1. Problem: Evaluate $$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \int_0^{2\cos\theta} d r d \theta$$. 2. Inner integral: $$\int_0^{2\cos\theta} d r = r \Big|_0^{2\cos\theta} = 2\cos\theta$$. 3. Outer integral: $$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 2\cos\theta d\theta = 2 \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos\theta d\theta$$. 4. Integral of cosine: $$\int \cos\theta d\theta = \sin\theta$$. 5. Evaluate: $$2 [\sin\theta]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = 2 (\sin(\frac{\pi}{2}) - \sin(-\frac{\pi}{2})) = 2 (1 - (-1)) = 2 \times 2 = 4$$. Final answer: $$4$$