Don't use casual calculus!

This is just an Archimedean spiral, with polar equation $r=\theta$ , and the area is $A=\frac{1}{2}\int_{0}^{\pi/2}{r^2}d\theta=\frac{{\pi}^3}{48}\approx 0.6459$ . The required answer is $\boxed{646}$ .

Clearly:by the substitution $x=r.cos(\theta)$ and $y=r.sin(\theta)$ yields : $r=\theta$ hence as $A$ is closed area we can write: $A=\frac{1}{2}\int_{0}^{pi/2}{r^2}dr=frac{pi^3}{48}$