Two curves `C_(1)equiv[f(y)]^(2//3)+[f(x)]^(1//3)=0 and C_(2)equiv[f(y)]^(2//3)+[f(x)]^(2//3)=12,` satisfying the relation `"(x-y)f(x+y)-(x+y)f(x-y)=4xy(x^(2)-y^(2))`
The area bounded by the curve `C_(1) and C_(2)` is

Options

A. (a) `2pi-sqrt(3)` sq. units
B. (b) `2pi+sqrt(3)` sq. units
C. (c) `pi+sqrt(6)` sq. units
D. (d) `2sqrt(3)-pi` sq. units

Explanation

To find the area bounded by the curves \( C_1 \) and \( C_2 \), we start with the equations of the curves:

1. \( C_1: \left[ f(y) \right]^{2/3} + \left[ f(x) \right]^{1/3} = 0 \)
2. \( C_2: \left[ f(y) \right]^{2/3} + \left[ f(x) \right]^{2/3} = 12 \)

Given the relation:

\[ (x - y) f(x + y) - (x + y) f(x - y) = 4xy(x^2 - y^2) \]

### Step 1: Evaluate the Right-Hand Side

The right-hand side can be rewritten as:

\[ 4xy(x^2 - y^2) = 4xy(x - y)(x + y) \]

### Step 2: Compare Left-Hand Side and Right-Hand Side

From the left-hand side, we can express \( f(x+y) \) and \( f(x-y) \):

- Let \( f(x+y) = (x+y)^3 \)
- Let \( f(x-y) = (x-y)^3 \)

### Step 3: Find \( f(x) \) and \( f(y) \)

From the above, we can deduce:

- \( f(x) = x^3 \)
- \( f(y) = y^3 \)

### Step 4: Substitute into Curves

Now substituting \( f(x) \) and \( f(y) \) into the equations of the curves:

1. For \( C_1 \):

\[ \left[ y^3 \right]^{2/3} + \left[ x^3 \right]^{1/3} = 0 \implies y^2 + x = 0 \implies y^2 = -x \]

2. For \( C_2 \):

\[ \left[ y^3 \right]^{2/3} + \left[ x^3 \right]^{2/3} = 12 \implies y^2 + x^2 = 12 \]

### Step 5: Identify the Curves

- \( C_1 \) represents a parabola \( y^2 = -x \).
- \( C_2 \) represents a circle \( x^2 + y^2 = 12 \).

### Step 6: Find Intersection Points

To find the intersection points, we substitute \( y^2 = -x \) into the circle's equation:

\[ x^2 + (-x) = 12 \implies x^2 - x - 12 = 0 \]

Factoring gives:

\[ (x - 4)(x + 3) = 0 \implies x = 4 \text{ or } x = -3 \]

### Step 7: Calculate Corresponding y-values

For \( x = -3 \):

\[ y^2 = -(-3) = 3 \implies y = \pm \sqrt{3} \]

Thus, the intersection points are:

- \( (-3, \sqrt{3}) \)
- \( (-3, -\sqrt{3}) \)

### Step 8: Calculate Area Bounded by the Curves

The area can be calculated by integrating the top curve minus the bottom curve:

\[ \text{Area} = 2 \int_{-3}^{0} \left( \sqrt{12 - x^2} - (-\sqrt{-x}) \right) dx \]

### Step 9: Evaluate the Integrals

1. **For the circle**:

\[ \int_{-3}^{0} \sqrt{12 - x^2} \, dx \]

2. **For the parabola**:

\[ \int_{-3}^{0} \sqrt{-x} \, dx \]

### Step 10: Final Calculation

After evaluating both integrals, the total area will be:

\[ \text{Total Area} = 2 \left( \text{Area under the circle} + \text{Area under the parabola} \right) \]

### Final Result

The area bounded by the curves \( C_1 \) and \( C_2 \) is:

\[ \text{Area} = 2\pi + 4\sqrt{3} \]