Area common to the circle `x^2+y^2=64` and the parabola `y^2=4x` is

Options

A. `(16)/(3)(4pi+sqrt(3))`
B. `(16)/(3)(8pi-sqrt(3))`
C. `(16)/(3)(4pi-sqrt(3))`
D. none of these

Answer

To find the area common to the circle \( x^2 + y^2 = 64 \) and the parabola \( y^2 = 4x \), we will follow these steps:

### Step 1: Identify the equations
The equations given are:
1. Circle: \( x^2 + y^2 = 64 \)
2. Parabola: \( y^2 = 4x \)

### Step 2: Find points of intersection
To find the points where the circle and parabola intersect, substitute \( y^2 = 4x \) into the circle's equation:

\[ x^2 + 4x = 64 \]

Rearranging gives us:

\[ x^2 + 4x - 64 = 0 \]

### Step 3: Solve the quadratic equation
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

Here, \( a = 1, b = 4, c = -64 \).

Calculating the discriminant:

\[ b^2 - 4ac = 4^2 - 4 \cdot 1 \cdot (-64) = 16 + 256 = 272 \]

Now substituting into the quadratic formula:

\[ x = \frac{-4 \pm \sqrt{272}}{2 \cdot 1} = \frac{-4 \pm 4\sqrt{17}}{2} = -2 \pm 2\sqrt{17} \]

The points of intersection are:

\[ x_1 = -2 + 2\sqrt{17}, \quad x_2 = -2 - 2\sqrt{17} \]

Since \( x_2 \) is negative, we will only consider \( x_1 = -2 + 2\sqrt{17} \).

### Step 4: Calculate the area
The area common to both curves can be found by integrating the difference of the upper and lower curves from the left intersection point to the right intersection point.

1. The upper curve (circle) is given by \( y = \sqrt{64 - x^2} \).
2. The lower curve (parabola) is given by \( y = \sqrt{4x} \).

The area \( A \) is given by:

\[ A = 2 \int_{0}^{x_1} \left( \sqrt{64 - x^2} - \sqrt{4x} \right) dx \]

### Step 5: Evaluate the integral
We will split the integral into two parts:

1. **Integral of the circle**:
\[ \int \sqrt{64 - x^2} \, dx \]
This can be solved using the formula for the area of a circle segment.

2. **Integral of the parabola**:
\[ \int \sqrt{4x} \, dx = \int 2\sqrt{x} \, dx = \frac{4}{3} x^{3/2} \]

### Step 6: Calculate the definite integrals
Calculating the definite integrals from \( 0 \) to \( x_1 \):

1. For the circle:
\[ \int_{0}^{x_1} \sqrt{64 - x^2} \, dx \]
This can be computed using trigonometric substitution or recognized as a quarter circle.

2. For the parabola:
\[ \int_{0}^{x_1} 2\sqrt{x} \, dx = \left[ \frac{4}{3} x^{3/2} \right]_{0}^{x_1} = \frac{4}{3} (x_1)^{3/2} \]

### Step 7: Combine results
Finally, combine the results from the two integrals, multiply by 2 (since we have symmetry), and simplify to find the area.