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The equation of the circle is $x^2 + y^2 = 16$. We can express $y$ in terms of $x$ as follows: $y^2 = 16 - x^2$ $y = \pm \sqrt{16 - x^2}$ Since we are finding the area enclosed, we consider both the positive and negative square roots.
The area of the region enclosed between the circle and the lines $x = -2$ and $x = 2$ can be found by integrating the function $y = \sqrt{16 - x^2}$ from $x = -2$ to $x = 2$ and multiplying the result by 2 (since we have both positive and negative y values). Area $A = 2 \int_{-2}^{2} \sqrt{16 - x^2} \, dx$
To evaluate the integral $\int_{-2}^{2} \sqrt{16 - x^2} \, dx$, we use the formula: $\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2} \sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1} \left( \frac{x}{a} \right) + C$ In our case, $a^2 = 16$, so $a = 4$. Thus, $\int_{-2}^{2} \sqrt{16 - x^2} \, dx = \left[ \frac{x}{2} \sqrt{16 - x^2} + \frac{16}{2} \sin^{-1} \left( \frac{x}{4} \right) \right]_{-2}^{2}$ $= \left[ \frac{2}{2} \sqrt{16 - 2^2} + 8 \sin^{-1} \left( \frac{2}{4} \right) \right] - \left[ \frac{-2}{2} \sqrt{16 - (-2)^2} + 8 \sin^{-1} \left( \frac{-2}{4} \right) \right]$ $= \left[ \sqrt{12} + 8 \sin^{-1} \left( \frac{1}{2} \right) \right] - \left[ -\sqrt{12} + 8 \sin^{-1} \left( -\frac{1}{2} \right) \right]$ $= \sqrt{12} + 8 \left( \frac{\pi}{6} \right) - \left( -\sqrt{12} + 8 \left( -\frac{\pi}{6} \right) \right)$ $= \sqrt{12} + \frac{4\pi}{3} + \sqrt{12} + \frac{4\pi}{3}$ $= 2\sqrt{12} + \frac{8\pi}{3} = 2(2\sqrt{3}) + \frac{8\pi}{3} = 4\sqrt{3} + \frac{8\pi}{3}$
Now, we multiply the result by 2 to get the total area: $A = 2 \left( 4\sqrt{3} + \frac{8\pi}{3} \right) = 8\sqrt{3} + \frac{16\pi}{3}$
Final Answer: $8\sqrt{3} + \frac{16\pi}{3}$