To find the volume of the region bounded by the plane $z = 0$ and the hyperboloid $z = \sqrt{10} - \sqrt{1 + x^2 + y^2}$ using cylindrical coordinates, follow these steps:

Step 1: Set Up in Cylindrical Coordinates

In cylindrical coordinates:

• $x = r \cos(\theta)$
• $y = r \sin(\theta)$
• $z = z$
• $r = \sqrt{x^2 + y^2}$

The equation of the hyperboloid becomes: $z = \sqrt{10} - \sqrt{1 + r^2}$

Step 2: Define the Boundaries

• The lower boundary is the plane $z = 0$.
• The upper boundary is given by $z = \sqrt{10} - \sqrt{1 + r^2}$.

Step 3: Determine the Limits of $r$

To find the limits of $r$, set $z = 0$: $0 = \sqrt{10} - \sqrt{1 + r^2} \Rightarrow \sqrt{1 + r^2} = \sqrt{10} \Rightarrow r^2 = 9 \Rightarrow r = 3$ So, $r$ ranges from 0 to 3.

Step 4: Set Up the Integral

The volume element in cylindrical coordinates is $dV = r \, dr \, d\theta \, dz$.

The volume is given by the triple integral: $V = \int_{0}^{2\pi} \int_{0}^{3} \int_{0}^{\sqrt{10} - \sqrt{1 + r^2}} r \, dz \, dr \, d\theta$

Step 5: Evaluate the Integral

Integrate with respect to $z$: $\int_{0}^{\sqrt{10} - \sqrt{1 + r^2}} r \, dz = r \left[ z \right]_{0}^{\sqrt{10} - \sqrt{1 + r^2}} = r (\sqrt{10} - \sqrt{1 + r^2})$

Now, integrate with respect to $r$: $\int_{0}^{3} r (\sqrt{10} - \sqrt{1 + r^2}) \, dr = \sqrt{10} \int_{0}^{3} r \, dr - \int_{0}^{3} r \sqrt{1 + r^2} \, dr$

Step 6: Solve the Integrals

1. For the first integral: $\sqrt{10} \int_{0}^{3} r \, dr = \sqrt{10} \left[ \frac{r^2}{2} \right]_{0}^{3} = \sqrt{10} \cdot \frac{9}{2} = \frac{9 \sqrt{10}}{2}$

2. For the second integral, use the substitution $u = 1 + r^2$, $du = 2r \, dr$: $\int_{0}^{3} r \sqrt{1 + r^2} \, dr = \frac{1}{2} \int_{1}^{10} \sqrt{u} \, du = \frac{1}{2} \left[ \frac{2}{3} u^{3/2} \right]_{1}^{10}$ Evaluating this gives: $\frac{1}{2} \left[ \frac{2}{3} (10^{3/2} - 1^{3/2}) \right] = \frac{1}{3} (20 \sqrt{10} - 1)$

Step 7: Combine the Results

Combine the results from both integrals: $V = \int_{0}^{2\pi} \left( \frac{9 \sqrt{10}}{2} - \frac{1}{3} (20 \sqrt{10} - 1) \right) d\theta$ $V = \int_{0}^{2\pi} \left( \frac{27 \sqrt{10}}{6} - \frac{20 \sqrt{10}}{6} + \frac{1}{3} \right) d\theta$ $V = \int_{0}^{2\pi} \left(\frac{7 \sqrt{10}}{6} + \frac{1}{3}\right) d\theta$

Step 8: Integrate with Respect to $\theta$

Integrating over $\theta$: $V = \left(\frac{7 \sqrt{10}}{6} + \frac{1}{3}\right) \int_{0}^{2\pi} d\theta = \left(\frac{7 \sqrt{10}}{6} + \frac{1}{3}\right) \cdot 2\pi$

Final Answer

$V = \left(\frac{7 \sqrt{10}}{6} + \frac{1}{3}\right) \cdot 2\pi = \frac{7 \sqrt{10}}{3}\pi + \frac{2\pi}{3}$

The volume of the region is $\frac{7 \sqrt{10} \pi}{3} + \frac{2\pi}{3}$.

Would you like more details on any step or have any questions?

Here are 5 related questions:

1. How would the problem change if the upper boundary was a different hyperboloid equation?
2. What if the radius limit changed to $r = 2$? How would the integral change?
3. How can the volume of a paraboloid bounded by a plane be calculated?
4. What are the differences between using cylindrical and spherical coordinates for this problem?
5. How would we evaluate this volume if the region was bounded by a sphere instead?

Tip: For triple integrals, always visualize the region first to set up the limits of integration correctly.