The area enclosed between the curves `|x| + |y| ge 2 and y ^(2) =4 (1- (x ^(2))/(9))` is :

Options

A. `(6pi -4 )` sq. units
B. `(6pi -8)` se. units
C. `(3pi-4)` se. units
D. `(3pi-2)` sq. units

Answer

To find the area enclosed between the curves \( |x| + |y| \geq 2 \) and \( y^2 = 4(1 - \frac{x^2}{9}) \), we will follow these steps:

### Step 1: Understand the curves
The first curve \( |x| + |y| = 2 \) represents a diamond shape (or square rotated by 45 degrees) with vertices at (2,0), (0,2), (-2,0), and (0,-2). The second curve \( y^2 = 4(1 - \frac{x^2}{9}) \) is an ellipse centered at the origin with semi-major axis 2 (along the y-axis) and semi-minor axis 3 (along the x-axis).

### Step 2: Determine the intersection points
To find the area enclosed between the curves, we first need to find the points where they intersect. We can substitute \( y \) from the ellipse equation into the diamond equation.

From the ellipse equation:
\[ y = \pm 2\sqrt{1 - \frac{x^2}{9}} \]

Substituting into the diamond equation \( |x| + |y| = 2 \):
\[ |x| + 2\sqrt{1 - \frac{x^2}{9}} = 2 \]

### Step 3: Solve for \( x \)
We can solve for \( x \) by isolating the square root:
\[ 2\sqrt{1 - \frac{x^2}{9}} = 2 - |x| \]
\[ \sqrt{1 - \frac{x^2}{9}} = 1 - \frac{|x|}{2} \]

Squaring both sides:
\[ 1 - \frac{x^2}{9} = \left(1 - \frac{|x|}{2}\right)^2 \]
Expanding the right side:
\[ 1 - \frac{x^2}{9} = 1 - |x| + \frac{x^2}{4} \]

### Step 4: Rearranging the equation
Rearranging gives:
\[ -\frac{x^2}{9} + |x| - \frac{x^2}{4} = 0 \]
Multiplying through by 36 to eliminate the fractions:
\[ -4x^2 + 36|x| - 9x^2 = 0 \]
Combining like terms:
\[ -13x^2 + 36|x| = 0 \]

### Step 5: Factor the equation
Factoring out \( |x| \):
\[ |x|(36 - 13|x|) = 0 \]
This gives us \( |x| = 0 \) or \( |x| = \frac{36}{13} \).

### Step 6: Find corresponding \( y \) values
For \( |x| = 0 \):
\[ y = \pm 2 \]
For \( |x| = \frac{36}{13} \):
Substituting back into the ellipse equation:
\[ y^2 = 4\left(1 - \frac{(\frac{36}{13})^2}{9}\right) \]
Calculating gives the corresponding \( y \) values.

### Step 7: Set up the double integral
The area \( A \) can be calculated using a double integral:
\[ A = 4 \int_0^2 \left(3\sqrt{4 - y^2} - (2 - y)\right) dy \]

### Step 8: Evaluate the integral
Calculating the integral:
1. Solve \( \int_0^2 3\sqrt{4 - y^2} dy \) using the formula for the area of a quarter circle.
2. Solve \( \int_0^2 (2 - y) dy \).

### Step 9: Combine results
Combine the results from the integrals to find the total area.

### Final Answer
The area enclosed between the curves is \( 6\pi - 8 \) square units.

---