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To solve the given double integral 

\[
\int_{-1}^{1} \int_{-1}^{1} \left(2 - x^2 - y^2 \right) \, dx \, dy,
\]

we will integrate the function \( 2 - x^2 - y^2 \) with respect to \(x\) first, and then with respect to \(y\).

1. **Integrate with respect to \(x\)**:
\[
\int_{-1}^{1} \left( \int_{-1}^{1} \left(2 - x^2 - y^2 \right) \, dx \right) \, dy
\]

First, consider the inner integral:
\[
\int_{-1}^{1} \left(2 - x^2 - y^2 \right) \, dx
\]

This can be broken down into three separate integrals:
\[
\int_{-1}^{1} 2 \, dx - \int_{-1}^{1} x^2 \, dx - \int_{-1}^{1} y^2 \, dx
\]

Evaluating each integral separately:

\[
\int_{-1}^{1} 2 \, dx = 2x \Bigg|_{-1}^{1} = 2(1) - 2(-1) = 4
\]

\[
\int_{-1}^{1} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{-1}^{1} = \frac{1}{3} - \left(-\frac{1}{3}\right) = \frac{2}{3}
\]

Since \( y^2 \) is a constant with respect to \( x \):
\[
\int_{-1}^{1} y^2 \, dx = y^2 \int_{-1}^{1} \, dx = y^2 \left[x\right]_{-1}^{1} = y^2 (1 - (-1)) = 2y^2
\]

Combining these results:
\[
\int_{-1}^{1} \left(2 - x^2 - y^2 \right) \, dx = 4 - \frac{2}{3} - 2y^2
\]

2. **Integrate with respect to \(y\)**:
\[
\int_{-1}^{1} \left(4 - \frac{2}{3} - 2y^2 \right) \, dy
\]

This integral can be broken down into three parts:
\[
\int_{-1}^{1} 4 \, dy - \int_{-1}^{1} \frac{2}{3} \, dy - \int_{-1}^{1} 2y^2 \, dy
\]

Evaluating each integral separately:

\[
\int_{-1}^{1} 4 \, dy = 4y \Bigg|_{-1}^{1} = 4(1) - 4(-1) = 8
\]

\[
\int_{-1}^{1} \frac{2}{3} \, dy = \frac{2}{3} y \Bigg|_{-1}^{1} = \frac{2}{3}(1) - \frac{2}{3}(-1) = \frac{4}{3}
\]

\[
\int_{-1}^{1} 2y^2 \, dy = 2 \left[\frac{y^3}{3}\right]_{-1}^{1} = 2 \left(\frac{1}{3} - \left(-\frac{1}{3}\right)\right) = 2 \cdot \frac{2}{3} = \frac{4}{3}
\]

Combining these results:
\[
\int_{-1}^{1} \left(4 - \frac{2}{3} - 2y^2 \right) \, dy = 8 - \frac{4}{3} - \frac{4}{3} = 8 - \frac{8}{3} = 8 - \frac{8}{3} = \frac{24}{3} - \frac{8}{3} = \frac{16}{3}
\]

Thus, the value of the double integral is:
\[
\boxed{\frac{16}{3}}
\]

Would you like further details on this solution or have any additional questions? Here are some related questions you might find interesting:

1. How do you evaluate double integrals over general regions?
2. What are the applications of double integrals in physics and engineering?
3. How do you convert a double integral to polar coordinates?
4. What is Fubini's theorem, and how is it used in evaluating double integrals?
5. Can you explain the concept of the Jacobian determinant in changing variables in multiple integrals?

**Tip:** When solving double integrals, it's often helpful to sketch the region of integration to visualize the bounds and understand the problem better.

Learn how to solve the double integral of 2 - x^2 - y^2 over the region [-1, 1]x[-1, 1] step-by-step. This problem involves integrating a polynomial function over a rectangular region using Fubini's theorem.

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