The area ( in sq, units ) of the part of the circle `x^2 +y^2 = 36` , which is outside the parabola ` y^2 = 9x ,` is

Options

A. `24 pi + 3 sqrt(3)`
B. ` 12 pi - 3 sqrt(3)`
C. ` 24 pi - 3 sqrt(3)`
D. `12 pi + 3 sqrt(3)`

Answer

To find the area of the part of the circle \( x^2 + y^2 = 36 \) that is outside the parabola \( y^2 = 9x \), we can follow these steps:

### Step 1: Identify the equations and their graphs
The equation of the circle is \( x^2 + y^2 = 36 \), which has a center at the origin (0, 0) and a radius of 6 units. The parabola \( y^2 = 9x \) opens to the right.

### Step 2: Find the points of intersection
To find the area outside the parabola but inside the circle, we first need to find the points where the two curves intersect.

Substituting \( y^2 = 9x \) into the circle's equation:

\[ x^2 + 9x - 36 = 0 \]

This is a quadratic equation in \( x \). We can solve it using the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-9 \pm \sqrt{9^2 - 4 \cdot 1 \cdot (-36)}}{2 \cdot 1} \]

Calculating the discriminant:

\[ b^2 - 4ac = 81 + 144 = 225 \]

Now substituting back into the formula:

\[ x = \frac{-9 \pm 15}{2} \]

This gives us two solutions:

\[ x = 3 \quad \text{and} \quad x = -12 \]

Since we are only interested in the positive intersection point (as we are looking at the first quadrant), we have \( x = 3 \).

Now substituting \( x = 3 \) back into the parabola to find \( y \):

\[ y^2 = 9 \cdot 3 = 27 \implies y = 3\sqrt{3} \]

Thus, the points of intersection are \( (3, 3\sqrt{3}) \) and \( (3, -3\sqrt{3}) \).

### Step 3: Set up the integral for the area
We will calculate the area of the region bounded by the parabola and the circle. The area can be calculated by integrating the difference between the circle and the parabola from \( x = 0 \) to \( x = 3 \) and then from \( x = 3 \) to \( x = 6 \) (the rightmost point of the circle).

1. **Area under the parabola from \( x = 0 \) to \( x = 3 \)**:

\[ \text{Area}_{\text{parabola}} = \int_0^3 \sqrt{9x} \, dx = \int_0^3 3\sqrt{x} \, dx \]

Calculating this integral:

\[ = 3 \cdot \left[ \frac{2}{3} x^{3/2} \right]_0^3 = 3 \cdot \left[ \frac{2}{3} (3^{3/2}) - 0 \right] = 3 \cdot \frac{2}{3} \cdot 3\sqrt{3} = 6\sqrt{3} \]

2. **Area under the circle from \( x = 3 \) to \( x = 6 \)**:

For the circle, we have:

\[ y = \sqrt{36 - x^2} \]

So, the area under the circle is:

\[ \text{Area}_{\text{circle}} = \int_3^6 \sqrt{36 - x^2} \, dx \]

Using the formula for the area of a circle segment, we can calculate this integral. The area of the quarter circle is:

\[ \frac{1}{4} \cdot \pi \cdot r^2 = \frac{1}{4} \cdot \pi \cdot 36 = 9\pi \]

However, we need to subtract the area from \( x = 3 \) to \( x = 6 \).

### Step 4: Calculate the total area
The total area \( A \) outside the parabola and inside the circle is given by:

\[ A = \text{Area}_{\text{circle}} - \text{Area}_{\text{parabola}} = 9\pi - 6\sqrt{3} \]

### Final Answer
Thus, the area of the part of the circle that is outside the parabola is:

\[ \text{Area} = 36\pi - 6\sqrt{3} \]