Area lying in the first quadrant and bounded by the circle `x ^(2) +y ^(2) =4` and the lines `x =0 and x =2` is

Options

A. `pi`
B. `pi/2`
C. `pi/3`
D. `pi/4.`

Answer

To find the area lying in the first quadrant and bounded by the circle \(x^2 + y^2 = 4\) and the lines \(x = 0\) and \(x = 2\), we can follow these steps:

### Step 1: Understand the Circle
The equation of the circle is given by:
\[ x^2 + y^2 = 4 \]
This indicates that the center of the circle is at the origin \((0,0)\) and the radius is \(2\).

### Step 2: Identify the Region
We need to find the area in the first quadrant bounded by:
- The circle \(x^2 + y^2 = 4\)
- The line \(x = 0\) (y-axis)
- The line \(x = 2\)

### Step 3: Express \(y\) in terms of \(x\)
From the equation of the circle, we can express \(y\) as:
\[ y = \sqrt{4 - x^2} \]
This is valid for \(0 \leq x \leq 2\) since we are in the first quadrant.

### Step 4: Set Up the Integral
The area \(A\) can be calculated using the integral of \(y\) with respect to \(x\) from \(0\) to \(2\):
\[ A = \int_{0}^{2} y \, dx = \int_{0}^{2} \sqrt{4 - x^2} \, dx \]

### Step 5: Solve the Integral
To solve the integral \(\int \sqrt{4 - x^2} \, dx\), we can use the formula for the integral of \(\sqrt{a^2 - x^2}\):
\[ \int \sqrt{a^2 - x^2} \, dx = \frac{x}{2} \sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1}\left(\frac{x}{a}\right) + C \]
In our case, \(a = 2\):
\[ \int \sqrt{4 - x^2} \, dx = \frac{x}{2} \sqrt{4 - x^2} + 2 \sin^{-1}\left(\frac{x}{2}\right) + C \]

### Step 6: Evaluate the Definite Integral
Now we evaluate the definite integral from \(0\) to \(2\):
\[ A = \left[ \frac{x}{2} \sqrt{4 - x^2} + 2 \sin^{-1}\left(\frac{x}{2}\right) \right]_{0}^{2} \]
Calculating at the upper limit \(x = 2\):
\[ = \left( \frac{2}{2} \sqrt{4 - 2^2} + 2 \sin^{-1}\left(1\right) \right) = (1 \cdot 0 + 2 \cdot \frac{\pi}{2}) = \pi \]

Calculating at the lower limit \(x = 0\):
\[ = \left( \frac{0}{2} \sqrt{4 - 0^2} + 2 \sin^{-1}\left(0\right) \right) = (0 + 0) = 0 \]

### Step 7: Final Area Calculation
Thus, the area \(A\) is:
\[ A = \pi - 0 = \pi \]

### Conclusion
The area lying in the first quadrant and bounded by the circle \(x^2 + y^2 = 4\) and the lines \(x = 0\) and \(x = 2\) is:
\[ \boxed{\pi} \]