Question

The area bounded between the curve $x^2 = y$ and the line $y = 4x$ is

• A. $\frac{32}{3}$ sq. units
• B. $\frac{8}{3}$ sq. units
• C. $\frac{1}{3}$ sq. units
• D. $\frac{16}{3}$ sq. units

Hint:

The area bounded between a standard parabola $x^2 = \text{v}y$ (or $y^2 = \text{u}x$) and the line $y = mx$ can be directly computed using the standard shortcut formula:
$$\text{Area} = \frac{8}{3} a^2 m^3 \quad \text{for } y^2 = 4ax \text{ and } y = mx$$ For the form $x^2 = 4ay$ and $y = mx$, the bounded area formulas simplify neatly to: $$\text{Area} = \frac{8}{3} a^2 m^3 \implies \text{Area} = \frac{m^3}{6k} \quad \text{where } y = kx^2$$ Here, $y = 1x^2$ ($k=1$) and $m = 4$: $$\text{Area} = \frac{4^3}{6(1)} = \frac{64}{6} = \frac{32}{3}\ \text{sq. units.}$$

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
Calculate the area bounded between the parabola x² = y and the line y = 4x.

Step 2: Key Formula or Approach:
Find intersection points by equating x² = 4x. The area between two curves is ∫ₐᵇ (upper curve - lower curve) dx.

Step 3: Detailed Explanation:
x² = 4x → x(x - 4) = 0 → x = 0, 4. On [0,4], y = 4x lies above y = x². Area = ∫₀⁴ (4x - x²) dx = [2x² - x³/3]₀⁴ = (32 - 64/3) - 0 = (96 - 64)/3 = 32/3 sq. units.

Step 4: Final Answer:
The area is 32/3 sq. units, option (A).