Using integration ,find the area of the region ` {(x,y) : y^(2) le 4x, 4x^(2) +4y^(2) le 9 } `

Answer

To find the area of the region defined by the inequalities \( y^2 \leq 4x \) and \( 4x^2 + 4y^2 \leq 9 \), we will follow these steps:

### Step 1: Identify the curves
1. The first inequality \( y^2 \leq 4x \) represents a parabola that opens to the right.
2. The second inequality \( 4x^2 + 4y^2 \leq 9 \) can be rewritten as \( x^2 + y^2 \leq \frac{9}{4} \), which represents a circle centered at the origin with a radius of \( \frac{3}{2} \).

### Step 2: Find the points of intersection
To find the area of the region, we first need to determine the points where these two curves intersect. We will set \( y^2 = 4x \) into the equation of the circle.

Substituting \( y^2 \) into the circle's equation:
\[ 4x^2 + 4(4x) \leq 9 \]
This simplifies to:
\[ 4x^2 + 16x - 9 = 0 \]
Dividing through by 4 gives:
\[ x^2 + 4x - \frac{9}{4} = 0 \]

### Step 3: Solve the quadratic equation
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1, b = 4, c = -\frac{9}{4} \):
\[ x = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot -\frac{9}{4}}}{2 \cdot 1} \]
\[ x = \frac{-4 \pm \sqrt{16 + 9}}{2} \]
\[ x = \frac{-4 \pm 5}{2} \]
Calculating the roots:
1. \( x = \frac{1}{2} \) (positive root)
2. \( x = -\frac{9}{2} \) (not relevant for our area calculation)

### Step 4: Determine the corresponding y-values
Using \( y^2 = 4x \):
For \( x = \frac{1}{2} \):
\[ y^2 = 4 \cdot \frac{1}{2} = 2 \implies y = \pm \sqrt{2} \]

### Step 5: Set up the integrals
The area can be found by integrating the upper curve minus the lower curve. The upper curve is given by the circle and the lower curve by the parabola.

1. From \( x = 0 \) to \( x = \frac{1}{2} \), the upper curve is \( y = 2\sqrt{x} \).
2. From \( x = \frac{1}{2} \) to \( x = \frac{3}{2} \), the upper curve is from the circle \( y = \sqrt{\frac{9}{4} - x^2} \).

### Step 6: Calculate the area
The total area \( A \) is given by:
\[ A = 2 \left( \int_0^{\frac{1}{2}} 2\sqrt{x} \, dx + \int_{\frac{1}{2}}^{\frac{3}{2}} \sqrt{\frac{9}{4} - x^2} \, dx \right) \]

### Step 7: Evaluate the integrals
1. For the first integral:
\[ \int 2\sqrt{x} \, dx = \frac{4}{3} x^{3/2} \bigg|_0^{\frac{1}{2}} = \frac{4}{3} \left( \frac{1}{2} \right)^{3/2} = \frac{4}{3} \cdot \frac{\sqrt{2}}{4} = \frac{\sqrt{2}}{3} \]

2. For the second integral, we use the formula for the area under a semicircle:
\[ \int \sqrt{a^2 - x^2} \, dx = \frac{x}{2} \sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1} \left( \frac{x}{a} \right) \]
Here, \( a = \frac{3}{2} \):
\[ \text{Evaluate from } \frac{1}{2} \text{ to } \frac{3}{2} \]

### Step 8: Combine results
Finally, we combine the results of the integrals to find the total area.