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To convert the given integral to spherical coordinates, we follow these steps:

### Step 1: Spherical Coordinates Relations
The spherical coordinates \((\rho, \theta, \phi)\) are related to Cartesian coordinates \((x, y, z)\) as:
\[
x = \rho \sin\phi \cos\theta, \quad y = \rho \sin\phi \sin\theta, \quad z = \rho \cos\phi,
\]
with the corresponding volume element:
\[
dx \, dy \, dz = \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta.
\]

### Step 2: Identifying the Boundaries
The given Cartesian integral is:
\[
\int_0^1 \int_0^{\sqrt{1-x^2}} \int_0^{\sqrt{4-x^2-y^2}} \frac{x^2 y}{\sqrt{x^2 + y^2}} \, dz \, dy \, dx.
\]

- **Outer integral (\(x\))**: \(x\) ranges from \(0\) to \(1\).
- **Middle integral (\(y\))**: For a fixed \(x\), \(y\) ranges from \(0\) to \(\sqrt{1-x^2}\).
- **Inner integral (\(z\))**: For fixed \(x\) and \(y\), \(z\) ranges from \(0\) to \(\sqrt{4-x^2-y^2}\).

These limits represent a region in the first octant of a sphere of radius \(2\).

### Step 3: Converting the Limits to Spherical Coordinates
- **\(\rho\)**: The radial distance ranges from \(0\) to \(2\), as the sphere has radius \(2\).
- **\(\theta\)**: The azimuthal angle ranges from \(0\) to \(\frac{\pi}{2}\), since the region is in the first octant.
- **\(\phi\)**: The polar angle ranges from \(0\) to \(\frac{\pi}{2}\), again due to the first-octant restriction.

### Step 4: Converting the Integrand
The integrand \(\frac{x^2 y}{\sqrt{x^2 + y^2}}\) becomes:
\[
x^2 y = (\rho \sin\phi \cos\theta)^2 (\rho \sin\phi \sin\theta),
\]
and \(\sqrt{x^2 + y^2} = \rho \sin\phi\). Substituting, the integrand in spherical coordinates is:
\[
\frac{x^2 y}{\sqrt{x^2 + y^2}} = \frac{(\rho \sin\phi \cos\theta)^2 (\rho \sin\phi \sin\theta)}{\rho \sin\phi} = \rho^2 \sin^2\phi \cos^2\theta \sin\theta.
\]

### Step 5: Volume Element
The volume element \(dx \, dy \, dz\) becomes \(\rho^2 \sin\phi \, d\rho \, d\phi \, d\theta\).

### Step 6: The Final Integral
Substituting everything into spherical coordinates:
\[
\int_0^1 \int_0^{\sqrt{1-x^2}} \int_0^{\sqrt{4-x^2-y^2}} \frac{x^2 y}{\sqrt{x^2 + y^2}} \, dz \, dy \, dx
\]
becomes:
\[
\int_0^{\frac{\pi}{2}} \int_0^{\frac{\pi}{2}} \int_0^2 \rho^4 \sin^3\phi \cos^2\theta \sin\theta \, d\rho \, d\phi \, d\theta.
\]

Let me know if you'd like additional details or have further questions!

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### Follow-Up Questions:
1. How are spherical coordinates derived from Cartesian coordinates?
2. What are the general integration limits for spherical coordinates?
3. Why does the volume element include \(\rho^2 \sin\phi\)?
4. How does the first-octant restriction influence the limits?
5. How would this integral change if the region was in the full sphere?

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### Tip:
Always carefully analyze the geometry of the region when converting integrals to spherical coordinates.

Convert the integral \( \int_0^1 \int_0^{\sqrt{1-x^2}} \int_0^{\sqrt{4-x^2-y^2}} \frac{x^2 y}{\sqrt{x^2 + y^2}} \, dz \, dy \, dx \) to spherical coordinates (do not evaluate it).

This problem explains how to convert a triple integral \( \int_0^1 \int_0^{\sqrt{1-x^2}} \int_0^{\sqrt{4-x^2-y^2}} \frac{x^2 y}{\sqrt{x^2 + y^2}} \, dz \, dy \, dx \) into spherical coordinates. Learn step-by-step how to transform the integrand and limits using spherical coordinates. Suitable for multivariable calculus students.

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