1. **Problem statement:** Find the volume of the solid bounded by the cone $$z = \sqrt{x^2 + y^2}$$ and the paraboloid $$z = 6 - x^2 - y^2$$ using cylindrical coordinates. 2. **Coordinate transformation:** In cylindrical coordinates, we use $$x = r\cos\theta$$, $$y = r\sin\theta$$, and $$z = z$$. 3. **Rewrite surfaces:** - Cone: $$z = r$$ (since $$\sqrt{x^2 + y^2} = r$$) - Paraboloid: $$z = 6 - r^2$$ 4. **Find intersection:** Set the two surfaces equal to find the boundary in $$r$$: $$r = 6 - r^2 \implies r^2 + r - 6 = 0$$ 5. **Solve quadratic:** $$r = \frac{-1 \pm \sqrt{1 + 24}}{2} = \frac{-1 \pm 5}{2}$$ Only positive root is valid for radius: $$r = 2$$ 6. **Volume integral setup:** The volume is given by $$V = \int_{\theta=0}^{2\pi} \int_{r=0}^2 \int_{z=r}^{6 - r^2} r \, dz \, dr \, d\theta$$ Note the extra $$r$$ in the integrand comes from the Jacobian of cylindrical coordinates. 7. **Integrate with respect to $$z$$:** $$\int_r^{6 - r^2} r \, dz = r \left[(6 - r^2) - r\right] = r(6 - r^2 - r) = 6r - r^3 - r^2$$ 8. **Integrate with respect to $$r$$:** $$\int_0^2 (6r - r^3 - r^2) dr = \left[3r^2 - \frac{r^4}{4} - \frac{r^3}{3}\right]_0^2 = 3(4) - \frac{16}{4} - \frac{8}{3} = 12 - 4 - \frac{8}{3} = 8 - \frac{8}{3} = \frac{24 - 8}{3} = \frac{16}{3}$$ 9. **Integrate with respect to $$\theta$$:** $$\int_0^{2\pi} \frac{16}{3} d\theta = \frac{16}{3} \times 2\pi = \frac{32\pi}{3}$$ **Final answer:** $$\boxed{\frac{32\pi}{3}}$$ This is the volume of the solid bounded by the given cone and paraboloid.