Q.Find the value of the following: Area of the region bounded by the curve $y^2 = 4x$, y-axis and the line $y = 3$ is (A) 2 (B) $\frac{9}{4}$ (C) $\frac{9}{3}$ (D) $\frac{9}{2}$

Concept: Area Under Parabola (integrating with respect to $y$).

Step 1: The curve $y^2 = 4x$ is a right-opening parabola. The region is bounded by the y-axis ($x=0$), the horizontal line $y=3$, and the parabola.

Step 2: Rewrite the parabola as $x = \frac{y^2}{4}$. The area is the horizontal strip area between $x=0$ and $x = y^2/4$, from $y=0$ to $y=3$.

Step 3: Integrate: …

When a curve is given as $y^2 = 4x$, the natural instinct is to solve for $y$ and integrate with respect to $x$. But here, the boundaries are the y-axis ($x = 0$) and the horizontal line $y = 3$. The region is bounded on the left by the y-axis, on the top by $y = 3$, and on the right by the parabola. If you try to integrate with respect to $x$, you'd have to split the region because the parabola gives two $y$ values for each $x$ — messy and unnecessary.

The cleaner approach: treat $x$ as a function of $y$. The parabola $y^2 = 4x$ can be rewritten as $x = \frac{y^2}{4}$. Now, for a given $y$, the horizontal distance from the y-axis to the curve is exactly $x(y)$. The region runs from $y = 0$ (the vertex of the parabola) to $y = 3$ (the given line). So the area is simply the integral of $x$ with respect to $y$ over that interval.

1. Rewrite the curve in terms of $y$.

   From $y^2 = 4x$, we get $x = \frac{y^2}{4}$. This expresses the horizontal distance from the y-axis to the parabola at a given $y$.

2. Set up the integral for area.

   The area between the y-axis (left boundary) and the curve (right boundary), from $y = 0$ to $y = 3$, is:

3. Evaluate the integral. Factor out the constant:

The antiderivative of $y^2$ is $\frac{y^3}{3}$, so:

$A = \frac{1}{4} \left[ \frac{y^3}{3} \right]_{0}^{3} = \frac{1}{4} \cdot \frac{27}{3} = \frac{1}{4} \cdot 9 = \frac{9}{4}$ …

Method: Area between a sideways parabola and the y-axis

Use this when a parabola opens sideways ($y^2 = 4ax$, i.e. $x = \tfrac{y^2}{4a}$) and the region is bounded by the $y$-axis and a horizontal line $y = k$ — integrate in $y$.

Steps

Step 1: Write $x$ as a function of $y$.

From $y^2 = 4ax$ get $x = \dfrac{y^2}{4a}$. This horizontal distance from the $y$-axis to the curve is the width of a horizontal strip.

Step 2: Set the $y$-limits.

The region runs from the vertex ($y = 0$) up to the given line $y = k$.

Step 3: Integrate the strip width in $y$. …

Common Mistakes

Mistake 1: Integrating with respect to $x$ against the wrong axis.

Why it's wrong: using $y = 2\sqrt{x}$ and computing $\int_0^{9/4} 2\sqrt{x}\,dx = \tfrac92$ measures the area between the parabola and the x-axis, not the $y$-axis — that is exactly the trap option (D). Correct approach: the boundary is the $y$-axis and the line $y=3$, so integrate $x=\tfrac{y^2}{4}$ in $y$, giving $\tfrac94$.

Mistake 2: Wrong limits for the horizontal strip. …