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): \( 4 \).

Answer 2

To find the area bounded by the curves \(x = y^2\) and \(x = 3 - 2y^2\), we first determine their points of intersection. Set the equations equal:

\(y^2 = 3 - 2y^2\)

Rearrange the terms:

\(3y^2 = 3\)

\(y^2 = 1\)

Thus, \(y = \pm 1\).

Substituting \(y = 1\) into either equation gives \(x = 1\), and for \(y = -1\), \(x = 1\) as well. Therefore, the curves intersect at \((1, 1)\) and \((1, -1)\).

To find the enclosed area, compute the integral from \(y = -1\) to \(y = 1\). The area \(A\) is:

\(A = \int_{-1}^{1} ((3 - 2y^2) - y^2) \, dy\)

Simplify the integrand:

\(A = \int_{-1}^{1} (3 - 3y^2) \, dy\)

Integrate:

\(A = \left[3y - y^3\right]_{-1}^{1}\)

Evaluate from \(-1\) to \(1\):

\(A = \left(3(1) - (1)^3\right) - \left(3(-1) - (-1)^3\right)\)

\(A = (3 - 1) - (-3 + 1)\)

\(A = 2 + 2 = 4\)

Thus, the area bounded by the curves is \(4\) square units.