1. **Problem statement:** Find the area of region S bounded by $f(x) = -\cos\left(\frac{\pi}{3} x\right)$, $g(x) = 2 - \frac{1}{4} x^2$, the y-axis ($x=0$), and the vertical line $x=2$. 2. **Formula and explanation:** The area between two curves $f(x)$ and $g(x)$ from $a$ to $b$ is given by: $$\text{Area} = \int_a^b \bigl(g(x) - f(x)\bigr) \, dx$$ where $g(x)$ is the upper function and $f(x)$ is the lower function. 3. **Identify bounds and functions:** Here, $a=0$, $b=2$, $g(x) = 2 - \frac{1}{4} x^2$, and $f(x) = -\cos\left(\frac{\pi}{3} x\right)$. 4. **Set up the integral:** $$\text{Area} = \int_0^2 \left(2 - \frac{1}{4} x^2 + \cos\left(\frac{\pi}{3} x\right)\right) dx$$ 5. **Integrate term-by-term:** $$\int_0^2 2 \, dx = 2x \Big|_0^2 = 4$$ $$\int_0^2 -\frac{1}{4} x^2 \, dx = -\frac{1}{4} \cdot \frac{x^3}{3} \Big|_0^2 = -\frac{1}{4} \cdot \frac{8}{3} = -\frac{2}{3}$$ $$\int_0^2 \cos\left(\frac{\pi}{3} x\right) dx = \left. \frac{3}{\pi} \sin\left(\frac{\pi}{3} x\right) \right|_0^2 = \frac{3}{\pi} \left( \sin\left(\frac{2\pi}{3}\right) - 0 \right) = \frac{3}{\pi} \cdot \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{2\pi}$$ 6. **Sum the results:** $$\text{Area} = 4 - \frac{2}{3} + \frac{3\sqrt{3}}{2\pi} \approx 4.16$$ **Final answer:** $$\boxed{\text{Area} \approx 4.16}$$