Solución

a. $T (u, v) = (2u + v, 3v)$

b. El área de $R$ es
$\displaystyle A(R) = \int_0^3\int_{y/3}^{(6-y)/3} dxdy = \int_0^1\int_0^{1-u}\bigg|\frac{\partial(x,y)}{\partial(u,v)}\bigg|dvdy = \int_0^1\int_0^{1-u} 6dvdu = 3$