Consider curves `S_(1): sqrt(|x|)+sqrt(|y|)=sqrt(a), S_(2): x^(2)+y^(2)=a^(2) and S_(3)": "|x|+|y|=a." If "alpha" is area bounded by "S_(1) and S_(2), beta" is area bounded by "S_(1) and S_(3) and gamma` is the area bounded by `S_(2) and S_(3),` then

Options

A. `alpha=a^(2)(pi-(2)/(3))`
B. `beta=(4a^(2))/(3)`
C. `gamma=2a^(2)(pi-1)`
D. the ratio in which `S_(3)` divides area between `S_(1) and S_(2)" is "4:3(pi-2)`

Explanation

To solve the problem, we need to find the areas bounded by the given curves \( S_1, S_2, \) and \( S_3 \). Let's break down the solution step by step.

### Step 1: Understand the curves
1. **Curve \( S_1 \)**: \( \sqrt{|x|} + \sqrt{|y|} = \sqrt{a} \)
2. **Curve \( S_2 \)**: \( x^2 + y^2 = a^2 \) (Circle)
3. **Curve \( S_3 \)**: \( |x| + |y| = a \) (Square)

### Step 2: Analyze \( S_1 \)
Assuming \( x \geq 0 \) and \( y \geq 0 \):
- The equation simplifies to \( \sqrt{x} + \sqrt{y} = \sqrt{a} \).
- Rearranging gives \( \sqrt{y} = \sqrt{a} - \sqrt{x} \) or \( y = (\sqrt{a} - \sqrt{x})^2 \).

### Step 3: Find the area bounded by \( S_1 \) and \( S_2 \) (denoted as \( \alpha \))
1. The area under \( S_1 \) in the first quadrant is given by:
\[ A_{S_1} = \int_0^a (\sqrt{a} - \sqrt{x})^2 \, dx \]
2. Expanding the integrand:
\[ = \int_0^a (a - 2\sqrt{a}\sqrt{x} + x) \, dx \]
3. Integrating term by term:
\[ = \left[ ax - \frac{4}{3} a^{3/2} \sqrt{x} + \frac{x^2}{2} \right]_0^a \]
\[ = \left[ a^2 - \frac{4}{3} a^{3/2} \cdot a^{1/2} + \frac{a^2}{2} \right] \]
\[ = a^2 - \frac{4}{3} a^2 + \frac{1}{2} a^2 = \frac{a^2}{6} \]

4. The area of \( S_2 \) in the first quadrant is:
\[ A_{S_2} = \frac{\pi a^2}{4} \]

5. Therefore, the area \( \alpha \) bounded by \( S_1 \) and \( S_2 \):
\[ \alpha = 4 \left( \frac{\pi a^2}{4} - \frac{a^2}{6} \right) = \pi a^2 - \frac{2a^2}{3} \]

### Step 4: Find the area bounded by \( S_1 \) and \( S_3 \) (denoted as \( \beta \))
1. The area of \( S_3 \) in the first quadrant is:
\[ A_{S_3} = \frac{a^2}{2} \]

2. Therefore, the area \( \beta \) bounded by \( S_1 \) and \( S_3 \):
\[ \beta = 4 \left( \frac{a^2}{2} - \frac{a^2}{6} \right) = 4 \left( \frac{3a^2}{6} - \frac{a^2}{6} \right) = \frac{4a^2}{3} \]

### Step 5: Find the area bounded by \( S_2 \) and \( S_3 \) (denoted as \( \gamma \))
1. The area \( \gamma \) bounded by \( S_2 \) and \( S_3 \):
\[ \gamma = 4 \left( \frac{\pi a^2}{4} - \frac{a^2}{2} \right) = \pi a^2 - 2a^2 \]

### Final Results
- \( \alpha = a^2 \left( \pi - \frac{2}{3} \right) \)
- \( \beta = \frac{4a^2}{3} \)
- \( \gamma = a^2 \left( \pi - 2 \right) \)