Question

The area of the region \(\{(x, y) : x^2 + y^2 \leq 2ax, y^2 > ax, x \geq 0, y \geq 0\} \text{ where } a > 0\) , is :

• A. \((\frac{\pi}{4}-\frac23)a^2 \) Sq.units
• B. \((\frac{\pi}{4}+\frac23)a^2 \) Sq.units
• C. \((\frac{\pi}{3}+\frac43)a^2 \) Sq.units
• D. \((\frac{\pi}{3}-\frac43)a^2 \) Sq.units

Answer 1

The Correct Option is A

We are given the region defined by the constraints:

• \(x^2 + y^2 \leq 2ax\)
• \(y^2 > ax\)
• \(x \geq 0\)
• \(y \geq 0\)

First, let's examine the equation \(x^2 + y^2 = 2ax\). This represents a circle with the following characteristics:

• Center at \((a, 0)\)
• Radius \(a\)

The condition \(x^2 + y^2 \leq 2ax\) means we are considering points inside or on this circle.

Next, consider the inequality \(y^2 > ax\). For a fixed \(x\), this represents the region above the parabola \(y = \sqrt{ax}\).

We aim to calculate the area of the region where both inequalities hold.

Let's analyze the geometric figures:

• The circle: Center at \((a, 0)\), radius \(a\).
• The parabola: Opens upwards, with \(x\) on the positive x-axis for \(y^2 = ax\).

We focus on the first quadrant \((x \geq 0, y \geq 0)\).

Intersection points between the circle \(x^2 + y^2 = 2ax\) and the parabola \(y^2 = ax\):

• Substitute \(y^2 = ax\) into \(x^2 + y^2 = 2ax\).
• This gives \(x^2 + ax = 2ax\), simplifying to \(x^2 = ax\), which leads to \(x(x-a) = 0\).
• Solutions: \(x = 0\) or \(x = a\).
• When \(x = a\), \(y^2 = ax = a^2\), so \(y = a\).

The curvilinear triangle formed by the region is bounded by:

• The sector from the circle in the first quadrant bounded from \(x = 0\) to \(x = a\).
• The area under the line \(y = \sqrt{ax}\), above \(x\)-axis from \(x = 0\) to \(x = a\).

For area calculations:

• Fraction of the circle's area: \(\frac{\pi a^2}{4}\) as it is a quarter circle
• Subtract the area under the parabola \(y = \sqrt{ax}\) for \(x = 0\) to \(x = a\):
• Integral of \(\sqrt{ax} \rightarrow \int_0^a\sqrt{ax} \, dx = \int_0^a ax^{0.5} \, dx = \frac{2}{3}a\left(x^{1.5}\right)\Big|_0^a = \frac{2}{3}a^2\)

Final area of the region \(= \frac{\pi a^2}{4} - \frac{2}{3}a^2 = \left(\frac{\pi}{4} - \frac{2}{3}\right)a^2\) square units.