1. **State the problem:** Convert the double integral $$\int_0^{\sqrt{2y - y^2}} \int_{-2}^{-\sqrt{2y - y^2}} 2x \, dx \, dy$$ from rectangular to polar coordinates. 2. **Recall the conversion formulas:** - $$x = r \cos \theta$$ - $$y = r \sin \theta$$ - The Jacobian determinant for polar coordinates is $$dx \, dy = r \, dr \, d\theta$$. 3. **Analyze the region of integration:** - The inner integral limits for $$x$$ are from $$-2$$ to $$-\sqrt{2y - y^2}$$. - The outer integral limits for $$y$$ are from $$0$$ to $$\sqrt{2y - y^2}$$. 4. **Rewrite the boundary curve:** - The expression $$\sqrt{2y - y^2}$$ can be rewritten as $$\sqrt{y(2 - y)}$$. - This describes a semicircle of radius 1 centered at $$y=1$$, but since the limits are from 0 to $$\sqrt{2y - y^2}$$, the region corresponds to a circle of radius 2. 5. **Convert the limits to polar:** - The radius $$r$$ goes from $$0$$ to $$-2 \sin \theta$$ (since $$y = r \sin \theta$$ and the boundary is $$y = 2 \sin \theta$$ but negative because of the limits). - The angle $$\theta$$ goes from $$0$$ to $$\pi$$ to cover the upper half-plane where $$y \geq 0$$. 6. **Convert the integrand:** - The integrand $$2x$$ becomes $$2r \cos \theta$$. 7. **Write the polar integral:** $$\int_0^{\pi} \int_0^{-2 \sin \theta} 2r \cos \theta \cdot r \, dr \, d\theta = \int_0^{\pi} \int_0^{-2 \sin \theta} 2r^2 \cos \theta \, dr \, d\theta$$. 8. **Compare with options:** - Option (c) is $$\int_0^{\pi} \int_0^{-2 \sin \theta} 2r^2 \cos \theta \, dr \, d\theta$$, which matches our result. **Final answer:** Option (c) $$\int_0^{\pi} \int_0^{-2 \sin \theta} 2r^2 \cos \theta \, dr \, d\theta$$.