Q.Find the area of the region bounded by the line $x = 2$ and the parabola $y^2 = 8x$.

Concept: Area Under a Parabola — we integrate the horizontal strips between the curve and the line.

The parabola $y^2 = 8x$ opens to the right. At $x = 2$, the $y$-coordinates are $y = \pm \sqrt{8 \cdot 2} = \pm 4$.

The region is symmetric about the $x$-axis, so we find the area in the upper half and double it.

Step 1: Express $x$ in terms of $y$:

$x = \frac{y^2}{8}$.

Step 2: For a fixed $y$, the horizontal strip runs from the parabola $x = y^2/8$ to the line $x = 2$. Strip length = $2 - \frac{y^2}{8}$.

Step 3: Integrate from $y = -4$ to $y = 4$, using symmetry:

Concept

The parabola $y^2=8x$ opens to the right with vertex at the origin; $x=\dfrac{y^2}{8}$. The vertical line $x=2$ closes off a region symmetric about the $x$-axis. Integrating with respect to $y$ (strip width = right boundary $-$ left boundary) is cleanest.

Solution

1. Intersection points. Set $\dfrac{y^2}{8}=2\Rightarrow y^2=16\Rightarrow y=\pm 4$. So $y$ runs from $-4$ to $4$.

2. Strip width. For a fixed $y$, the region runs from the parabola $x=\dfrac{y^2}{8}$ to the line $x=2$, width $2-\dfrac{y^2}{8}$.

3. Set up and use symmetry (integrand is even):

4. Evaluate.

5. Check (integrating in $x$). $A=2\displaystyle\int_0^2\sqrt{8x}\,dx=2\sqrt{8}\cdot\frac{2}{3}x^{3/2}\Big|_0^2=\frac{4\sqrt8}{3}\,(2\sqrt2)=\frac{32}{3}.$ Both methods agree.

Method: Area by horizontal strips (integrating with respect to $y$)

When a sideways parabola $y^2 = 4ax$ is closed off by a vertical line $x = c$, integrating in $y$ is cleaner than in $x$ because each horizontal strip has two clean $x$-boundaries.

Steps

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

From $y^2 = 4ax$ write $x = \dfrac{y^2}{4a}$ — the left boundary of a horizontal strip. The vertical line $x = c$ is the right boundary.

Step 2: Find the $y$-limits.

Set $\dfrac{y^2}{4a} = c$ to get $y = \pm\sqrt{4ac}$: the strip heights range symmetrically about the $x$-axis.

Step 3: Use symmetry and integrate the strip width.

The region is symmetric about the $x$-axis, so

using $\int y^2\,dy = \dfrac{y^3}{3}$. Substitute the limit $y_0$ to finish. (You can cross-check by integrating $2\sqrt{4ax}$ in $x$ from $0$ to $c$.)

Common Mistakes

Mistake 1: Forgetting the region is symmetric and dropping the factor of $2$

The line $x=2$ cuts the parabola at $y=+4$ and $y=-4$, so the region lies both above and below the x-axis. Why it's wrong: integrating only the upper half gives $\frac{16}{3}$, half the true answer. Correct approach: double the upper-half area, or integrate $y$ from $-4$ to $4$, giving $\frac{32}{3}$.

Mistake 2: Using the wrong strip length

With horizontal strips, the width is (right boundary $-$ left boundary) $=2-\frac{y^2}{8}$. Why it's wrong: writing $\frac{y^2}{8}-2$ makes the length negative and the area wrong. Correct approach: the line $x=2$ is to the right of the parabola $x=\frac{y^2}{8}$, so subtract parabola from line.

Mistake 3: Wrong limits at $x=2$

At $x=2$, $y^2=8(2)=16$ so $y=\pm4$. Why it's wrong: students take $y=\pm\sqrt{8}$ or forget to substitute $x=2$. Correct approach: plug $x=2$ into $y^2=8x$ to get $y=\pm4$.