^3 \, dA 
$$
$R$ is the rectangle $[0,6] \times [1,2]$, which means:
$$
\{ (u,v) \, \vert \, 0 \leq u \leq 6 \, , \, 1 \leq v \leq 2 \}
$$
That is both type I and type II. Using the type-I formulation we have
$$
\int_{0}^{6} \int_1^{2} 2v^3  \, dv \, du =
\left ( \int_{0}^{6} \, du \right ) 
\left (  \int_1^{2} 2v^3  \, dv \right ) =
$$
$$
\left ( u \Bigl \vert_{0}^{6} \right )
\left ( {{v^4} \over {2}} \Bigl \vert_{1}^{2} \right )
=
(6-0) \cdot ( {{2^4-1^4} \over {2}} )
=
6 \cdot ( {{15} \over {2}} )= 45 \quad .
$$

{\bf Ans.} $45$ (type number).

{\bf Comment}: About \%60 got it perfectly. Almost everyone
set-it up correctly, but a few people got mixed up with
the Jacobian, and for some reason got $2v-1$. Other people, as usual,
forgot their multiplication table facts. Some people got confused
and took $y=y+v$ rather than $y=v^2$. Don't confuse the $x$ and the $y$!


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