Answer
$\frac{\pi}{4}(\cos{1}-cos{9})$
Work Step by Step
We begin with the integral in rectangular coordinates:
$$\iint_{D}\sin\,(x^2+y^2)\,dA $$
Using the substitutions for polar coordinates and adding the bounds, we get:
$$\int_{0}^{\pi/2}\bigg(\int_{1}^{3}\sin (r^2)\,r\, dr\bigg)d\theta$$
Using the substitution $u=r^2$, we can rewrite the integral as:
$$\int_{0}^{\pi/2}\bigg(\int_{1}^{9}\frac{1}{2}\,\sin(u)\,du\bigg)d\theta$$
Solving, we get:
$$\int_{0}^{\pi/2}\bigg[-\frac{1}{2}\,\cos u \bigg]_{1}^{9}d\theta\\=\int_{0}^{\pi/2}\frac{1}{2}(\cos{1}-cos{9})d\theta\\=\frac{1}{2}\bigg[(\cos{1}-cos{9})\theta\bigg]_{0}^{\pi/2}\\=\frac{\pi}{4}(\cos{1}-cos{9})$$
