Use cylindrical coordinates to find the volume of the region bounded by the plane zequals0 and the hyperboloid zequalsStartRoot 10 EndRootminusStartRoot 1 plus x squared plus y squared EndRoot.
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Part 1
The volume is 
  
enter your response here.
​(Type an exact​ answer, using pi as​ needed.)

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.

Use cylindrical coordinates to find the volume of the region bounded by the plane z = 0 and the hyperboloid z = √10 - √(1 + x² + y²).

This problem demonstrates how to calculate the volume of a region bounded by the plane z = 0 and the hyperboloid z = √10 - √(1 + x² + y²) using cylindrical coordinates. The solution involves setting up a triple integral and applying substitution techniques, suitable for students studying multivariable calculus.

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