Match List-I with List-II List-I (Region)List-II (Area in sq. units)…

The correct answer is Option (3) → (A)-(IV), (B)-(III), (C)-(I), (D)-(II)

In each item the line is perpendicular to the axis of the parabola, so the region is symmetric about that axis and is found by one integral across it.

(A) Region bounded by the parabola \(y^{2} = 3x\) and the line x = 3

This matches (IV) 12.

The halves are \(y = \pm\sqrt{3}\,\sqrt{x}\), for x from 0 to 3:

    \( \displaystyle A = 2\sqrt{3}\int_{0}^{3}x^{1/2}\,dx = 2\sqrt{3}\times\frac{2}{3}\times3\sqrt{3} = 12 \)

(B) Region bounded by the parabola \(x^{2} = 2y\) and the line y = 2

This matches (III) \(\displaystyle\frac{16}{3}\).

The parabola is \(\displaystyle y = \frac{x^{2}}{2}\), which meets y = 2 at x = ±2, and the region lies under the line:

    \( \displaystyle A = \int_{-2}^{2}\left(2-\frac{x^{2}}{2}\right)dx = 2\left[2x-\frac{x^{3}}{6}\right]_{0}^{2} = 2\left(4-\frac{4}{3}\right) = \frac{16}{3} \)

(C) Region bounded by the parabola \(y^{2} = -9x\) and the line x = −1

This matches (I) 4.

The parabola opens to the left, \(\displaystyle x = -\frac{y^{2}}{9}\), and meets x = −1 at y = ±3. For each y the width from the line to the curve is \(\displaystyle 1-\frac{y^{2}}{9}\):

    \( \displaystyle A = \int_{-3}^{3}\left(1-\frac{y^{2}}{9}\right)dy = 2\left[y-\frac{y^{3}}{27}\right]_{0}^{3} = 2(3-1) = 4 \)

(D) Region bounded by the parabola \(x^{2} = -y\) and the line y = −9

This matches (II) 36.

The parabola opens downwards, \(y = -x^{2}\), and meets y = −9 at x = ±3. The height from the line up to the curve is \(-x^{2}-(-9) = 9-x^{2}\):

    \( \displaystyle A = \int_{-3}^{3}(9-x^{2})\,dx = 2\left[9x-\frac{x^{3}}{3}\right]_{0}^{3} = 2(27-9) = 36 \)