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

The correct answer is Option (1) → (A)-(IV), (B)-(III), (C)-(II), (D)-(I)

Each region lies between a curve x = g(y) and the y-axis (or a vertical line), so it is found by integrating in y.

(A) Region bounded by the curve \(x = 4-y^{2}\) and the y-axis

This matches (IV) \(\displaystyle\frac{32}{3}\).

The curve meets the y-axis where \(4-y^{2} = 0\), that is y = ±2:

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

    \( \displaystyle = \frac{16}{3}-\left(-\frac{16}{3}\right) = \frac{32}{3} \)

(B) Region bounded by the line x = 2y, the y-axis and the line y = 3

This matches (III) 9.

The line meets the y-axis at the origin, so y runs from 0 to 3:

    \( \displaystyle A = \int_{0}^{3}2y\,dy = \left[y^{2}\right]_{0}^{3} = 9 \)

(C) Region bounded by the curve \(x = 3y-y^{2}\) and the y-axis

This matches (II) \(\displaystyle\frac{9}{2}\).

Since \(3y-y^{2} = y(3-y)\), the curve meets the y-axis at y = 0 and y = 3 and lies to the right of it in between:

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

    \( \displaystyle = \frac{27}{2}-9 = \frac{9}{2} \)

(D) Region bounded by the curve x = |y| and the line x = 2

This matches (I) 4.

The curve meets x = 2 at y = ±2. For each y the width of the region is \(2-|y|\), and the region is symmetric about the x-axis:

    \( \displaystyle A = 2\int_{0}^{2}(2-y)\,dy = 2\left[2y-\frac{y^{2}}{2}\right]_{0}^{2} \)

    \(= 2(4-2) = 4\)