https://file.aiquickdraw.com/mathbot/file/1730083591777wxt9gx4k.jpg

To solve this problem, let's approach both parts (a) and (b) as requested, using Cartesian coordinates \((x, y, z)\) and cylindrical coordinates \((r, \theta, z)\) respectively. 

### Step 1: Analyzing the Boundaries

The region shown is bounded by:
1. A paraboloid: \( z = x^2 + y^2 \)
2. An upper surface: \( z = 8 - x^2 - y^2 \)

These surfaces intersect where \( x^2 + y^2 = 4 \) (since \( 8 - x^2 - y^2 = x^2 + y^2 \)), or equivalently, \( r = 2 \) in cylindrical coordinates.

### (a) Volume Calculation Using \( x, y, z \) Coordinates

1. **Set up the limits for \( z \):** The volume is bounded between the two surfaces, so \( z \) ranges from the lower surface \( z = x^2 + y^2 \) to the upper surface \( z = 8 - x^2 - y^2 \).
2. **Set up the limits for \( x \) and \( y \):** The region is a circle in the \( xy \)-plane with radius 2, so \( x^2 + y^2 \leq 4 \).

The volume \( V \) is given by the integral:
\[
V = \iint_{x^2 + y^2 \leq 4} \int_{z = x^2 + y^2}^{z = 8 - x^2 - y^2} dz \, dx \, dy
\]

Evaluating the \( z \)-integral first:
\[
V = \iint_{x^2 + y^2 \leq 4} \left(8 - x^2 - y^2 - (x^2 + y^2)\right) \, dx \, dy
\]
\[
= \iint_{x^2 + y^2 \leq 4} \left(8 - 2(x^2 + y^2)\right) \, dx \, dy
\]

This can be simplified further by switching to polar coordinates.

### (b) Volume Calculation Using Cylindrical Coordinates

In cylindrical coordinates, \( x^2 + y^2 = r^2 \), so the equations become:
1. \( z = r^2 \)
2. \( z = 8 - r^2 \)

The volume \( V \) is given by:
\[
V = \int_{0}^{2\pi} \int_{0}^{2} \int_{r^2}^{8 - r^2} r \, dz \, dr \, d\theta
\]

1. **Integrate with respect to \( z \):**
   \[
   V = \int_{0}^{2\pi} \int_{0}^{2} \left[ r \cdot z \right]_{z = r^2}^{z = 8 - r^2} \, dr \, d\theta
   \]
   \[
   = \int_{0}^{2\pi} \int_{0}^{2} r \left(8 - r^2 - r^2\right) \, dr \, d\theta
   \]
   \[
   = \int_{0}^{2\pi} \int_{0}^{2} r \left(8 - 2r^2\right) \, dr \, d\theta
   \]

2. **Integrate with respect to \( r \):**
   \[
   V = \int_{0}^{2\pi} \int_{0}^{2} \left(8r - 2r^3\right) \, dr \, d\theta
   \]
   \[
   = \int_{0}^{2\pi} \left[ 4r^2 - \frac{r^4}{2} \right]_{0}^{2} \, d\theta
   \]
   \[
   = \int_{0}^{2\pi} \left(4 \cdot 4 - \frac{16}{2}\right) \, d\theta
   \]
   \[
   = \int_{0}^{2\pi} (16 - 8) \, d\theta
   \]
   \[
   = \int_{0}^{2\pi} 8 \, d\theta
   \]
   \[
   = 8 \cdot 2\pi = 16\pi
   \]

### Answer:
The volume of the region is \( V = 16\pi \).

Would you like further details on any of the steps, or do you have any questions?

### Related Questions
1. How do we determine the limits for \( r \) and \( z \) when using cylindrical coordinates?
2. Why is polar conversion useful in calculating the volume of solids with circular symmetry?
3. Can the volume be found using spherical coordinates instead? If so, how would it differ?
4. How can we interpret the bounds of integration in a physical context?
5. Why is it simpler to integrate in cylindrical coordinates for solids with rotational symmetry?

### Tip
When dealing with symmetric regions around the \( z \)-axis, cylindrical coordinates often simplify the integration process by leveraging symmetry.

Calculate the volume of the region shown in the figure using (a) Cartesian coordinates and (b) cylindrical coordinates.

Learn how to calculate the volume of a region bounded by a paraboloid and an inverted paraboloid using Cartesian and cylindrical coordinates. This problem covers essential calculus concepts, including triple integrals, volume calculation, and coordinate transformations. Suitable for advanced calculus students.

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