Use spherical coordinate systems to evaluate: \[ \int_{-3}^{3} \int_{-\sqrt{9-y^{2}}}^{\sqrt{9-y^{2}}} \int_{-\sqrt{9-x^{2}-y^{2}}}^{\sqrt{9-x^{2}-y^{2}}} \sqrt{x^{2}+y^{2}+z^{2}} d z d x d y \]

To evaluate the given integral using spherical coordinates, we first understand the limits of integration. The integral is defined over a region that resembles a sphere of radius 3, centered at the origin. In spherical coordinates, we can represent \( x, y, z \) in terms of the spherical coordinates \( \rho, \theta, \phi \) as follows: \[ x = \rho \sin \phi \cos \theta, \quad y = \rho \sin \phi \sin \theta, \quad z = \rho \cos \phi \] where: - \( \rho \) is the radius, - \( \phi \) is the polar angle (angle from the positive z-axis), - \( \theta \) is the azimuthal angle (angle in the xy-plane from the positive x-axis). ### Jacobian Transformation The Jacobian for the transformation from Cartesian to spherical coordinates is given by \( \rho^2 \sin \phi \). Therefore, the volume element \( dV \) in spherical coordinates becomes: \[ dV = \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta \] ### Integral Setup We need to convert the integral into spherical coordinates. The expression \( \sqrt{x^2 + y^2 + z^2} \) simplifies to \( \rho \) in spherical coordinates. The limits of integration for \( \rho \) are from 0 to 3 (the radius of the sphere), for \( \phi \) from 0 to \( \pi \) (the full range from the positive z-axis to the negative z-axis), and for \( \theta \) from 0 to \( 2\pi \) (full rotation around the z-axis). Thus, the integral becomes: \[ \int_0^{2\pi} \int_0^{\pi} \int_0^{3} \rho \cdot \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta \] This simplifies to: \[ \int_0^{2\pi} d\theta \int_0^{\pi} \sin \phi \, d\phi \int_0^{3} \rho^3 \, d\rho \] ### Evaluation of Each Integral 1. **Integral over \( \rho \)**: \[ \int_0^{3} \rho^3 \, d\rho = \left[\frac{\rho^4}{4}\right]_0^{3} = \frac{81}{4} \] 2. **Integral over \( \phi \)**: \[ \int_0^{\pi} \sin \phi \, d\phi = [-\cos \phi]_0^{\pi} = -(-1 - 1) = 2 \] 3. **Integral over \( \theta \)**: \[ \int_0^{2\pi} d\theta = 2\pi \] ### Combine Results Now we multiply the results of these integrals together: \[ \int_{-3}^{3} \int_{-\sqrt{9-y^2}}^{\sqrt{9-y^2}} \int_{-\sqrt{9-x^2-y^2}}^{\sqrt{9-x^2-y^2}} \sqrt{x^2+y^2+z^2} \, dz \, dx \, dy = \frac{81}{4} \cdot 2 \cdot 2\pi \] Calculating this gives: \[ \frac{81}{4} \cdot 4\pi = 81\pi \] Thus, the final result of the integral is: \[ \boxed{81\pi} \]