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Thm 5.3:  Suppose $T: \R^2 \rightarrow \R^2 \in C^1$ 

$T(u, v) = (T_1(u, v), T_2(u, v)) =  (x(u, v), y(u, v))$


If $D^* \subset \R^2$ and if $f: T(D^*) \rightarrow \R$ is integrable, then

$\int \int_{T(D^*)} f(x, y) dx dy =$
\f

\rightline{$ \int \int_{D^*} f( (x(u, v), y(u, v))) abs(det(DT(u, v))) du 
dv$}


Ex:  Evaluate $\int \int_{T(D^*)} xy dx dy $ 

where $T(D^*)$ = the parallelogram where the vectors (4, 1) and (2, 3) form two of its sides.


\vskip 60pt

Find $D^*$ and $T: \R^2 \rightarrow \R^2$ such that  $T(D^*)$ is the parallelogram described above.


Note $T$ is a linear transformation.

$T: \R^2 \rightarrow \R^2$, $T(u, v) = \left(\matrix{a & b \cr c & d}\right)$
$\left(\matrix{u\cr v}\right)$

\eject

$ T(1, 0) = \left(\matrix{a & b \cr c & d}\right)$
$\left(\matrix{1\cr 0}\right)
= \left(\matrix{4\cr 1}\right)
$ 

$T(0, 1) = \left(\matrix{a & b \cr c & d}\right) \left(\matrix{0\cr 1}\right)
= \left(\matrix{2\cr 3}\right)
$ 


Hence 
$T: \R^2 \rightarrow \R^2$, 

$T(u, v) = \left(\matrix{4 & 2 \cr 1 & 3}\right)$
$ \left(\matrix{u\cr v}\right) = \left(\matrix{4u + 2v\cr u + 3v}\right)$
$ = \left(\matrix{x(u, v) \cr y(u, v) }\right)$

Let $x = 4u + 2v$ and $y = u + 3v$

$DT = \left(\matrix{4 & 2 \cr 1 & 3}\right)$
and $abs(det \left(\matrix{4 & 2 \cr 1 & 3}\right) ) = 12 - 2 = 10$


$\int \int_{T(D^*)} (xy) dx dy $ 

\rightline{= $ \int \int_{D^*} f( (x(u, v), y(u, v))) abs(det(DT(u, v))) 
du dv$}

$ =\int \int_{D^*} (4u + 2v)(u + 3v)(10) du dv$

= $10 \int_0^1 \int_0^1 (4u^2 + 14uv + 6v^2) du dv$

= $10 \int_0^1 ({4 \over 3}u^3 + 7u^2v + 6uv^2|_0^1 dv$

= $10 \int_0^1 ({4 \over 3} + 7v + 6v^2) dv$

= $10 ({4 \over 3}v + {7 \over 2}v^2 + 2v^3)|_0^1 $

= $10 ({4 \over 3} + {7 \over 2} + 2) $


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