The area of the region lying above x - axis, and included between the circle `x^(2)+y^(2)=2ax` & the parabola `y^(2)=ax, a gt 0` is

Options

A. `8pia^(2)`
B. `a^(2)((pi)/(4)-(2)/(3))`
C. `(16pia^(2))/(9)`
D. `pi((27)/(8)+3a^(2))`

Answer

To find the area of the region lying above the x-axis and included between the circle \(x^2 + y^2 = 2ax\) and the parabola \(y^2 = ax\) (where \(a > 0\)), we can follow these steps:

### Step 1: Rewrite the equations
1. **Circle**: The equation \(x^2 + y^2 = 2ax\) can be rewritten as:
\[ (x - a)^2 + y^2 = a^2 \]
This represents a circle with center at \((a, 0)\) and radius \(a\).

2. **Parabola**: The equation \(y^2 = ax\) can be rewritten as:
\[ y = \sqrt{ax} \quad \text{(considering only the positive root since we are above the x-axis)} \]

### Step 2: Find the points of intersection
To find the points of intersection between the circle and the parabola, we substitute \(y^2 = ax\) into the circle's equation:
\[ x^2 + ax = 2ax \]
This simplifies to:
\[ x^2 - ax = 0 \]
Factoring gives:
\[ x(x - a) = 0 \]
Thus, the solutions are \(x = 0\) and \(x = a\).

### Step 3: Set up the area calculation
The area we want to find is the area above the x-axis between \(x = 0\) and \(x = a\). This area can be calculated as:
\[ \text{Area} = \text{Area of the quarter circle} - \text{Area under the parabola} \]

### Step 4: Calculate the area of the quarter circle
The area of a full circle is \(\pi r^2\). For our quarter circle:
\[ \text{Area of quarter circle} = \frac{1}{4} \pi a^2 \]

### Step 5: Calculate the area under the parabola
To find the area under the parabola from \(x = 0\) to \(x = a\), we use the integral:
\[ \text{Area under the parabola} = \int_0^a \sqrt{ax} \, dx \]
Substituting \(y = \sqrt{ax}\):
\[ = \sqrt{a} \int_0^a x^{1/2} \, dx \]
The integral of \(x^{1/2}\) is:
\[ \int x^{1/2} \, dx = \frac{x^{3/2}}{3/2} = \frac{2}{3} x^{3/2} \]
Evaluating from \(0\) to \(a\):
\[ = \sqrt{a} \cdot \left[ \frac{2}{3} a^{3/2} - 0 \right] = \frac{2}{3} a^2 \]

### Step 6: Combine the areas
Now we can find the required area:
\[ \text{Required Area} = \frac{1}{4} \pi a^2 - \frac{2}{3} a^2 \]
To combine these, we need a common denominator:
\[ = \frac{3\pi a^2}{12} - \frac{8 a^2}{12} = \frac{(3\pi - 8) a^2}{12} \]

### Final Result
Thus, the area of the region lying above the x-axis and included between the circle and the parabola is:
\[ \text{Area} = \frac{(3\pi - 8) a^2}{12} \]

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