CUET UG 2025 — Mathematics Calculus

The curves are $3y^2 = ax$3y2=ax, $y = a$y=a, $a > 0$a>0, and the $y$y-axis.

The parabola $3y^2 = ax$3y2=ax can be rewritten as:

$$x = \frac{3y^2}{a}$$x=a3y2​

This parabola opens to the right and passes through the origin $(0, 0)$(0,0).

---

The bounded region has:

Left boundary: $y$y-axis where $x = 0$x=0

Right boundary: the parabola where $x = \frac{3y^2}{a}$x=a3y2​

Bottom: $y = 0$y=0

Top: the line $y = a$y=a

---

At any height $y$y, the width of the region is:

$$\frac{3y^2}{a} - 0 = \frac{3y^2}{a}$$a3y2​−0=a3y2​

The area is:

$$\text{Area} = \int_0^a \frac{3y^2}{a} \, dy$$Area=∫0a​a3y2​dy

$$= \frac{3}{a} \int_0^a y^2 \, dy$$=a3​∫0a​y2dy

$$= \frac{3}{a} \left[\frac{y^3}{3}\right]_0^a$$=a3​[3y3​]0a​

$$= \frac{3}{a} \left(\frac{a^3}{3} - 0\right)$$=a3​(3a3​−0)

$$= \frac{3}{a} \cdot \frac{a^3}{3}$$=a3​⋅3a3​

$$= \frac{a^3}{a}$$=aa3​

$$= a^2$$=a2

---

Therefore, the area of the region is $a^2$a2 square units.