Let `Delta` be the area of the region `{(x,y)in R^(2):x^(2)+y^(2)le21,y^(2)le4x,xge1}`.Then `(1)/(2)(Delta-21sin^(-1)((2)/(sqrt(7))))` is equal to

Options

A. `sqrt(3)-(2)/(3)`
B. `2sqrt(3)-(2)/(3)`
C. `2sqrt(3)-(1)/(3),`
D. `sqrt(3)-(4)/(3)`

Answer

To solve the problem, we need to find the area \( \Delta \) of the region defined by the inequalities:

1. \( x^2 + y^2 \leq 21 \) (a circle of radius \( \sqrt{21} \))
2. \( y^2 \leq 4x \) (a parabola)
3. \( x \geq 1 \) (a vertical line)

### Step 1: Understand the region defined by the inequalities

- The first inequality \( x^2 + y^2 \leq 21 \) describes a circle centered at the origin with radius \( \sqrt{21} \).
- The second inequality \( y^2 \leq 4x \) describes a parabola that opens to the right.
- The third inequality \( x \geq 1 \) restricts the region to the right of the line \( x = 1 \).

### Step 2: Find the points of intersection

To find the points where the circle intersects the parabola, we substitute \( y^2 = 4x \) into the circle's equation:

\[ x^2 + 4x \leq 21 \]

Rearranging gives:

\[ x^2 + 4x - 21 = 0 \]

Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

\[ x = \frac{-4 \pm \sqrt{16 + 84}}{2} = \frac{-4 \pm \sqrt{100}}{2} = \frac{-4 \pm 10}{2} \]

This gives us:

\[ x = 3 \quad \text{(valid since } x \geq 1\text{)} \quad \text{and} \quad x = -7 \quad \text{(not valid)} \]

Now, substituting \( x = 3 \) back into \( y^2 = 4x \):

\[ y^2 = 12 \implies y = \pm 2\sqrt{3} \]

The points of intersection are \( (3, 2\sqrt{3}) \) and \( (3, -2\sqrt{3}) \).

### Step 3: Set up the area integral

The area \( \Delta \) can be calculated by integrating the area under the parabola from \( x = 1 \) to \( x = 3 \) and then from \( x = 3 \) to \( x = \sqrt{21} \) under the circle.

1. **From \( x = 1 \) to \( x = 3 \)**:
The upper curve is \( y = 2\sqrt{x} \) and the lower curve is \( y = -2\sqrt{x} \).

The area is:

\[ A_1 = \int_1^3 (2\sqrt{x} - (-2\sqrt{x})) \, dx = \int_1^3 4\sqrt{x} \, dx \]

Evaluating this integral:

\[ A_1 = 4 \cdot \left[ \frac{2}{3} x^{3/2} \right]_1^3 = \frac{8}{3} (3\sqrt{3} - 1) \]

2. **From \( x = 3 \) to \( x = \sqrt{21} \)**:
The upper curve is the circle \( y = \sqrt{21 - x^2} \) and the lower curve is \( y = -\sqrt{21 - x^2} \).

The area is:

\[ A_2 = \int_3^{\sqrt{21}} (\sqrt{21 - x^2} - (-\sqrt{21 - x^2})) \, dx = \int_3^{\sqrt{21}} 2\sqrt{21 - x^2} \, dx \]

Evaluating this integral using the formula for the area under a circle:

\[ A_2 = 2 \left[ \frac{x}{2} \sqrt{21 - x^2} + \frac{21}{2} \sin^{-1}\left(\frac{x}{\sqrt{21}}\right) \right]_3^{\sqrt{21}} \]

### Step 4: Combine the areas

The total area \( \Delta \) is:

\[ \Delta = A_1 + A_2 \]

### Step 5: Compute the final expression

We need to compute:

\[ \frac{1}{2} \left( \Delta - 21 \sin^{-1}\left(\frac{2}{\sqrt{7}}\right) \right) \]

After calculating \( A_1 \) and \( A_2 \) and substituting back, we will simplify to find the final answer.