Area (in sq. units) of the region bounded by the curve $y^2 = 4x$, $y$-axis and the line $y = 3$ is

The region is bounded by the curve $y^2 = 4x$, the $y$-axis, and the line $y = 3$.

The curve $y^2 = 4x$ is a parabola opening to the right.

Rewriting: $x = \frac{y^2}{4}$

For any $y$-value, $x$ equals $y$-squared divided by 4.

When $y = 0$: $x = 0$ (parabola passes through origin)

When $y = 2$: $x = \frac{4}{4} = 1$

When $y = 3$: $x = \frac{9}{4} = 2.25$

---

The region has:

Left boundary: $y$-axis ($x = 0$)

Right boundary: The parabola ($x = \frac{y^2}{4}$)

Top boundary: Line $y = 3$

Bottom boundary: Origin ($y = 0$)

The area is between the $y$-axis and the parabola, from $y = 0$ to $y = 3$.

---

Since $x$ is expressed in terms of $y$, integrate with respect to $y$:

$\text{Area} = \int_0^3 x \, dy$

$= \int_0^3 \frac{y^2}{4} \, dy$

$= \frac{1}{4} \int_0^3 y^2 \, dy$

$= \frac{1}{4} \left[\frac{y^3}{3}\right]_0^3$

$= \frac{1}{4} \times \frac{3^3}{3}$

$= \frac{1}{4} \times \frac{27}{3}$

$= \frac{1}{4} \times 9$

$= \frac{9}{4}$

---

Therefore, the area of the region is $\frac{9}{4}$ square units.