Question

The area (in square units) bounded by the curves \( x = y^2 \) and \( x = 3 - 2y^2 \) is:

• A. \( 8 \)
• B. \( \frac{8}{3} \)
• C. \( 4 \)
• D. \( 6 \)

Hint:

For finding the area between curves, always set up the integral as \( \int (\text{upper function} - \text{lower function}) \, dy \). Finding points of intersection is crucial for setting the limits of integration.

Answer 1

The Correct Option is C

Step 1: Finding Points of Intersection The given curves are: \[ x = y^2 \] \[ x = 3 - 2y^2. \] To find the points of intersection, set both equations equal to each other: \[ y^2 = 3 - 2y^2. \] Solving for \( y^2 \): \[ y^2 + 2y^2 = 3. \] \[ 3y^2 = 3. \] \[ y^2 = 1 \Rightarrow y = \pm 1. \] Thus, the curves intersect at \( y = -1 \) and \( y = 1 \).
Step 2: Setting up the Area Integral The area between the curves is given by: \[ A = \int_{-1}^{1} \left[ (3 - 2y^2) - y^2 \right] dy. \] Simplifying the integrand: \[ A = \int_{-1}^{1} \left( 3 - 3y^2 \right) dy. \] \[ A = \int_{-1}^{1} 3 \,dy - \int_{-1}^{1} 3y^2 \,dy. \]
Step 3: Evaluating the Integral First integral: \[ \int_{-1}^{1} 3 \,dy = 3[y]_{-1}^{1} = 3(1 - (-1)) = 3(2) = 6. \] Second integral: \[ \int_{-1}^{1} 3y^2 \,dy = 3 \int_{-1}^{1} y^2 \,dy. \] Using the standard formula: \[ \int y^n dy = \frac{y^{n+1}}{n+1}, \] we get: \[ \int y^2 dy = \frac{y^3}{3}. \] Evaluating from \( -1 \) to \( 1 \): \[ \left[ \frac{y^3}{3} \right]_{-1}^{1} = \frac{1^3}{3} - \frac{(-1)^3}{3} = \frac{1}{3} - \left(-\frac{1}{3}\right) = \frac{2}{3}. \] Thus, \[ \int_{-1}^{1} 3y^2 dy = 3 \times \frac{2}{3} = 2. \]
Step 4: Final Calculation \[ A = 6 - 2 = 4. \] Thus, the correct answer is option (3): \( 4 \).