The area common to the circle `x^2 + y^2 = 64` and the parabola `y^2 = 12x` is equal to

Answer

To find the area common to the circle \( x^2 + y^2 = 64 \) and the parabola \( y^2 = 12x \), we will follow these steps:

### Step 1: Identify the equations
The given equations are:
1. Circle: \( x^2 + y^2 = 64 \)
2. Parabola: \( y^2 = 12x \)

### Step 2: Find the points of intersection
To find the points of intersection, substitute \( y^2 \) from the parabola into the circle's equation:
\[ x^2 + 12x = 64 \]
Rearranging gives:
\[ x^2 + 12x - 64 = 0 \]
Now, we can solve this quadratic equation using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 1, b = 12, c = -64 \):
\[ x = \frac{-12 \pm \sqrt{12^2 - 4 \cdot 1 \cdot (-64)}}{2 \cdot 1} \]
\[ x = \frac{-12 \pm \sqrt{144 + 256}}{2} \]
\[ x = \frac{-12 \pm \sqrt{400}}{2} \]
\[ x = \frac{-12 \pm 20}{2} \]
Calculating the two possible values:
1. \( x = \frac{8}{2} = 4 \)
2. \( x = \frac{-32}{2} = -16 \)

### Step 3: Determine the limits for integration
The relevant intersection point for our area calculation is \( x = 4 \) (since we are interested in the area in the first quadrant). The limits for integration will be from \( x = 0 \) to \( x = 4 \).

### Step 4: Set up the area integral
The area common to both curves can be calculated by integrating the upper curve minus the lower curve. The upper curve from \( x = 0 \) to \( x = 4 \) is the parabola, and from \( x = 4 \) to \( x = 8 \) is the circle.

1. For the parabola \( y = \sqrt{12x} \):
\[ \text{Area}_1 = \int_0^4 \sqrt{12x} \, dx \]

2. For the circle \( y = \sqrt{64 - x^2} \):
\[ \text{Area}_2 = \int_4^8 \sqrt{64 - x^2} \, dx \]

### Step 5: Calculate the integrals
**Integral for the parabola:**
\[ \text{Area}_1 = \int_0^4 \sqrt{12x} \, dx = \int_0^4 2\sqrt{3} \sqrt{x} \, dx = 2\sqrt{3} \cdot \frac{2}{3} x^{3/2} \bigg|_0^4 \]
\[ = \frac{4\sqrt{3}}{3} (4^{3/2} - 0) = \frac{4\sqrt{3}}{3} \cdot 8 = \frac{32\sqrt{3}}{3} \]

**Integral for the circle:**
\[ \text{Area}_2 = \int_4^8 \sqrt{64 - x^2} \, dx \]
Using the formula for the area of a circle segment:
\[ = \frac{1}{2} \left( 64 \sin^{-1}\left(\frac{x}{8}\right) + x\sqrt{64 - x^2} \right) \bigg|_4^8 \]
Calculating this gives:
\[ = \frac{1}{2} \left( 64 \left(\frac{\pi}{2}\right) + 8 \cdot 0 - \left( 64 \left(\frac{\pi}{6}\right) + 4\sqrt{48} \right) \right) \]
\[ = 32\pi - \left( \frac{32\pi}{3} + 16\sqrt{3} \right) = 32\pi - \frac{32\pi}{3} - 16\sqrt{3} \]
\[ = \frac{96\pi - 32\pi}{3} - 16\sqrt{3} = \frac{64\pi}{3} - 16\sqrt{3} \]

### Step 6: Combine the areas
The total area common to both curves is:
\[ \text{Total Area} = \text{Area}_1 + \text{Area}_2 = \frac{32\sqrt{3}}{3} + \left( \frac{64\pi}{3} - 16\sqrt{3} \right) \]
\[ = \frac{32\sqrt{3}}{3} - \frac{48\sqrt{3}}{3} + \frac{64\pi}{3} = \frac{64\pi - 16\sqrt{3}}{3} \]

### Final Answer
The area common to the circle and the parabola is:
\[ \frac{64\pi - 16\sqrt{3}}{3} \text{ square units} \]