1. **State the problem:** Evaluate the triple integral $$\int_{-1}^1 \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \int_{x^2+y^2}^{2 - x^2 - y^2} (x^2 + y^2)^{3/2} \, dz \, dy \, dx.$$\n\n2. **Convert to cylindrical coordinates:** Recall the transformations: $$x = r \cos \theta, \quad y = r \sin \theta, \quad z = z,$$ and $$x^2 + y^2 = r^2.$$ The volume element is $$dV = dz \, r \, dr \, d\theta.$$\n\n3. **Determine the limits:** The projection on the xy-plane is the disk $$x^2 + y^2 \leq 1,$$ so $$r$$ goes from $$0$$ to $$1,$$ and $$\theta$$ from $$0$$ to $$2\pi.$$ The limits for $$z$$ are from $$r^2$$ to $$2 - r^2.$$\n\n4. **Rewrite the integral:**\n$$\int_0^{2\pi} \int_0^1 \int_{r^2}^{2 - r^2} (r^2)^{3/2} \, dz \, r \, dr \, d\theta = \int_0^{2\pi} \int_0^1 \int_{r^2}^{2 - r^2} r^3 \, dz \, r \, dr \, d\theta = \int_0^{2\pi} \int_0^1 \int_{r^2}^{2 - r^2} r^4 \, dz \, dr \, d\theta.$$\n\n5. **Integrate with respect to $$z$$:**\n$$\int_{r^2}^{2 - r^2} r^4 \, dz = r^4 \left[(2 - r^2) - r^2\right] = r^4 (2 - 2r^2) = 2r^4 (1 - r^2).$$\n\n6. **Integral reduces to:**\n$$\int_0^{2\pi} \int_0^1 2r^4 (1 - r^2) \, dr \, d\theta = 2 \int_0^{2\pi} d\theta \int_0^1 r^4 (1 - r^2) \, dr.$$\n\n7. **Integrate with respect to $$r$$:**\n$$\int_0^1 r^4 (1 - r^2) \, dr = \int_0^1 (r^4 - r^6) \, dr = \left[ \frac{r^5}{5} - \frac{r^7}{7} \right]_0^1 = \frac{1}{5} - \frac{1}{7} = \frac{7 - 5}{35} = \frac{2}{35}.$$\n\n8. **Integrate with respect to $$\theta$$:**\n$$2 \int_0^{2\pi} \frac{2}{35} \, d\theta = \frac{4}{35} \times 2\pi = \frac{8\pi}{35}.$$\n\n**Final answer:** $$\boxed{\frac{8\pi}{35}}.$$