Draw a rough sketch of the following region and find the area enclosed by the region, using method of integration :
`{(x,y): y ^(2) le 5x, 5x ^(2) + 5y ^(2)le 36}.`

Answer

To find the area enclosed by the region defined by the inequalities \( y^2 \leq 5x \) and \( 5x^2 + 5y^2 \leq 36 \), we will follow these steps:

### Step 1: Understand the curves
1. **Curve 1:** \( y^2 = 5x \) represents a rightward-opening parabola.
2. **Curve 2:** \( 5x^2 + 5y^2 = 36 \) can be simplified to \( x^2 + y^2 = \frac{36}{5} \), which is a circle centered at the origin with radius \( r = \sqrt{\frac{36}{5}} = \frac{6}{\sqrt{5}} \).

### Step 2: Sketch the region
- Draw the parabola \( y^2 = 5x \).
- Draw the circle \( x^2 + y^2 = \frac{36}{5} \).
- Identify the region where both inequalities hold true.

### Step 3: Find the points of intersection
To find the points where the curves intersect, we set \( y^2 = 5x \) into the circle's equation:

\[ x^2 + 5x = \frac{36}{5} \]

Rearranging gives:

\[ 5x^2 + 5x - 36 = 0 \]

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

Here, \( a = 5 \), \( b = 5 \), and \( c = -36 \):

\[ x = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 5 \cdot (-36)}}{2 \cdot 5} \]
\[ x = \frac{-5 \pm \sqrt{25 + 720}}{10} \]
\[ x = \frac{-5 \pm \sqrt{745}}{10} \]

Calculating \( \sqrt{745} \approx 27.34 \):

\[ x \approx \frac{-5 + 27.34}{10} \approx 2.234 \quad \text{(taking the positive root)} \]

### Step 5: Find corresponding y-values
Substituting \( x \approx 2.234 \) back into \( y^2 = 5x \):

\[ y^2 = 5(2.234) \approx 11.17 \implies y \approx \pm \sqrt{11.17} \approx \pm 3.34 \]

Thus, the points of intersection are approximately \( (2.234, 3.34) \) and \( (2.234, -3.34) \).

### Step 6: Set up the integral for area
The area can be calculated by integrating the difference between the upper and lower curves from the leftmost intersection point to the rightmost intersection point. The area \( A \) is given by:

\[ A = 2 \int_0^{2.234} \left( \sqrt{5x} - \sqrt{\frac{36}{5} - x^2} \right) dx \]

### Step 7: Evaluate the integral
1. **Integral of \( \sqrt{5x} \)**:
\[ \int \sqrt{5x} \, dx = \frac{2}{3} \cdot \sqrt{5} \cdot x^{3/2} \]

2. **Integral of \( \sqrt{\frac{36}{5} - x^2} \)**:
Using the formula for the area of a circle segment, we can evaluate this integral.

### Step 8: Calculate the area
After evaluating both integrals, substitute the limits and simplify to find the total area.

### Final Area Calculation
The final area will be a numerical value after evaluating the integrals.

---