91Ó°ÊÓ

To evaluate a double integral effectively, understanding the region of integration is crucial. In this exercise, the given region \( D \) is defined by a set of inequalities: \( y \geq x \), \( y \geq -x \), and \( y \leq 2-x^2 \). These inequalities describe an area in the coordinate plane where all conditions are true simultaneously.
To build a visual image of this region, we start by sketching several key boundaries. The lines \( y=x \) and \( y=-x \) form a V-shape through the origin. Simultaneously, the curve described by \( y=2-x^2 \) is a downward-facing parabola. **Intersection Points**
- By determining where the parabola intersects the V-shape, we find the limits of integration.
- The points \( (x, y) = (-1, 1) \) and \( (1, 1) \) define the "height" of the region relative to the x-axis.
Understanding the interplay between these curves helps in dividing the region \( D \) efficiently for better integration handling.

Once the region of integration is clarified, the next step is to set up and evaluate the integral properly. In this case, we divided the region \( D \) into two parts based on the x-axis: one where \( x < 0 \) and another where \( x > 0 \). This division exploits the symmetry of the problem across the y-axis and makes the integration simpler to handle.
### Setting Up the Integrals- **For \( x < 0 \):**
\ \( \int_{-1}^{0} \int_{x}^{2-x^2} y^2 \, dy \, dx \)
- **For \( x > 0 \):**
\ \( \int_{0}^{1} \int_{x}^{2-x^2} y^2 \, dy \, dx \)
These integrals compiled reflect the vertical "slices" from the lower boundary \( y = x \) to the upper \( y = 2 - x^2 \), accurately covering the entire region \( D \) as prescribed by the boundaries identified earlier.
Integrating with respect to \( y \), both integrals take the form \ \( \int_{x}^{2-x^2} y^2 \, dy \), which simplifies to evaluating the expression \( \left[\frac{y^3}{3}\right]_{x}^{2-x^2} \). The resulting expression then needs to be integrated with respect to \( x \). Further solving these integrations, and summing the results provides the total integral over \( D \).

A fascinating feature of this problem is the parabola \( y = 2 - x^2 \), which serves as a boundary. Recognizing how this parabola interacts with other boundaries helps in determining the limits for integration. As a downward-facing parabola, it creates a natural upper limit that smoothly curves over the V-shape boundaries \( y = x \) and \( y = -x \).
### Understanding the Parabola's Role- The curvature means that as \( x \) gets larger, the range for \( y \) narrows consistently.
- The symmetry around the y-axis allows the problem-solving approach to be segmented by the sign of \( x \), making integration manageable.
The parabola not only defines the maximum extent of the region in terms of \( y \), but it also impacts where the boundaries sit on the x-axis. Solving \( 2-x^2 = x \) and \( 2-x^2 = -x \) gives critical points \(-1\) and \(1\). These points are essential for accurately determining the region of integration where the parabola encapsulates the area. Thus, understanding the geometry relative to the parabola enables correct problem-solving steps when evaluating the integral.