Answer
$\iint_R\sin(x^2+y^2)dA=\frac{\pi(\cos 1-\cos 9)}{4}$
Work Step by Step
In the first quadrant, the circles with center the origin and radii 1 and 3 have the equations $r=1$ and $r=3$ where $0\leq \theta\leq \pi/2$
Then, the region $R$ can be written $R=\{(r,\theta)|1\leq r\leq 3, 0\leq \theta\leq \pi/2\}$.
Evaluate the given integral:
$\iint_R \sin(x^2+y^2)dA=\int_1^3\int_0^{\pi/2}\sin (r^2)\cdot rd\theta dr$
$=\int_1^3\int_0^{\pi/2}r\sin(r^2)d\theta dr$
$=\int_1^3r\sin(r^2)\theta]_0^{\pi/2}dr$
$=\int_1^3\frac{\pi r\sin(r^2)}{2}dr$
$=-\frac{\pi\cos(r^2)}{4}]_1^3$
$=\frac{\pi(\cos 1-\cos 9)}{4}$
