1. **Problem statement:** Evaluate the double integral $$\iint_R \tan^{-1}\left(\frac{y}{x}\right) \, dA$$ where $R$ is the region inside the circle $$x^2 + (y-1)^2 = 1$$ and outside the circle $$x^2 + y^2 = 3$$. 2. **Understanding the region $R$:** - The first circle is centered at $(0,1)$ with radius $1$. - The second circle is centered at $(0,0)$ with radius $\sqrt{3}$. - Region $R$ is the set of points inside the first circle but outside the second. 3. **Key observation about the integrand:** The integrand is $$\tan^{-1}\left(\frac{y}{x}\right)$$ which is the angle $\theta$ in polar coordinates, but note that $\tan^{-1}(y/x)$ is the principal value of the arctangent function. 4. **Change of variables and symmetry:** - The function $\tan^{-1}(y/x)$ is an odd function with respect to reflection about the $y$-axis because $$\tan^{-1}\left(\frac{y}{-x}\right) = -\tan^{-1}\left(\frac{y}{x}\right).$$ - The region $R$ is symmetric about the $y$-axis because both circles are symmetric about $y$-axis. 5. **Integral over symmetric region of an odd function:** - Since the integrand is odd in $x$ and the region $R$ is symmetric about the $y$-axis, the integral over $R$ cancels out. 6. **Conclusion:** $$\iint_R \tan^{-1}\left(\frac{y}{x}\right) \, dA = 0.$$