Question

The area of the region bounded by the curves \( x = y^2 - 2 \) and \( x = y \) is:

• A. \( \frac{9}{4} \)
• B. 9
• C. \( \frac{9}{2} \)
• D. \( \frac{9}{7} \)

Hint:

For finding the area between curves, set up an integral with the difference of the functions. Ensure the limits of integration are the points where the curves intersect. Simplify the integrand before computing the area.

Answer 1

The Correct Option is C

Given curves \( x = y^2 - 2 \) and \( x = y \), the points of intersection are \( (-2, 0) \) and \( (2, 2) \). To find the area, we integrate the difference between the two functions over the range from \( y = -2 \) to \( y = 2 \): \[ A = \int_{-1}^{2} y \, dy - \int_{-1}^{2} (y^2 - 2) \, dy \]
\[= \left[\frac{y^2}{2} - \frac{y^3}{3} + 2y\right]_{-1}^{2} = \left(\frac{4}{2} - \frac{8}{3} + 4\right) - \left(\frac{1}{2} + \frac{1}{3} - 2\right)\]
\[= \frac{10}{3} + \frac{7}{6} = \frac{27}{6} = \frac{9}{2} \] Thus, the area is \( \frac{9}{7} \).

Answer 2

Step 1: Given Curves
The given curves are: \[ x = y^2 - 2 \quad \text{and} \quad x = y \]
Step 2: Find the Points of Intersection
To find the points of intersection, set the equations equal to each other: \[ y^2 - 2 = y \] Rearrange the equation: \[ y^2 - y - 2 = 0 \] Factor the quadratic equation: \[ (y - 2)(y + 1) = 0 \] Thus, the points of intersection are \( y = 2 \) and \( y = -1 \).
Step 3: Set up the Integral for the Area
The area between the curves is given by: \[ \text{Area} = \int_{y_1}^{y_2} \left( x_{\text{right}} - x_{\text{left}} \right) dy \] Here, the right curve is \( x = y \), and the left curve is \( x = y^2 - 2 \). Thus, the area is: \[ \text{Area} = \int_{-1}^{2} \left( y - (y^2 - 2) \right) dy \]
Step 4: Simplify the Integral
Simplify the integrand: \[ y - (y^2 - 2) = y - y^2 + 2 \] Thus, the area becomes: \[ \text{Area} = \int_{-1}^{2} (-y^2 + y + 2) \, dy \]
Step 5: Compute the Integral
Now, compute the integral: \[ \text{Area} = \int_{-1}^{2} (-y^2 + y + 2) \, dy \] Break the integral into separate terms: \[ \text{Area} = \int_{-1}^{2} -y^2 \, dy + \int_{-1}^{2} y \, dy + \int_{-1}^{2} 2 \, dy \] 1. For \( \int_{-1}^{2} -y^2 \, dy \): \[ \int_{-1}^{2} -y^2 \, dy = -\left[ \frac{y^3}{3} \right]_{-1}^{2} = -\left( \frac{2^3}{3} - \frac{(-1)^3}{3} \right) = -\left( \frac{8}{3} + \frac{1}{3} \right) = -\frac{9}{3} = -3 \] 2. For \( \int_{-1}^{2} y \, dy \): \[ \int_{-1}^{2} y \, dy = \left[ \frac{y^2}{2} \right]_{-1}^{2} = \frac{2^2}{2} - \frac{(-1)^2}{2} = \frac{4}{2} - \frac{1}{2} = \frac{3}{2} \] 3. For \( \int_{-1}^{2} 2 \, dy \): \[ \int_{-1}^{2} 2 \, dy = 2 \left[ y \right]_{-1}^{2} = 2 \left( 2 - (-1) \right) = 2 \times 3 = 6 \]
Step 6: Add the Results
Now, add the results of the integrals: \[ \text{Area} = -3 + \frac{3}{2} + 6 = 3 + \frac{3}{2} = \frac{6}{2} + \frac{3}{2} = \frac{9}{2} \]
Step 7: Conclusion
The area of the region bounded by the curves is: \[ \boxed{\frac{9}{2}} \]