Question

The area of the region bounded by the curves \(x + 3y^2 = 0\) and \(x + 4y^2 = 1\) is equal to:

• A. \(\frac{1}{3}\)
• B. \(\frac{2}{3}\)
• C. \(\frac{4}{3}\)
• D. \(\frac{5}{3}\)

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The curves are parabolas opening to the left. It is easier to integrate with respect to \(y\). We find the points of intersection and integrate the horizontal distance between the curves.

Step 2: Key Formula or Approach:
Area \(= \int_{y_1}^{y_2} [x_{right}(y) - x_{left}(y)] dy\).

Step 3: Detailed Explanation:
\(x_1 = -3y^2\) and \(x_2 = 1 - 4y^2\).
Intersection: \(-3y^2 = 1 - 4y^2 \implies y^2 = 1 \implies y = \pm 1\).
Area \(= \int_{-1}^{1} (x_2 - x_1) dy = \int_{-1}^{1} (1 - 4y^2 + 3y^2) dy = \int_{-1}^{1} (1 - y^2) dy\).
Area \(= 2 \int_{0}^{1} (1 - y^2) dy = 2 [ y - \frac{y^3}{3} ]_0^1 = 2 [ 1 - \frac{1}{3} ] = \frac{4}{3}\).

Step 4: Final Answer:
The area is \(4/3\).