Answer
$\cos (1)-\cos (2)$
Work Step by Step
The domain $D$ for given region can be expressed as: $1 \leq y \leq 2$ and $y \leq x \leq 2y$
The iterated integral can be calculated as:
$\iint_{D} f(x,y) d A=\int_1^2 \int_{y}^{2y} \dfrac{\sin y}{y} \ dx \ dy\\=\int_1^2 (\dfrac{x \sin y}{y})_{y}^{2y} dy \\=\int_1^2 [2 \sin y -\sin y ) \ dy \\=\int_1^2 \sin y \ dy \\=[-\cos y]_1^2 \\=\cos (1)-\cos (2)$
