Question

Area of the region \((x, y) : x^2 + (y - 2)^2 \leq 4, \, x^2 \geq 2y\) is:

• A. \( \frac{8}{3} \)
• B. \( 2\pi - \frac{16}{3} \)
• C. \( \pi - \frac{8}{3} \)
• D. \( \pi \)

Hint:

The area of a region bounded by a circle and a parabola can be calculated by solving their equations simultaneously and then using integration to find the area between the curves.

Answer 1

The Correct Option is C

We are given the equations of a circle and a parabola: \[ x^2 + (y - 2)^2 \leq 4 \quad \text{and} \quad x^2 \geq 2y \] The first equation represents a circle centered at \( (0, 2) \) with radius 2, and the second equation represents a parabola opening upwards.
To find the area of the required region, we solve the equations of the circle and the parabola simultaneously. From the circle equation: \[ x^2 + (y - 2)^2 = 4 \quad \Rightarrow \quad y = 2 \pm \sqrt{4 - x^2} \] From the parabola equation: \[ x^2 = 2y \quad \Rightarrow \quad y = \frac{x^2}{2} \] Now, equate the two expressions for \(y\): \[ 2 + \sqrt{4 - x^2} = \frac{x^2}{2} \] Solving for \(x\), we find the points of intersection as \( x = 2 \) and \( x = -2 \). Therefore, the region lies between \( x = -2 \) and \( x = 2 \). Now, the area is given by the integral: \[ \text{Area} = \int_{-2}^{2} \left( \sqrt{4 - x^2} - \frac{x^2}{2} \right) \, dx \] We can break this up into two separate integrals: \[ \text{Area} = \int_{-2}^{2} \sqrt{4 - x^2} \, dx - \int_{-2}^{2} \frac{x^2}{2} \, dx \] The first integral represents the area of the upper half of the circle, and the second integral is the area under the parabola.
We know that the area of the semicircle is \( \pi r^2 = \pi \times 2^2 / 2 = 2\pi \).
The second integral is a simple polynomial, which gives \( \frac{16}{3} \).
Thus, the total area is: \[ \text{Area} = 2\pi - \frac{16}{3} \] Thus, the required area is: \[ \boxed{\pi - \frac{8}{3}} \]