1. **State the problem:** Find the volume of the solid $Q$ bounded above by the surface $$y = 4 - x^2 - z^2$$ and below by the $xz$-plane (where $y=0$). 2. **Set up the integral:** The volume is the integral of $1$ over the region $Q$. Since $y$ ranges from $0$ to $4 - x^2 - z^2$, and the projection on the $xz$-plane is the region where $4 - x^2 - z^2 \geq 0$, i.e., $x^2 + z^2 \leq 4$. 3. **Use cylindrical coordinates:** Let $x = r\cos\theta$, $z = r\sin\theta$, and $y = y$. The Jacobian for cylindrical coordinates in $x,z$ is $r$. 4. **Bounds:** - $r$ from $0$ to $2$ (since $x^2 + z^2 \leq 4$) - $\theta$ from $0$ to $2\pi$ - $y$ from $0$ to $4 - r^2$ 5. **Volume integral:** $$V = \int_0^{2\pi} \int_0^2 \int_0^{4 - r^2} r \, dy \, dr \, d\theta$$ 6. **Integrate with respect to $y$:** $$\int_0^{4 - r^2} r \, dy = r[y]_0^{4 - r^2} = r(4 - r^2)$$ 7. **Integral reduces to:** $$V = \int_0^{2\pi} \int_0^2 r(4 - r^2) \, dr \, d\theta$$ 8. **Integrate with respect to $r$:** $$\int_0^2 r(4 - r^2) \, dr = \int_0^2 (4r - r^3) \, dr = \left[2r^2 - \frac{r^4}{4}\right]_0^2 = \left(2 \times 4 - \frac{16}{4}\right) = 8 - 4 = 4$$ 9. **Integral reduces to:** $$V = \int_0^{2\pi} 4 \, d\theta = 4[\theta]_0^{2\pi} = 4 \times 2\pi = 8\pi$$ **Final answer:** $$\boxed{8\pi}$$ This is the volume of the solid bounded by the paraboloid and the $xz$-plane.