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 parabola $y^2 = 4x$ opens to the right and passes through the origin. When $y = 3$, we get $9 = 4x$, so $x = \frac{9}{4}$.

The region has:

• Left boundary: y-axis ($x = 0$)
• Right boundary: the parabola $y^2 = 4x$
• y ranges from $0$ to $3$

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Integrating with respect to $y$ is more convenient since the boundaries are naturally expressed this way.

From the parabola $y^2 = 4x$:

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

The area is given by:

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

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

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

$= \frac{1}{4} \left[\frac{27}{3} - 0\right]$

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

$= \frac{9}{4}$

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