The area of the region \(\{(x, y) : x^2 + y^2 \le 2ax,\ y^2 \ge ax,\ x \ge 0,\ y \ge 0\}\) \(a \gt 0\), is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\left(\displaystyle\frac{\pi}{4} - \frac{2}{3}\right)a^2\) sq. units
\(x^2 + y^2 \le 2ax\) is the disc \((x - a)^2 + y^2 \le a^2\), centre (a, 0), radius a. In the first quadrant, \(y^2 \ge ax\) means \(y \ge \sqrt{ax}\), the points on or above the parabola.
The circle and the parabola meet where
\(x^2 + ax = 2ax\)
\(x(x - a) = 0\)
\(x = 0 \text{ or } x = a\)
For \(0 \le x \le a\), \(2ax - x^2 \ge ax\), so the circle is above the parabola. For \(x \gt a\), \(2ax - x^2 \lt ax\), and no point is inside the circle and above the parabola. So the region lies between the two curves from x = 0 to x = a:
\( \displaystyle \text{Area} = \int_{0}^{a}\sqrt{2ax - x^2}\,dx - \int_{0}^{a}\sqrt{ax}\,dx \)
The first integral is the part of the disc above the x-axis between x = 0 and the centre x = a, that is a quarter of the circle:
\( \displaystyle \int_{0}^{a}\sqrt{a^2 - (x - a)^2}\,dx = \frac{\pi a^2}{4} \)
The second integral:
\( \displaystyle \int_{0}^{a}\sqrt{a}\,x^{1/2}\,dx = \sqrt{a}\times\frac{2}{3}a^{3/2} = \frac{2a^2}{3} \)
So
\( \displaystyle \text{Area} = \frac{\pi a^2}{4} - \frac{2a^2}{3} = \left(\frac{\pi}{4} - \frac{2}{3}\right)a^2 \) sq. units