1. **Stating the problem:** Find the arc length of the entire curve given by the polar equation $$r = 2a \cos \theta$$. 2. **Formula for arc length in polar coordinates:** The arc length $$S$$ of a curve given by $$r(\theta)$$ from $$\theta = \alpha$$ to $$\theta = \beta$$ is $$ S = \int_{\alpha}^{\beta} \sqrt{r(\theta)^2 + \left(\frac{dr}{d\theta}\right)^2} \, d\theta $$ 3. **Identify the interval:** Since the problem asks for the entire curve of $$r = 2a \cos \theta$$, which is a circle, the interval for $$\theta$$ is $$0$$ to $$2\pi$$. 4. **Calculate $$\frac{dr}{d\theta}$$:** $$ r = 2a \cos \theta \\ \frac{dr}{d\theta} = -2a \sin \theta $$ 5. **Substitute into the arc length formula:** $$ S = \int_0^{2\pi} \sqrt{(2a \cos \theta)^2 + (-2a \sin \theta)^2} \, d\theta = \int_0^{2\pi} \sqrt{4a^2 \cos^2 \theta + 4a^2 \sin^2 \theta} \, d\theta $$ 6. **Simplify inside the square root:** $$ \sqrt{4a^2 (\cos^2 \theta + \sin^2 \theta)} = \sqrt{4a^2 \cdot 1} = 2a $$ 7. **Integral becomes:** $$ S = \int_0^{2\pi} 2a \, d\theta = 2a \times (2\pi - 0) = 4\pi a $$ 8. **Final answer:** The arc length of the entire curve $$r = 2a \cos \theta$$ is $$ \boxed{4 \pi a} $$