Find the area of the circle `x^2+y^2=16 which is exterior to the parabola y^2=6x` by using integration.

Explanation

To find the area of the circle \( x^2 + y^2 = 16 \) that is exterior to the parabola \( y^2 = 6x \) using integration, we will follow these steps:

### Step 1: Identify the equations
The equation of the circle is:
\[ x^2 + y^2 = 16 \]
The equation of the parabola is:
\[ y^2 = 6x \]

### Step 2: Find the intersection points
To find the intersection points between the circle and the parabola, we can substitute \( y^2 \) from the parabola's equation into the circle's equation:
\[ x^2 + 6x = 16 \]
Rearranging gives:
\[ x^2 + 6x - 16 = 0 \]
Now we will solve this quadratic equation using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 1, b = 6, c = -16 \):
\[ x = \frac{-6 \pm \sqrt{6^2 - 4 \cdot 1 \cdot (-16)}}{2 \cdot 1} = \frac{-6 \pm \sqrt{36 + 64}}{2} = \frac{-6 \pm \sqrt{100}}{2} = \frac{-6 \pm 10}{2} \]
This gives us:
\[ x = 2 \quad \text{and} \quad x = -8 \]
Since \( x = -8 \) is not valid in the context of the parabola, we take \( x = 2 \).

### Step 3: Find the corresponding \( y \) values
Substituting \( x = 2 \) back into the parabola's equation:
\[ y^2 = 6 \cdot 2 = 12 \implies y = \pm 2\sqrt{3} \]
Thus, the intersection points are \( (2, 2\sqrt{3}) \) and \( (2, -2\sqrt{3}) \).

### Step 4: Set up the integrals for the area
We will find the area between the circle and the parabola from \( x = 0 \) to \( x = 2 \). The area \( A \) can be expressed as:
\[ A = 2 \left( \int_0^2 \sqrt{16 - x^2} \, dx - \int_0^2 \sqrt{6x} \, dx \right) \]
Here, \( \sqrt{16 - x^2} \) is the upper half of the circle and \( \sqrt{6x} \) is the upper half of the parabola.

### Step 5: Calculate the first integral
The integral \( \int_0^2 \sqrt{16 - x^2} \, dx \) can be solved using the formula for the area of a quarter circle:
\[ \int_0^2 \sqrt{16 - x^2} \, dx = \frac{1}{4} \pi r^2 = \frac{1}{4} \pi (4^2) = 4\pi \]

### Step 6: Calculate the second integral
Now, we calculate \( \int_0^2 \sqrt{6x} \, dx \):
\[ \int_0^2 \sqrt{6x} \, dx = \sqrt{6} \int_0^2 x^{1/2} \, dx = \sqrt{6} \left[ \frac{x^{3/2}}{3/2} \right]_0^2 = \sqrt{6} \cdot \frac{2^{3/2}}{3/2} = \frac{2\sqrt{6} \cdot 2\sqrt{2}}{3} = \frac{4\sqrt{12}}{3} = \frac{8\sqrt{3}}{3} \]

### Step 7: Combine the areas
Now substituting back into the area formula:
\[ A = 2 \left( 4\pi - \frac{8\sqrt{3}}{3} \right) = 8\pi - \frac{16\sqrt{3}}{3} \]

### Final Result
The area of the circle that is exterior to the parabola is:
\[ A = 8\pi - \frac{16\sqrt{3}}{3} \]