Question

The area (in square units) of the smaller region lying above the X-axis and bounded between the circle \[ x^2 + y^2 = 2ax \] and the parabola \[ y^2 = ax \]

• A. \( 2a^2 \left(\frac{\pi}{4} - \frac{2}{3} \right) \)
• B. \( a^2 \left(\frac{\pi}{4} - \frac{2}{3} \right) \)
• C. \( a^2 \left(\frac{\pi}{4} + \frac{2}{3} \right) \)
• D. \( a^2 \left(\frac{\pi^2}{4} - \frac{1}{3} \right) \)

Hint:

When finding the area enclosed between curves, always check their points of intersection and integrate the difference of the upper and lower functions.

Answer 1

The Correct Option is B

We are given two curves:
\[ x^2 + y^2 = 2ax \quad \text{(Circle)} \]
\[ y^2 = ax \quad \text{(Parabola)} \]
We need to find the area of the smaller region above the x-axis bounded by these curves.

Step 1: Rearranging the Equations
Circle Equation:
\[ x^2 + y^2 = 2ax \]
Rearranging,
\[ x^2 - 2ax + y^2 = 0 \]
Completing the square for the \(x\)-terms,
\[ (x - a)^2 + y^2 = a^2 \]
This is a circle with center \( (a, 0) \) and radius \( a \).

 Parabola Equation:
The given parabola is:
\[ y^2 = ax \]

Step 2: Intersection Points
Equating the circle and parabola:
\[ (x - a)^2 + y^2 = a^2 \]
Since \(y^2 = ax\),
\[ (x - a)^2 + ax = a^2 \]
Expanding the square:
\[ x^2 - 2ax + a^2 + ax = a^2 \]
\[ x^2 - ax = 0 \]
\[ x(x - a) = 0 \]
Thus, \(x = 0\) or \(x = a\).

Step 3: Area Between the Curves
The area bounded between two curves is given by:
\[ \text{Area} = \int_{0}^{a} [\text{Upper Curve} - \text{Lower Curve}] \, dx \]
From the given equations:
- Upper curve (circle) → \(y = \sqrt{a^2 - (x - a)^2} \)
- Lower curve (parabola) → \(y = \sqrt{ax} \)
\[ \text{Area} = \int_{0}^{a} \left[ \sqrt{a^2 - (x - a)^2} - \sqrt{ax} \right] \, dx \]

Step 4: Evaluating the Integrals
1. Integral for the circle:
\[ \int_{0}^{a} \sqrt{a^2 - (x - a)^2} \, dx = \frac{a^2\pi}{4} \]
2. Integral for the parabola:
\[ \int_{0}^{a} \sqrt{ax} \, dx = \int_{0}^{a} \sqrt{a} \sqrt{x} \, dx = a^{1/2} \int_{0}^{a} x^{1/2} \, dx = a^{1/2} \left[ \frac{2}{3} x^{3/2} \right]_{0}^{a} = a^{1/2} \cdot \frac{2}{3} a^{3/2} = \frac{2}{3} a^2 \]

Step 5: Computing the Final Area
\[ \text{Area} = \frac{a^2 \pi}{4} - \frac{2a^2}{3} \]
\[ = a^2 \left( \frac{\pi}{4} - \frac{2}{3} \right) \]

Step 6: Final Answer
\[ \boxed{a^2 \left( \frac{\pi}{4} - \frac{2}{3} \right)} \]

Final Answer: (B) \( a^2 \left( \frac{\pi}{4} - \frac{2}{3} \right) \)