Using integration,
(i) find the area of the first quadrant of the circle :
`x ^(2) +y ^(2) =4`
(ii) find the area of the circle :
`x ^(2) + y ^(2) =4.`

Answer

To solve the given problem, we will break it down into two parts as specified in the question.

### Part (i): Finding the area of the first quadrant of the circle defined by the equation \(x^2 + y^2 = 4\).

1. **Identify the Circle's Radius**:
The equation \(x^2 + y^2 = 4\) represents a circle centered at the origin (0, 0) with a radius of \(r = \sqrt{4} = 2\).

2. **Set Up the Integral**:
We need to find the area in the first quadrant. The area can be expressed as an integral of the function \(y\) in terms of \(x\). From the equation of the circle, we can express \(y\) as:
\[ y = \sqrt{4 - x^2} \]
The limits for \(x\) in the first quadrant are from 0 to 2.

3. **Write the Integral**:
The area \(A\) in the first quadrant can be calculated using the integral:
\[ A = \int_{0}^{2} \sqrt{4 - x^2} \, dx \]

4. **Evaluate the Integral**:
To solve the integral, we can use a trigonometric substitution. Let:
\[ x = 2 \sin(\theta) \quad \Rightarrow \quad dx = 2 \cos(\theta) \, d\theta \]
The limits change as follows:
- When \(x = 0\), \(\theta = 0\)
- When \(x = 2\), \(\theta = \frac{\pi}{2}\)

Substituting into the integral:
\[ A = \int_{0}^{\frac{\pi}{2}} \sqrt{4 - (2 \sin(\theta))^2} \cdot 2 \cos(\theta) \, d\theta \]
Simplifying:
\[ = \int_{0}^{\frac{\pi}{2}} \sqrt{4(1 - \sin^2(\theta))} \cdot 2 \cos(\theta) \, d\theta \]
\[ = \int_{0}^{\frac{\pi}{2}} \sqrt{4 \cos^2(\theta)} \cdot 2 \cos(\theta) \, d\theta \]
\[ = \int_{0}^{\frac{\pi}{2}} 4 \cos^2(\theta) \, d\theta \]

Using the identity \(\cos^2(\theta) = \frac{1 + \cos(2\theta)}{2}\):
\[ = 4 \int_{0}^{\frac{\pi}{2}} \frac{1 + \cos(2\theta)}{2} \, d\theta \]
\[ = 2 \left[ \theta + \frac{1}{2} \sin(2\theta) \right]_{0}^{\frac{\pi}{2}} \]
Evaluating the limits:
\[ = 2 \left[ \frac{\pi}{2} + 0 - (0 + 0) \right] = \pi \]

5. **Final Area of the First Quadrant**:
Therefore, the area of the first quadrant of the circle is:
\[ A = \pi \text{ square units} \]

### Part (ii): Finding the area of the entire circle.

1. **Use the Area of the First Quadrant**:
Since the circle is symmetric in all quadrants, the area of the entire circle can be found by multiplying the area of the first quadrant by 4:
\[ \text{Total Area} = 4 \times A = 4 \times \pi = 4\pi \text{ square units} \]

### Summary of Solutions:
- **Area of the first quadrant**: \( \pi \) square units.
- **Area of the entire circle**: \( 4\pi \) square units.