Question

Area bounded between the curves \[ x = -2y^2 \] \[ x = 1 - 4y^2 \] is

• A. \(\frac{\sqrt{2}}{3}\)
• B. \(\frac{2\sqrt{2}}{3}\)
• C. \(\frac{2}{3}\)
• D. \(\frac{3\sqrt{3}}{3}\)

Hint:

If curves are expressed as \(x=f(y)\), integrate with respect to \(y\) using \(A=\int(x_{right}-x_{left})dy\).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The curves are parabolas opening towards the negative x-axis. To find the area bounded between them, we first find their points of intersection and then integrate the difference of the functions (Right curve minus Left curve) with respect to \( y \).
Step 2: Key Formula or Approach:
1. Points of intersection: Set \( x_1 = x_2 \). 2. Area \( A = \int_{y_1}^{y_2} (x_{right} - x_{left}) \, dy \).
Step 3: Detailed Explanation:
1. Find intersection points: \[ -2y^2 = 1 - 4y^2 \] \[ 2y^2 = 1 \implies y^2 = \frac{1}{2} \implies y = \pm \frac{1}{\sqrt{2}} \] 2. Determine which curve is to the right: At \( y = 0 \), \( x_1 = 0 \) and \( x_2 = 1 \). Since \( 1>0 \), \( x = 1 - 4y^2 \) is the right curve. 3. Set up the integral: \[ A = \int_{-1/\sqrt{2}}^{1/\sqrt{2}} [(1 - 4y^2) - (-2y^2)] \, dy \] \[ A = \int_{-1/\sqrt{2}}^{1/\sqrt{2}} (1 - 2y^2) \, dy \] 4. Evaluate the integral (using symmetry): \[ A = 2 \int_{0}^{1/\sqrt{2}} (1 - 2y^2) \, dy \] \[ A = 2 \left[ y - \frac{2y^3}{3} \right]_0^{1/\sqrt{2}} \] \[ A = 2 \left[ \frac{1}{\sqrt{2}} - \frac{2}{3} \left( \frac{1}{2\sqrt{2}} \right) \right] \] \[ A = 2 \left[ \frac{1}{\sqrt{2}} - \frac{1}{3\sqrt{2}} \right] = 2 \left[ \frac{3 - 1}{3\sqrt{2}} \right] \] \[ A = 2 \left[ \frac{2}{3\sqrt{2}} \right] = \frac{4}{3\sqrt{2}} = \frac{2\sqrt{2}}{3} \]
Step 4: Final Answer:
The area bounded between the curves is \( \frac{2\sqrt{2}}{3} \).