The parabola `y^2 = 4x` divides the area of the circle `x^2 + y^2 = 5` in two parts. The area of the smaller part is equal to:

Options

A. `2/3 + sqrt 5 sin^(-1)(2/(sqrt 5))`
B. `1/3 + sqrt 5 sin^(-1)(2/(sqrt 5))`
C. `2/3 + 5 sin^(-1)(2/(sqrt 5))`
D. `1/3 + 5 sin^(-1)(2/(sqrt 5))`

Answer

To find the area of the smaller part of the circle divided by the parabola, we will follow these steps:

### Step 1: Identify the equations
The equations we are dealing with are:
- Circle: \( x^2 + y^2 = 5 \)
- Parabola: \( y^2 = 4x \)

### Step 2: Find the points of intersection
To find the points where the parabola intersects the circle, we substitute \( y^2 = 4x \) into the equation of the circle.

Substituting gives:
\[ x^2 + 4x - 5 = 0 \]

### Step 3: Solve the quadratic equation
Now we solve the quadratic equation:
\[ x^2 + 4x - 5 = 0 \]
Factoring gives:
\[ (x + 5)(x - 1) = 0 \]
Thus, the solutions are:
\[ x = -5 \quad \text{and} \quad x = 1 \]

### Step 4: Find the corresponding y-values
Using \( x = 1 \) in the parabola's equation to find \( y \):
\[ y^2 = 4(1) = 4 \implies y = 2 \text{ or } y = -2 \]
So the points of intersection are \( (1, 2) \) and \( (1, -2) \).

### Step 5: Set up the integrals for area calculation
We need to find the area between the parabola and the circle from \( x = 0 \) to \( x = 1 \) (the intersection point).

The area under the parabola from \( x = 0 \) to \( x = 1 \):
\[ A_1 = \int_0^1 2\sqrt{x} \, dx \]

The area under the circle from \( x = 0 \) to \( x = 1 \):
\[ A_2 = \int_0^1 \sqrt{5 - x^2} \, dx \]

### Step 6: Calculate the area under the parabola
Calculating \( A_1 \):
\[ A_1 = \int_0^1 2\sqrt{x} \, dx = 2 \cdot \left[ \frac{2}{3} x^{3/2} \right]_0^1 = 2 \cdot \frac{2}{3} = \frac{4}{3} \]

### Step 7: Calculate the area under the circle
Calculating \( A_2 \):
\[ A_2 = \int_0^1 \sqrt{5 - x^2} \, dx \]
Using the formula for the area of a quarter circle:
\[ A_2 = \frac{1}{4} \cdot \pi \cdot 5 = \frac{5\pi}{4} \]

### Step 8: Find the area between the curves
The area of the smaller part is:
\[ A = A_2 - A_1 = \frac{5\pi}{4} - \frac{4}{3} \]

### Step 9: Final calculation
To combine these areas, we need a common denominator:
\[ A = \frac{15\pi}{12} - \frac{16}{12} = \frac{15\pi - 16}{12} \]

### Conclusion
Thus, the area of the smaller part is:
\[ \frac{15\pi - 16}{12} \]