The least integer which is greater than or equal to the area of region in `x-y` plane satisfying `x ^(6) -x ^(2) +y ^(2) le 0` is:

Answer

To solve the problem of finding the least integer greater than or equal to the area of the region in the \( x-y \) plane satisfying the inequality \( x^6 - x^2 + y^2 \leq 0 \), we will follow these steps:

### Step 1: Rewrite the inequality
We start by rewriting the inequality:
\[ x^6 - x^2 + y^2 \leq 0 \]
This can be rearranged to:
\[ y^2 \leq x^2 - x^6 \]
This implies:
\[ y = \pm \sqrt{x^2 - x^6} \]

### Step 2: Determine the domain for \( x \)
Next, we need to find the values of \( x \) for which \( x^2 - x^6 \geq 0 \). Factoring gives us:
\[ x^2(1 - x^4) \geq 0 \]
This means \( x^2 \geq 0 \) (which is always true) and \( 1 - x^4 \geq 0 \). Therefore:
\[ x^4 \leq 1 \implies -1 \leq x \leq 1 \]

### Step 3: Find the area under the curve
The area under the curve can be calculated by integrating \( y \) from \( 0 \) to \( 1 \) and then multiplying by 4 (due to symmetry about both axes):
\[ \text{Area} = 4 \int_0^1 \sqrt{x^2 - x^6} \, dx \]

### Step 4: Simplify the integrand
We can simplify the integrand:
\[ \sqrt{x^2 - x^6} = \sqrt{x^2(1 - x^4)} = x \sqrt{1 - x^4} \]
Thus, the area becomes:
\[ \text{Area} = 4 \int_0^1 x \sqrt{1 - x^4} \, dx \]

### Step 5: Use substitution for integration
Let \( t = x^2 \), then \( dt = 2x \, dx \) or \( dx = \frac{dt}{2\sqrt{t}} \). Changing the limits:
- When \( x = 0 \), \( t = 0 \)
- When \( x = 1 \), \( t = 1 \)

The integral becomes:
\[ \text{Area} = 4 \int_0^1 \sqrt{1 - t^2} \frac{dt}{2} = 2 \int_0^1 \sqrt{1 - t^2} \, dt \]

### Step 6: Evaluate the integral
The integral \( \int_0^1 \sqrt{1 - t^2} \, dt \) represents the area of a quarter circle with radius 1:
\[ \int_0^1 \sqrt{1 - t^2} \, dt = \frac{\pi}{4} \]
Thus:
\[ \text{Area} = 2 \cdot \frac{\pi}{4} = \frac{\pi}{2} \]

### Step 7: Find the least integer greater than or equal to the area
The area we calculated is \( \frac{\pi}{2} \approx 1.57 \). The least integer greater than or equal to this value is:
\[ \lceil \frac{\pi}{2} \rceil = 2 \]

### Final Answer
The least integer which is greater than or equal to the area of the region is:
\[ \boxed{2} \]