Area of the region bounded by the curve $y^2 = 4x$, $y$-axis and the line $y = 3$ is equal to

The region is bounded by:

• $y^2 = 4x$: A parabola opening to the right
• $y$-axis: The line $x = 0$
• $y = 3$: A horizontal line at height 3

The region is a slice between the $y$-axis (left edge) and the parabola (right edge), from $y = 0$ to $y = 3$.

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From the parabola equation $y^2 = 4x$, we express $x$ in terms of $y$:

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

The region extends from $x = 0$ (the $y$-axis) to $x = \frac{y^2}{4}$ (the parabola) for each value of $y$ from $0$ to $3$.

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The area is given by:

$\text{Area} = \int_{0}^{3} \left(\frac{y^2}{4} - 0\right) dy$

$= \int_{0}^{3} \frac{y^2}{4} dy$

$= \frac{1}{4} \int_{0}^{3} y^2 \, dy$

$= \frac{1}{4} \left[\frac{y^3}{3}\right]_{0}^{3}$

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

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

$= \frac{9}{4}$

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