Using integration, find the area of the region bounded between :
(i) the line `x =2` and the parabola `y ^(2) = 8x`
(ii) the line `x=3` and the parabola `y ^(2) =4x.`

Answer

To find the area of the regions bounded by the given curves, we will solve each part step by step.

### Part (i): Area between the line \( x = 2 \) and the parabola \( y^2 = 8x \)

1. **Identify the curves**:
The parabola \( y^2 = 8x \) can be rewritten as \( y = \pm \sqrt{8x} \). The line \( x = 2 \) is a vertical line.

2. **Find the points of intersection**:
To find the area, we need to determine the limits of integration. The line \( x = 2 \) intersects the parabola at:
\[ y^2 = 8(2) = 16 \implies y = \pm 4 \]
Thus, the points of intersection are \( (2, 4) \) and \( (2, -4) \).

3. **Set up the integral**:
The area \( A \) between the line and the parabola from \( x = 0 \) to \( x = 2 \) is given by:
\[ A = 2 \int_{0}^{2} \sqrt{8x} \, dx \]
(We multiply by 2 because the parabola is symmetric about the x-axis.)

4. **Simplify the integral**:
We can simplify \( \sqrt{8x} = 2\sqrt{2} \sqrt{x} \):
\[ A = 2 \int_{0}^{2} 2\sqrt{2} \sqrt{x} \, dx = 4\sqrt{2} \int_{0}^{2} x^{1/2} \, dx \]

5. **Integrate**:
The integral of \( x^{1/2} \) is:
\[ \int x^{1/2} \, dx = \frac{x^{3/2}}{3/2} = \frac{2}{3} x^{3/2} \]
Therefore,
\[ A = 4\sqrt{2} \cdot \left[ \frac{2}{3} x^{3/2} \right]_{0}^{2} \]

6. **Evaluate the limits**:
Plugging in the limits:
\[ A = 4\sqrt{2} \cdot \left( \frac{2}{3} (2^{3/2}) - 0 \right) = 4\sqrt{2} \cdot \left( \frac{2}{3} \cdot 2\sqrt{2} \right) = \frac{16}{3} \]

### Final Area for Part (i):
\[ A = \frac{16}{3} \text{ square units} \]

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### Part (ii): Area between the line \( x = 3 \) and the parabola \( y^2 = 4x \)

1. **Identify the curves**:
The parabola \( y^2 = 4x \) can be rewritten as \( y = \pm 2\sqrt{x} \). The line \( x = 3 \) is a vertical line.

2. **Find the points of intersection**:
The line \( x = 3 \) intersects the parabola at:
\[ y^2 = 4(3) = 12 \implies y = \pm 2\sqrt{3} \]
Thus, the points of intersection are \( (3, 2\sqrt{3}) \) and \( (3, -2\sqrt{3}) \).

3. **Set up the integral**:
The area \( A \) between the line and the parabola from \( x = 0 \) to \( x = 3 \) is given by:
\[ A = 2 \int_{0}^{3} 2\sqrt{x} \, dx \]
(Again, we multiply by 2 because of symmetry.)

4. **Simplify the integral**:
Thus,
\[ A = 4 \int_{0}^{3} \sqrt{x} \, dx \]

5. **Integrate**:
The integral of \( \sqrt{x} \) is:
\[ \int \sqrt{x} \, dx = \frac{x^{3/2}}{3/2} = \frac{2}{3} x^{3/2} \]
Therefore,
\[ A = 4 \cdot \left[ \frac{2}{3} x^{3/2} \right]_{0}^{3} \]

6. **Evaluate the limits**:
Plugging in the limits:
\[ A = 4 \cdot \left( \frac{2}{3} (3^{3/2}) - 0 \right) = 4 \cdot \left( \frac{2}{3} \cdot 3\sqrt{3} \right) = \frac{8\sqrt{3}}{3} \]

### Final Area for Part (ii):
\[ A = \frac{8\sqrt{3}}{3} \text{ square units} \]

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