Find the area of the region lying in the first quadrant and included between the curves
`x^(2)+y^(2)=3a^(2).x^(2)=2ay` and `y^(2)=2ax.a gt0`

Answer

To find the area of the region lying in the first quadrant and included between the curves \( x^2 + y^2 = 3a^2 \), \( x^2 = 2ay \), and \( y^2 = 2ax \), we will follow these steps:

### Step 1: Identify the Curves
1. **Circle**: The equation \( x^2 + y^2 = 3a^2 \) represents a circle centered at the origin with radius \( \sqrt{3}a \).
2. **Parabola 1**: The equation \( x^2 = 2ay \) represents a parabola opening upwards.
3. **Parabola 2**: The equation \( y^2 = 2ax \) represents a parabola opening to the right.

### Step 2: Find Points of Intersection
We need to find the points where these curves intersect in the first quadrant.

#### Intersection of Circle and Parabola \( y^2 = 2ax \):
Substituting \( y^2 = 2ax \) into the circle's equation:
\[ x^2 + 2ax = 3a^2 \]
Rearranging gives:
\[ x^2 + 2ax - 3a^2 = 0 \]
This is a quadratic equation in \( x \). Using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-2a \pm \sqrt{(2a)^2 + 4 \cdot 3a^2}}{2} \]
\[ x = \frac{-2a \pm \sqrt{4a^2 + 12a^2}}{2} = \frac{-2a \pm 4a}{2} \]
This gives:
\[ x = a \quad \text{(valid in the first quadrant)} \quad \text{and} \quad x = -3a \quad \text{(not valid)} \]
Now substituting \( x = a \) back into \( y^2 = 2ax \):
\[ y^2 = 2a(a) = 2a^2 \implies y = \sqrt{2}a \]
Thus, the point of intersection is \( (a, \sqrt{2}a) \).

#### Intersection of Circle and Parabola \( x^2 = 2ay \):
Substituting \( x^2 = 2ay \) into the circle's equation:
\[ 2ay + y^2 = 3a^2 \]
Rearranging gives:
\[ y^2 + 2ay - 3a^2 = 0 \]
Using the quadratic formula:
\[ y = \frac{-2a \pm \sqrt{(2a)^2 + 4 \cdot 3a^2}}{2} = \frac{-2a \pm 4a}{2} \]
This gives:
\[ y = a \quad \text{(valid in the first quadrant)} \quad \text{and} \quad y = -3a \quad \text{(not valid)} \]
Now substituting \( y = a \) back into \( x^2 = 2ay \):
\[ x^2 = 2a(a) = 2a^2 \implies x = \sqrt{2}a \]
Thus, the point of intersection is \( (\sqrt{2}a, a) \).

### Step 3: Calculate Areas
We need to find the area between the curves from \( x = a \) to \( x = \sqrt{2}a \).

#### Area under the Circle:
The equation of the circle can be rewritten as:
\[ y = \sqrt{3a^2 - x^2} \]
The area under the circle from \( x = a \) to \( x = \sqrt{2}a \) is given by:
\[ \text{Area}_{\text{circle}} = \int_a^{\sqrt{2}a} \sqrt{3a^2 - x^2} \, dx \]

#### Area under the Parabola \( x^2 = 2ay \):
The equation can be rewritten as:
\[ y = \frac{x^2}{2a} \]
The area under this parabola from \( x = a \) to \( x = \sqrt{2}a \) is given by:
\[ \text{Area}_{\text{parabola}} = \int_a^{\sqrt{2}a} \frac{x^2}{2a} \, dx \]

### Step 4: Total Area Calculation
The total area between the curves is:
\[ \text{Total Area} = \text{Area}_{\text{circle}} - \text{Area}_{\text{parabola}} \]

### Step 5: Evaluate the Integrals
1. **Area under the Circle**:
\[ \int_a^{\sqrt{2}a} \sqrt{3a^2 - x^2} \, dx \]
This can be solved using trigonometric substitution or standard integral formulas.

2. **Area under the Parabola**:
\[ \int_a^{\sqrt{2}a} \frac{x^2}{2a} \, dx = \frac{1}{2a} \left[ \frac{x^3}{3} \right]_a^{\sqrt{2}a} \]

### Final Result
After evaluating both integrals and subtracting, we will arrive at the final area of the region lying in the first quadrant between the curves.