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We are given the solid \( E \) which is bounded by three surfaces:

1. \( S_1: z = 8 \)
2. \( S_2: z = \sqrt{3x^2 + 3y^2} \)
3. \( S_3: z = 2 \)

The task is to set up the triple integral in spherical coordinates to calculate the volume of the solid \( E \).

In spherical coordinates, the relations between Cartesian and spherical coordinates are:
\[
x = \rho \sin\phi \cos\theta, \quad y = \rho \sin\phi \sin\theta, \quad z = \rho \cos\phi
\]
Where:
- \( \rho \) is the radial distance,
- \( \phi \) is the polar angle (angle from the positive \( z \)-axis),
- \( \theta \) is the azimuthal angle (angle in the \( xy \)-plane from the positive \( x \)-axis).

### Step-by-step setup:

1. **Bounds for \( \theta \):**
   Since the solid is symmetric around the \( z \)-axis, \( \theta \) will range from \( 0 \) to \( 2\pi \).

2. **Bounds for \( \phi \):**
   The bounds for \( z \) are given by \( z = 8 \) (top surface) and \( z = 2 \) (bottom surface), which correspond to angles \( \phi \). Using \( z = \rho \cos\phi \), we can calculate the corresponding bounds in \( \phi \).

   - For \( z = 8 \), we get \( \cos\phi = \frac{8}{\rho} \).
   - For \( z = 2 \), we get \( \cos\phi = \frac{2}{\rho} \).

3. **Bounds for \( \rho \):**
   The radial distance \( \rho \) is bounded by the surfaces. The surface \( S_2 \) corresponds to \( z = \sqrt{3x^2 + 3y^2} \), which simplifies in spherical coordinates to \( \rho = 2 \) when solving for \( \rho \sin\phi = 2 \).

Thus, the triple integral for the volume \( V \) will be of the form:
\[
V = \int_0^{2\pi} \int_{\phi_1}^{\phi_2} \int_{\rho_1(\phi)}^{\rho_2(\phi)} \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta
\]

Now, let me analyze the options provided in the image and compare it with the correct setup.After reviewing the problem and the spherical coordinates setup, let's carefully compare the integral bounds.

1. **\( \theta \)-bound**:
   The azimuthal angle \( \theta \) for the solid, being rotationally symmetric around the \( z \)-axis, should range from \( 0 \) to \( 2\pi \). This confirms that all options showing \( 0 \) to \( 2\pi \) for \( \theta \) are correct.

2. **\( \phi \)-bound**:
   The polar angle \( \phi \) is determined by the surfaces of the solid. For the upper surface \( z = 8 \), \( \cos \phi = \frac{8}{\rho} \), and for the lower surface \( z = 2 \), \( \cos \phi = \frac{2}{\rho} \). The angles corresponding to these \( z \)-values determine the limits for \( \phi \). Based on the equation for \( z \) and spherical coordinates, we expect bounds from \( \frac{\pi}{6} \) to \( \frac{\pi}{3} \), which is shown correctly in the options A and E.

3. **\( \rho \)-bound**:
   The radial coordinate \( \rho \) is bounded between the surfaces. From the equation \( z = \sqrt{3x^2 + 3y^2} \), the surface at \( z = 2 \) gives \( \rho = 2 \), which means the upper limit for \( \rho \) should be 8, the maximum height for the solid. Both A and E correctly display this.

### Conclusion:
The correct triple integral in spherical coordinates is represented by **option A**:
\[
V = \int_0^{2\pi} \int_{\frac{\pi}{3}}^{\frac{\pi}{6}} \int_0^8 \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta
\]
This matches the correct volume calculation setup for the given solid.

Would you like a detailed explanation of any part, or do you have further questions?

---

Here are 5 related questions for practice:
1. How do you convert Cartesian coordinates into spherical coordinates for any point in space?
2. What is the physical significance of the polar angle \( \phi \) in spherical coordinates?
3. Can you derive the limits of \( \rho \) for a different surface equation in spherical coordinates?
4. How does the triple integral change if the symmetry around the \( z \)-axis is not present?
5. How would you set up the integral if the solid was bounded by a different function of \( z \)?

**Tip**: Always check the geometry of the problem carefully when converting between coordinate systems, especially for the limits of integration.

Plantee la integral triple en coordenadas esféricas que permite calcular el volumen del sólido E limitado por las superficies: S1: z = 8, S2: z = √(3x² + 3y²), S3: z = 2.

Learn how to calculate the volume of a solid bounded by surfaces using spherical coordinates. This problem explores the triple integral setup for a solid defined by surfaces z = 8, z = √(3x² + 3y²), and z = 2, and demonstrates the use of spherical coordinate transformations.

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