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Let’s go through each part of this question in detail.

### 1. Expressing the Triple Integral in Spherical Coordinates

The original integral is:
\[
\int_0^2 \int_0^{\sqrt{4 - x^2}} \int_{-\sqrt{4 - x^2 - y^2}}^{\sqrt{4 - x^2 - y^2}} \sqrt{x^2 + y^2 + z^2} \, dz \, dy \, dx
\]

#### Step 1: Converting to Spherical Coordinates

In spherical coordinates, we have:
- \( x = \rho \sin \phi \cos \theta \)
- \( y = \rho \sin \phi \sin \theta \)
- \( z = \rho \cos \phi \)
- \( \sqrt{x^2 + y^2 + z^2} = \rho \)
- The volume element \( dV = \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta \).

#### Step 2: Setting Up the Bounds

From the given limits:
- \( x \) ranges from \( 0 \) to \( 2 \),
- \( y \) from \( 0 \) to \( \sqrt{4 - x^2} \),
- \( z \) from \( -\sqrt{4 - x^2 - y^2} \) to \( \sqrt{4 - x^2 - y^2} \).

These limits describe a region in the first octant of a sphere with radius 2. In spherical coordinates:
- \( \rho \) ranges from \( 0 \) to \( 2 \),
- \( \phi \) ranges from \( 0 \) to \( \frac{\pi}{2} \),
- \( \theta \) ranges from \( 0 \) to \( \frac{\pi}{2} \).

#### Step 3: Writing the Integral in Spherical Coordinates

Substituting into spherical coordinates, we get:
\[
\int_0^{\pi/2} \int_0^{\pi/2} \int_0^2 \rho \cdot \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta = \int_0^{\pi/2} \int_0^{\pi/2} \int_0^2 \rho^3 \sin \phi \, d\rho \, d\phi \, d\theta
\]

#### Answer for Part 1

The correct choice is:
**(d) \( \int_0^{\pi/2} \int_0^{\pi/2} \int_0^2 \rho^3 \sin \phi \, d\rho \, d\phi \, d\theta \)**

---

### 2. Evaluating the Integral

Now, we’ll evaluate the integral:
\[
\int_0^{\pi/2} \int_0^{\pi/2} \int_0^2 \rho^3 \sin \phi \, d\rho \, d\phi \, d\theta
\]

#### Step 1: Integrate with Respect to \( \rho \)
\[
\int_0^{\pi/2} \int_0^{\pi/2} \left( \int_0^2 \rho^3 \, d\rho \right) \sin \phi \, d\phi \, d\theta
\]
The inner integral with respect to \( \rho \) is:
\[
\int_0^2 \rho^3 \, d\rho = \left[ \frac{\rho^4}{4} \right]_0^2 = \frac{2^4}{4} = \frac{16}{4} = 4
\]
So the integral simplifies to:
\[
4 \int_0^{\pi/2} \int_0^{\pi/2} \sin \phi \, d\phi \, d\theta
\]

#### Step 2: Integrate with Respect to \( \phi \)
Now we have:
\[
4 \int_0^{\pi/2} \left( \int_0^{\pi/2} \sin \phi \, d\phi \right) d\theta
\]
The integral with respect to \( \phi \) is:
\[
\int_0^{\pi/2} \sin \phi \, d\phi = \left[ -\cos \phi \right]_0^{\pi/2} = 1
\]
So now we have:
\[
4 \int_0^{\pi/2} 1 \, d\theta = 4 \cdot \frac{\pi}{2} = 2\pi
\]

#### Answer for Part 2

The value of the integral is:
\[
2\pi
\]

---

Would you like further clarification on any of the steps? 

Here are 5 additional questions for exploration:
1. How are spherical coordinates defined in three-dimensional space?
2. What are some typical applications of triple integrals in physics?
3. How would the integral change if the region were a full sphere instead of the first octant?
4. Can we solve this problem using cylindrical coordinates instead of spherical coordinates?
5. Why is the \( \sin \phi \) term necessary in the spherical coordinate volume element?

**Tip:** When converting integrals to spherical coordinates, always ensure that both the integrand and the limits are adapted to match the spherical geometry and the volume element \( \rho^2 \sin \phi \).

Express the triple integral \( \int_0^2 \int_0^{\sqrt{4 - x^2}} \int_{-\sqrt{4 - x^2 - y^2}}^{\sqrt{4 - x^2 - y^2}} \sqrt{x^2 + y^2 + z^2} \, dz \, dy \, dx \) in spherical coordinates and evaluate the integral.

Learn how to convert and evaluate the triple integral \( \int_0^2 \int_0^{\sqrt{4 - x^2}} \int_{-\sqrt{4 - x^2 - y^2}}^{\sqrt{4 - x^2 - y^2}} \sqrt{x^2 + y^2 + z^2} \, dz \, dy \, dx \) using spherical coordinates. This problem explores spherical transformations, integration limits, and volume elements, suitable for Calculus III students.

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