Calculate the area enclosed by the curve `4lex^(2)+y^(2)le2(|x|+|y|)`.

Answer

To calculate the area enclosed by the curve \( 4x^2 + y^2 \leq 2(|x| + |y|) \), we will follow these steps:

### Step 1: Rewrite the Inequality
We start with the inequality:
\[ 4x^2 + y^2 \leq 2(|x| + |y|) \]
This can be rewritten as:
\[ 4x^2 + y^2 - 2|x| - 2|y| \leq 0 \]

### Step 2: Analyze the Cases for Absolute Values
Since we have absolute values, we need to consider four cases based on the signs of \(x\) and \(y\):
1. \(x \geq 0\), \(y \geq 0\)
2. \(x \geq 0\), \(y < 0\)
3. \(x < 0\), \(y \geq 0\)
4. \(x < 0\), \(y < 0\)

### Step 3: Case 1: \(x \geq 0\), \(y \geq 0\)
For this case, \( |x| = x \) and \( |y| = y \):
\[ 4x^2 + y^2 - 2x - 2y \leq 0 \]

### Step 4: Rearranging the Equation
Rearranging gives:
\[ 4x^2 + y^2 - 2x - 2y = 0 \]
This represents a conic section. To analyze it, we can complete the square.

### Step 5: Completing the Square
Completing the square for \(x\) and \(y\):
\[ 4(x^2 - \frac{1}{2}x) + (y^2 - 2y) = 0 \]
\[ 4\left(x - \frac{1}{4}\right)^2 - 4\left(\frac{1}{4}\right)^2 + (y - 1)^2 - 1 = 0 \]
\[ 4\left(x - \frac{1}{4}\right)^2 + (y - 1)^2 = 1 \]

### Step 6: Identify the Center and Radius
This is an ellipse centered at \( \left(\frac{1}{4}, 1\right) \) with semi-major axis \( \frac{1}{2} \) along the \(x\)-axis and semi-minor axis \( 1 \) along the \(y\)-axis.

### Step 7: Area of the Ellipse
The area \(A\) of an ellipse is given by the formula:
\[ A = \pi \times a \times b \]
where \(a\) and \(b\) are the semi-major and semi-minor axes, respectively. Here, \(a = \frac{1}{2}\) and \(b = 1\):
\[ A = \pi \times \frac{1}{2} \times 1 = \frac{\pi}{2} \]

### Step 8: Considering All Quadrants
Since the original inequality is symmetric in all four quadrants, the total area enclosed by the curve is:
\[ \text{Total Area} = 4 \times \frac{\pi}{2} = 2\pi \]

### Final Answer
Thus, the area enclosed by the curve \(4x^2 + y^2 \leq 2(|x| + |y|)\) is:
\[ \boxed{2\pi} \]