The image shows a double integral of the form:

$\int_{-3}^{3} \int_{\sqrt{9 - x^2}}^{\sqrt{9 - x^2}} x^2 \, y^2 \, dy \, dx$

subject to the constraint $x^2 + y^2 = 9$, which describes a circle of radius 3 centered at the origin in the $xy$-plane.

This integral setup suggests that polar coordinates might be suitable for solving, as the limits and region of integration are circular.

Steps to Solve in Polar Coordinates

1. Convert to Polar Coordinates: In polar coordinates, $x = r \cos \theta$ and $y = r \sin \theta$, and the area element $dx \, dy$ becomes $r \, dr \, d\theta$.

2. Set Up the Bounds in Polar Coordinates:

   • The radius $r$ varies from 0 to 3 (from the center of the circle to the boundary).
   • The angle $\theta$ varies from $0$ to $2\pi$ to cover the entire circle.

3. Transform the Integrand: The integrand $x^2 y^2$ becomes $(r \cos \theta)^2 (r \sin \theta)^2 = r^4 \cos^2 \theta \sin^2 \theta$.

4. Rewrite the Integral: $\int_{0}^{2\pi} \int_{0}^{3} r^4 \cos^2 \theta \sin^2 \theta \cdot r \, dr \, d\theta$

   Simplifying, we have: $\int_{0}^{2\pi} \int_{0}^{3} r^5 \cos^2 \theta \sin^2 \theta \, dr \, d\theta$

5. Evaluate the Integral: This will involve separating the $r$-integral and the trigonometric components.

Would you like to proceed with detailed steps for each part of this solution?

Here are some related questions for further exploration:

1. How does converting to polar coordinates simplify circular region integrals?
2. What are common transformations for integrals in circular or radial regions?
3. How does the Jacobian affect the area element when changing to polar coordinates?
4. Why is it beneficial to separate the integrals by variables in polar coordinates?
5. How do symmetry properties impact the evaluation of integrals over circular regions?

Tip: When integrating over circular regions, always consider polar coordinates, as they often simplify the integration limits and calculations.