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 $\rho^2 \sin \phi$.