Question

The area of the region included between the parabolas $y^2 = 8x$ and $x^2 = 8y$ is

• A. $\frac{128}{3}$ sq. units
• B. $\frac{64}{3}$ sq. units
• C. $\frac{32}{3}$ sq. units
• D. $\frac{16}{3}$ sq. units

Hint:

Whenever you encounter the area bounded between two symmetric orthogonal parabolas $y^2 = kx$ and $x^2 = ky$, the enclosed area is always exactly equal to $\frac{1}{3}k^2$. Here, $k = 8$, so the area is $\frac{8^2}{3} = \frac{64}{3}$ in under 5 seconds!

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The problem asks for the area enclosed between two intersecting symmetric parabolas: a horizontal parabola $y^2 = 8x$ and a vertical parabola $x^2 = 8y$.

Step 2: Key Formula or Approach:
The standard area enclosed between two standard parabolas of the form $y^2 = 4ax$ and $x^2 = 4by$ is given by the elegant definite integration formula shortcut:
$$\text{Area} = \frac{16}{3}ab$$ Alternatively, it can be computed by finding the points of intersection and evaluating $\int_{x_1}^{x_2} (y_{\text{upper}} - y_{\text{lower}})\, dx$.

Step 3: Detailed Explanation:
Let's compare the given equations with the standard forms to find $a$ and $b$:
From $y^2 = 8x$, we get $4a = 8 \implies a = 2$.
From $x^2 = 8y$, we get $4b = 8 \implies b = 2$.
Let's verify the intersection bounds manually to ensure consistency:
$$\left(\frac{x^2}{8}\right)^2 = 8x \implies \frac{x^4}{64} = 8x \implies x(x^3 - 512) = 0$$ This gives the intersection limits from $x = 0$ to $x = 8$.
Substituting our derived parameters $a = 2$ and $b = 2$ directly into the area shortcut formula yields:
$$\text{Area} = \frac{16}{3} \times 2 \times 2$$ $$\text{Area} = \frac{16 \times 4}{3} = \frac{64}{3}\text{ sq. units}$$ This matches option (B).

Step 4: Final Answer:
The area of the region included between the parabolas is $\frac{64}{3}$ sq. units, which corresponds to option (B).