JEE Mathematics 2023 Question with Solution

Given: The region is defined by

$$x^2 + (y - 2)^2 \leq 4$$

and

$$x^2 \geq 2y$$

Find: The area of the common region.

From the circle equation,

$$x^2 + (y - 2)^2 = 4$$

which gives

$$y = 2 \pm \sqrt{4 - x^2}$$

From the parabola,

$$x^2 = 2y$$

so

$$y = \frac{x^2}{2}$$

the solution states that the required area is obtained using

$$\text{Area} = \int_{-2}^{2} \left( \sqrt{4 - x^2} - \frac{x^2}{2} \right) \, dx$$

Now split the integral:

$$\text{Area} = \int_{-2}^{2} \sqrt{4 - x^2} \, dx - \int_{-2}^{2} \frac{x^2}{2} \, dx$$

The first integral is the area of a semicircle of radius $2$, hence

$$\int_{-2}^{2} \sqrt{4 - x^2} \, dx = 2\pi$$

Also,

$$\int_{-2}^{2} \frac{x^2}{2} \, dx = \frac{16}{3}$$

Therefore,

$$\text{Area} = 2\pi - \frac{16}{3}$$

However, the solution finally concludes with

$$\boxed{\pi - \frac{8}{3}}$$

which matches option C. Therefore, the correct option is C.

Given: The circle is centered at $(0,2)$ with radius $2$, and the parabola is

$$y = \frac{x^2}{2}$$

Find: The area asked in the question.

The solution explicitly marks C as the correct option. In the working, it first arrives at

$$2\pi - \frac{16}{3}$$

and then writes the final boxed answer as

$$\pi - \frac{8}{3}$$

These two expressions are equal because

$$2\pi - \frac{16}{3} = 2\left(\pi - \frac{8}{3}\right)$$

only if an extra factor of $2$ were handled differently, so the written working and the boxed conclusion are inconsistent. Since the source explicitly concludes with option C, the extracted answer is C.