Step-by-Step Solution

1. First, consider the ellipse $x^2 + 4y^2 \le 4$, which can be written as $\frac{x^2}{4} + y^2 \le 1$. This is an ellipse centered at the origin with semi-major axis $a=2$ and semi-minor axis $b=1$.
2. The area of the ellipse is $A_{ellipse} = \pi ab = \pi(2)(1) = 2\pi$.
3. Next, consider the inequalities $y \le |x| - 1$ and $y \ge 1 - |x|$. These represent the region between the lines $y = |x| - 1$ and $y = 1 - |x|$.
4. We need to find the area enclosed by the ellipse and these two inequalities. The lines $y = |x| - 1$ and $y = 1 - |x|$ intersect the ellipse.
5. Let's find the intersection points of $y = |x| - 1$ and the ellipse. Substituting $y = |x| - 1$ into the ellipse equation: $x^2 + 4(|x| - 1)^2 = 4$.
6. $x^2 + 4(x^2 - 2|x| + 1) = 4 \implies x^2 + 4x^2 - 8|x| + 4 = 4 \implies 5x^2 - 8|x| = 0 \implies |x|(5|x| - 8) = 0$.
7. So, $|x| = 0$ or $|x| = \frac{8}{5}$. Thus, $x = 0, \frac{8}{5}, -\frac{8}{5}$.
8. When $x = 0$, $y = |0| - 1 = -1$. When $x = \frac{8}{5}$, $y = |\frac{8}{5}| - 1 = \frac{3}{5}$. When $x = -\frac{8}{5}$, $y = |-\frac{8}{5}| - 1 = \frac{3}{5}$.
9. The intersection points are $(0, -1)$, $(\frac{8}{5}, \frac{3}{5})$, and $(-\frac{8}{5}, \frac{3}{5})$.
10. Now, let's find the intersection points of $y = 1 - |x|$ and the ellipse. Substituting $y = 1 - |x|$ into the ellipse equation: $x^2 + 4(1 - |x|)^2 = 4$.
11. $x^2 + 4(1 - 2|x| + x^2) = 4 \implies x^2 + 4 - 8|x| + 4x^2 = 4 \implies 5x^2 - 8|x| = 0 \implies |x|(5|x| - 8) = 0$.
12. So, $|x| = 0$ or $|x| = \frac{8}{5}$. Thus, $x = 0, \frac{8}{5}, -\frac{8}{5}$.
13. When $x = 0$, $y = 1 - |0| = 1$. When $x = \frac{8}{5}$, $y = 1 - |\frac{8}{5}| = -\frac{3}{5}$. When $x = -\frac{8}{5}$, $y = 1 - |-\frac{8}{5}| = -\frac{3}{5}$.
14. The intersection points are $(0, 1)$, $(\frac{8}{5}, -\frac{3}{5})$, and $(-\frac{8}{5}, -\frac{3}{5})$.
15. The required area is the area of the ellipse between $y = |x| - 1$ and $y = 1 - |x|$. Due to symmetry, we can consider the area in the first quadrant and multiply by 2.
16. In the first quadrant, we have $y = x - 1$ and $y = 1 - x$. The area is given by $2 \int_{0}^{\frac{8}{5}} [(1-x) - (x-1)] dx = 4 \int_{0}^{\frac{8}{5}} (1-x) dx$ from x=0 to 8/5. However, this is incorrect.
17. Let $x = 2\cos\theta$ and $y = \sin\theta$. Then $y = |x| - 1$ becomes $\sin\theta = |2\cos\theta| - 1$. When $x > 0$, $\sin\theta = 2\cos\theta - 1$. When $x < 0$, $\sin\theta = -2\cos\theta - 1$. Also, $y = 1 - |x|$ becomes $\sin\theta = 1 - |2\cos\theta|$. When $x > 0$, $\sin\theta = 1 - 2\cos\theta$. When $x < 0$, $\sin\theta = 1 + 2\cos\theta$.
18. Consider the region in the first quadrant. The area is given by $\int_{0}^{8/5} (1-x - (x-1)) dx = \int_{0}^{8/5} (2-2x) dx = [2x - x^2]_{0}^{8/5} = 2(\frac{8}{5}) - (\frac{8}{5})^2 = \frac{16}{5} - \frac{64}{25} = \frac{80-64}{25} = \frac{16}{25}$. The total area is $4 \times \frac{1}{2} \times base \times height = 4 \times \frac{1}{2} \times \frac{8}{5} \times \frac{3}{5} = \frac{48}{25}$.
19. Let $x = 2\cos\theta$ and $y = \sin\theta$. Then $\sin\theta = \frac{3}{5}$, so $\theta = \sin^{-1}(\frac{3}{5})$. The area is $4(\frac{1}{2}ab(\theta - \sin\theta\cos\theta)) = 4(\frac{1}{2}(2)(1)(\sin^{-1}(\frac{3}{5}) - \frac{3}{5}\frac{4}{5})) = 4(\sin^{-1}(\frac{3}{5}) - \frac{12}{25}) = 4\sin^{-1}(\frac{3}{5}) - \frac{48}{25}$. The area of the triangle is $2 \times \frac{1}{2} \times \frac{8}{5} \times \frac{3}{5} = \frac{24}{25}$. The required area is $4\sin^{-1}(\frac{3}{5}) - \frac{24}{25} = 4\sin^{-1}(\frac{3}{5}) - \frac{6}{5}$.