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• Notes

Use the given transformation to evaluate the integral.
$\iint_{R}(x-3 y) d A$ . where $R$ is the triangular region with
vertices $(0,0),(2,1),$ and $(1,2) ; \quad x=2 u+v, y=u+2 v$

$x=2 u+v \quad , \quad[y=u+2 v) * 2$

$2 y=2 u+4 v$

$x=2 u+v$
$2 y=2 u+4 v$
$x-2 y=-3 v$

$v=\frac{1}{3}[2 y-x]$

$u=y-2 y$
$u=y-2 \frac{1}{3}[2 y-x]$
$u=-\frac{1}{3} y+\frac{2}{3} x$

$(1)(0,0) \Rightarrow(0,0)$
$v=\frac{1}{3}[0-0]=0$
$u=0$
$(2)(1,2) \Rightarrow(0,1)$
$v=\frac{1}{3}[2 \times 2-1]=1$
$u=\frac{-1}{3}\times 2+\frac{2}{3} \times 1=0$

$(3)(2,1)\Rightarrow(1,0)$

$\iint_{R\rightarrow(x, y)} f(x, y) d A=\iint_{S\rightarrow (u, v)} f\left[x(u, v), y(u, v)\right] | \frac{\partial(x, y)}{\partial(u, v)} | d v d u$

$\left|\begin{array}{ll}{\frac{\partial x}{\partial u}} & {\frac{\partial x}{\partial v}} \\ {\frac{\partial y}{\partial u}} & {\frac{\partial y}{\partial v}}\end{array}\right|=\left|\begin{array}{ll}{2} & {1} \\ {1} & {2}\end{array}\right|=2 * 2-1 * 1=4-1=3$

$\int_{R} \int(x-3 y) d x d y =\int_{0}^{1} \int_{0}^{1-u}(2 u+v)-3(u+2 v) * 3 d v d u$

$\int_{0}^{1} 3\left[-u v-5 \frac{(1-u)^{2}}{2}\right] d u=-3$