To find the volume of the solid bounded above by $z = 1 - x^2 - y^2$ and below by $z = 0$, we can use cylindrical coordinates. The equation $z = 1 - x^2 - y^2$ describes an upper surface of a paraboloid with its vertex at $(0, 0, 1)$ and opening downward.

Step 1: Convert to cylindrical coordinates

In cylindrical coordinates: $x = r \cos \theta, \quad y = r \sin \theta, \quad z = z$ and $x^2 + y^2 = r^2$.

Thus, the equation of the surface becomes: $z = 1 - r^2$

Step 2: Volume integral setup

The volume is given by: $V = \int \int \int \, dz \, r \, dr \, d\theta$

• The bounds for $z$ are $0 \leq z \leq 1 - r^2$.
• The bounds for $r$ are $0 \leq r \leq 1$ (the maximum $r$ occurs when $1 - r^2 = 0$).
• The bounds for $\theta$ are $0 \leq \theta \leq 2\pi$.

Step 3: Evaluate the integral

The integral becomes: $V = \int_0^{2\pi} \int_0^1 \int_0^{1 - r^2} r \, dz \, dr \, d\theta$

Step 3.1: Integrate with respect to $z$:

$\int_0^{1 - r^2} dz = [z]_0^{1 - r^2} = 1 - r^2$ So the integral reduces to: $V = \int_0^{2\pi} \int_0^1 (1 - r^2) r \, dr \, d\theta$

Step 3.2: Integrate with respect to $r$:

$\int_0^1 (1 - r^2) r \, dr = \int_0^1 \big( r - r^3 \big) \, dr = \left[ \frac{r^2}{2} - \frac{r^4}{4} \right]_0^1 = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}$

Step 3.3: Integrate with respect to $\theta$:

$\int_0^{2\pi} d\theta = 2\pi$

Step 4: Combine the results

$V = 2\pi \cdot \frac{1}{4} = \frac{\pi}{2}$

Final Answer:

The volume of the solid is: $\boxed{\frac{\pi}{2}}$

Would you like me to explain any step further?

---

1. What is the geometric interpretation of this problem?
2. Can we solve this using Cartesian coordinates?
3. How would the setup differ if the paraboloid opened upward?
4. What would happen if the region were not symmetric?
5. How can we verify the volume result?

Tip: Cylindrical coordinates are particularly useful for problems with rotational symmetry!