1. **State the problem:** We want to evaluate the double integral $$\iint_R \arctan\left(\frac{y}{x}\right) dA$$ where the region $$R = \{(x,y) \mid 1 \leq x^2 + y^2 \leq 4, 0 \leq y \leq x\}$$. 2. **Convert to polar coordinates:** Recall that $$x = r\cos\theta$$, $$y = r\sin\theta$$, and $$dA = r dr d\theta$$. 3. **Rewrite the region in polar coordinates:** - The condition $$1 \leq x^2 + y^2 \leq 4$$ becomes $$1 \leq r^2 \leq 4$$ or $$1 \leq r \leq 2$$. - The inequality $$0 \leq y \leq x$$ translates to $$0 \leq r\sin\theta \leq r\cos\theta$$. - Since $$r > 0$$, divide by $$r$$: $$0 \leq \sin\theta \leq \cos\theta$$. - This implies $$0 \leq \theta \leq \frac{\pi}{4}$$ because in this interval $$\sin\theta \leq \cos\theta$$ and both are nonnegative. 4. **Rewrite the integrand:** $$\arctan\left(\frac{y}{x}\right) = \arctan\left(\frac{r\sin\theta}{r\cos\theta}\right) = \arctan(\tan\theta) = \theta$$ for $$\theta \in [0, \frac{\pi}{4}]$$. 5. **Set up the integral in polar coordinates:** $$\int_{\theta=0}^{\pi/4} \int_{r=1}^2 \theta \cdot r \, dr \, d\theta$$. 6. **Integrate with respect to $$r$$:** $$\int_1^2 r \, dr = \left[\frac{r^2}{2}\right]_1^2 = \frac{4}{2} - \frac{1}{2} = \frac{3}{2}$$. 7. **Integral reduces to:** $$\int_0^{\pi/4} \theta \cdot \frac{3}{2} \, d\theta = \frac{3}{2} \int_0^{\pi/4} \theta \, d\theta$$. 8. **Integrate with respect to $$\theta$$:** $$\int_0^{\pi/4} \theta \, d\theta = \left[\frac{\theta^2}{2}\right]_0^{\pi/4} = \frac{(\pi/4)^2}{2} = \frac{\pi^2}{32}$$. 9. **Final answer:** $$\frac{3}{2} \times \frac{\pi^2}{32} = \frac{3\pi^2}{64}$$. Thus, the value of the integral is $$\boxed{\frac{3\pi^2}{64}}$$.