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, we will follow these steps:

### Step 1: Identify the curves
We have two inequalities:
1. \( y^2 \leq 6ax \) (which represents a parabola)
2. \( x^2 + y^2 \leq 16a^2 \) (which represents a circle)

### Step 2: Sketch the curves
- The equation \( y^2 = 6ax \) is a rightward-opening parabola with its vertex at the origin (0,0).
- The equation \( x^2 + y^2 = 16a^2 \) is a circle centered at the origin with a radius of \( 4a \).

### Step 3: Find the intersection points
To find the intersection points of the parabola and the circle, we substitute \( y^2 = 6ax \) into the circle's equation:

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

Rearranging gives us:

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

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

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

Calculating the discriminant:

\[ D = (6a)^2 - 4(1)(-16a^2) = 36a^2 + 64a^2 = 100a^2 \]

Now, substituting into the quadratic formula:

\[ x = \frac{-6a \pm 10a}{2} \]

This gives us:

1. \( x = 2a \) (valid intersection point)
2. \( x = -8a \) (not valid in the first quadrant)

### Step 5: Find corresponding y-values
Substituting \( x = 2a \) back into the parabola equation:

\[ y^2 = 6a(2a) = 12a^2 \implies y = \pm \sqrt{12a^2} = \pm 2\sqrt{3}a \]

Thus, the intersection points are \( (2a, 2\sqrt{3}a) \) and \( (2a, -2\sqrt{3}a) \).

### Step 6: Set up the area calculation
The area of the region can be calculated by finding the area under the parabola from \( x = 0 \) to \( x = 2a \) and the area under the circle from \( x = 2a \) to \( x = 4a \).

#### Area under the parabola:
\[ A_1 = \int_0^{2a} \sqrt{6ax} \, dx \]

#### Area under the circle:
The equation of the circle gives \( y = \sqrt{16a^2 - x^2} \), so:
\[ A_2 = \int_{2a}^{4a} \sqrt{16a^2 - x^2} \, dx \]

### Step 7: Calculate \( A_1 \)
Calculating \( A_1 \):
\[ A_1 = \int_0^{2a} \sqrt{6a} \sqrt{x} \, dx = \sqrt{6a} \int_0^{2a} x^{1/2} \, dx \]
\[ = \sqrt{6a} \left[ \frac{2}{3} x^{3/2} \right]_0^{2a} = \sqrt{6a} \cdot \frac{2}{3} (2a)^{3/2} \]
\[ = \frac{4\sqrt{6}}{3} a^2 \cdot 2\sqrt{2} = \frac{8\sqrt{12}}{3} a^2 = \frac{8\sqrt{3}}{3} a^2 \]

### Step 8: Calculate \( A_2 \)
Calculating \( A_2 \):
Using the formula for the area under a semicircle:
\[ A_2 = \int_{2a}^{4a} \sqrt{16a^2 - x^2} \, dx = \frac{1}{2} \left( \text{Area of the circle} \right) - \text{Area of the triangle} \]
The area of the circle is \( \pi (4a)^2 = 16\pi a^2 \).

### Step 9: Total Area
The total area \( A \) is given by:
\[ A = 2A_1 + A_2 \]

### Final Answer
The area of the region is:
\[ A = 2 \left( \frac{8\sqrt{3}}{3} a^2 \right) + A_2 \]