Step 1: Understanding the Concept:
To find the area bounded by a left curve $x = f(y)$ and a right curve $x = g(y)$, we integrate with respect to $y$. The area is given by the integral of $(\text{Right Curve} - \text{Left Curve})$ evaluated between their points of intersection.
Step 2: Key Formula or Approach:
The area $A$ is given by:
\[ A = \int_{y_1}^{y_2} [g(y) - f(y)] \, dy \]
where $g(y)$ is the curve on the right and $f(y)$ is the curve on the left.
Step 3: Detailed Explanation:
First, find the points of intersection by setting the equations equal to each other:
\[ y^2 - 2 = y \]
\[ y^2 - y - 2 = 0 \]
Factor the quadratic equation:
\[ (y - 2)(y + 1) = 0 \]
Thus, the points of intersection are at $y = -1$ and $y = 2$.
Next, determine which curve is on the right (has larger $x$ values) in the interval $(-1, 2)$. Pick a test point, say $y = 0$:
For the line $x = y$, $x = 0$.
For the parabola $x = y^2 - 2$, $x = -2$.
Since $0>-2$, the line $x = y$ is the right curve, and the parabola $x = y^2 - 2$ is the left curve.
Set up the integral for the area:
\[ A = \int_{-1}^{2} [y - (y^2 - 2)] \, dy \]
\[ A = \int_{-1}^{2} (-y^2 + y + 2) \, dy \]
Integrate term by term:
\[ A = \left[ -\frac{y^3}{3} + \frac{y^2}{2} + 2y \right]_{-1}^{2} \]
Evaluate the definite integral at the upper limit ($y = 2$):
\[ \left( -\frac{8}{3} + \frac{4}{2} + 4 \right) = -\frac{8}{3} + 2 + 4 = 6 - \frac{8}{3} = \frac{10}{3} \]
Evaluate at the lower limit ($y = -1$):
\[ \left( -\frac{(-1)^3}{3} + \frac{(-1)^2}{2} + 2(-1) \right) = \frac{1}{3} + \frac{1}{2} - 2 = \frac{5}{6} - \frac{12}{6} = -\frac{7}{6} \]
Subtract the lower limit value from the upper limit value:
\[ A = \frac{10}{3} - \left( -\frac{7}{6} \right) = \frac{20}{6} + \frac{7}{6} = \frac{27}{6} \]
Simplify the fraction:
\[ A = \frac{9}{2} \]
Step 4: Final Answer:
The area of the region bounded by the curves is $\frac{9}{2}$. The correct option is (C).