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:
The problem asks for the area of the region bounded between a vertical parabola opening upwards, $x^2 = y$, and a linear line passing through the origin, $y = 4x$.

Step 2: Key Formula or Approach:
1. Find the points of intersection by setting the two equations equal to each other. 2. The area bounded between an upper curve $y_2 = f(x)$ and a lower curve $y_1 = g(x)$ from $x = a$ to $x = b$ is given by the definite integral: $$\text{Area} = \int_{a}^{b} (y_2 - y_1) \, dx$$

Step 3: Detailed Explanation:
First, let's find the intersection points of $x^2 = y$ and $y = 4x$: $$x^2 = 4x \implies x^2 - 4x = 0$$ $$x(x - 4) = 0 \implies x = 0 \quad \text{or} \quad x = 4$$ The corresponding $y$-coordinates are: For $x = 0$, $y = 0 \implies (0, 0)$ For $x = 4$, $y = 16 \implies (4, 16)$ In the interval $[0, 4]$, the line $y = 4x$ lies above the parabola $y = x^2$. Thus, the bounded area is: $$\text{Area} = \int_{0}^{4} (4x - x^2) \, dx$$ Integrating term by term: $$\text{Area} = \left[ 4 \cdot \frac{x^2}{2} - \frac{x^3}{3} \right]_{0}^{4} = \left[ 2x^2 - \frac{x^3}{3} \right]_{0}^{4}$$ Substitute the limits: $$\text{Area} = \left( 2(4)^2 - \frac{(4)^3}{3} \right) - (0 - 0)$$ $$\text{Area} = 32 - \frac{64}{3} = \frac{96 - 64}{3} = \frac{32}{3}$$

Step 4: Final Answer:
The area bounded between the curves is $\frac{32}{3}$ sq. units, which corresponds to option (A).