Find the area bounded between `y^2 ge 9x` and `x^2+y^2 le 36`

Answer

To find the area bounded between the curves \( y^2 \geq 9x \) (a parabola) and \( x^2 + y^2 \leq 36 \) (a circle), we will follow these steps:

### Step 1: Identify the curves
1. The equation \( y^2 = 9x \) represents a rightward-opening parabola.
2. The equation \( x^2 + y^2 = 36 \) represents a circle centered at the origin with a radius of 6.

### Step 2: Find the intersection points
To find the area bounded between these two curves, we first need to find their points of intersection. We substitute \( y^2 \) from the parabola into the circle's equation:

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

### Step 3: Solve the quadratic equation
Now we solve the quadratic equation \( x^2 + 9x - 36 = 0 \) using the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Here, \( a = 1, b = 9, c = -36 \):

\[ x = \frac{-9 \pm \sqrt{9^2 - 4 \cdot 1 \cdot (-36)}}{2 \cdot 1} \]
\[ = \frac{-9 \pm \sqrt{81 + 144}}{2} \]
\[ = \frac{-9 \pm \sqrt{225}}{2} \]
\[ = \frac{-9 \pm 15}{2} \]

Calculating the two possible values for \( x \):

1. \( x = \frac{6}{2} = 3 \)
2. \( x = \frac{-24}{2} = -12 \) (not relevant since we are looking for the first quadrant)

Thus, the intersection point in the first quadrant is \( (3, 0) \).

### Step 4: Set up the area integral
We will calculate the area in the first quadrant and then multiply by 2 for symmetry. The area can be calculated as:

\[ \text{Area} = \int_0^3 (y_{\text{circle}} - y_{\text{parabola}}) \, dx \]

Where:
- \( y_{\text{circle}} = \sqrt{36 - x^2} \)
- \( y_{\text{parabola}} = \sqrt{9x} \)

### Step 5: Calculate the area
The area becomes:

\[ \text{Area} = \int_0^3 \left( \sqrt{36 - x^2} - \sqrt{9x} \right) \, dx \]

#### Step 5.1: Calculate the integral for the circle
Using the formula for the area under a quarter circle:

\[ \int_0^3 \sqrt{36 - x^2} \, dx = \frac{1}{2} \left( 36 \cdot \frac{\pi}{2} \right) = 18\pi \]

#### Step 5.2: Calculate the integral for the parabola
Using the substitution:

\[ \int_0^3 \sqrt{9x} \, dx = \int_0^3 3\sqrt{x} \, dx = 3 \cdot \left[ \frac{2}{3} x^{3/2} \right]_0^3 = 3 \cdot \left[ \frac{2}{3} (3^{3/2}) \right] = 3 \cdot \left[ \frac{2}{3} \cdot 3\sqrt{3} \right] = 2\sqrt{3} \]

### Step 6: Combine the areas
Now we can combine the areas:

\[ \text{Area}_{\text{first quadrant}} = 18\pi - 2\sqrt{3} \]

### Step 7: Multiply by 2 for symmetry
Since the area is symmetric, we multiply by 2:

\[ \text{Total Area} = 2(18\pi - 2\sqrt{3}) = 36\pi - 4\sqrt{3} \]

### Final Answer
The area bounded between the curves \( y^2 \geq 9x \) and \( x^2 + y^2 \leq 36 \) is:

\[ \boxed{36\pi - 4\sqrt{3}} \]