Draw a rough sketch of the region `{(x,y) : y^(2)le 6` a x and `x^(2)+y^(2)le 16 a^(2)}`. Also, find the area of the region sketched using method of integration.

Answer

To solve the problem step by step, we will first sketch the region defined by the inequalities and then calculate the area using integration.

### Step 1: Understand the equations

We have two inequalities:
1. \( y^2 \leq 6x \) (which represents a parabola)
2. \( x^2 + y^2 \leq 16a^2 \) (which represents a circle)

### Step 2: Sketch the curves

1. **Parabola**: The equation \( y^2 = 6x \) can be rewritten as \( y = \pm \sqrt{6x} \). This parabola opens to the right and intersects the x-axis at the origin (0,0).

2. **Circle**: The equation \( x^2 + y^2 = 16a^2 \) represents a circle with a radius of \( 4a \) centered at the origin (0,0).

### Step 3: Find points of intersection

To find the points where the parabola intersects the circle, we substitute \( y^2 = 6x \) into the circle's equation:

\[ x^2 + 6x = 16a^2 \]

This simplifies to:

\[ x^2 + 6x - 16a^2 = 0 \]

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

Here, \( a = 1, b = 6, c = -16a^2 \):

\[ x = \frac{-6 \pm \sqrt{6^2 - 4 \cdot 1 \cdot (-16a^2)}}{2 \cdot 1} \]
\[ x = \frac{-6 \pm \sqrt{36 + 64a^2}}{2} \]
\[ x = \frac{-6 \pm \sqrt{36 + 64a^2}}{2} \]

### Step 4: Calculate the area

The area between the curves can be found by integrating the difference of the two functions from the left intersection point to the right intersection point.

1. **Area under the parabola** from \( x = 0 \) to \( x = 2a \):

\[ \text{Area}_\text{parabola} = \int_0^{2a} \sqrt{6x} \, dx \]

2. **Area under the circle** from \( x = 2a \) to \( x = 4a \):

\[ \text{Area}_\text{circle} = \int_{2a}^{4a} \sqrt{16a^2 - x^2} \, dx \]

### Step 5: Evaluate the integrals

1. **Integrating the parabola**:

\[ \int_0^{2a} \sqrt{6x} \, dx = \int_0^{2a} \sqrt{6} \cdot x^{1/2} \, dx = \sqrt{6} \cdot \left[ \frac{2}{3} x^{3/2} \right]_0^{2a} = \sqrt{6} \cdot \frac{2}{3} (2a)^{3/2} \]

2. **Integrating the circle**:

Using the formula for the area of a circle segment:

\[ \int_{2a}^{4a} \sqrt{16a^2 - x^2} \, dx = \text{Area of segment} = \frac{1}{2} \left( 16a^2 \cdot \theta - 16a^2 \cdot \sin(\theta) \right) \]

Where \( \theta \) is the angle corresponding to the arc.

### Final Area Calculation

Combining both areas gives the total area of the region bounded by the parabola and the circle.