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To compute the given surface integral \( \int_\sigma x^2 z \, dS \), let's go step by step. The surface \( \sigma \) is part of the sphere \( x^2 + y^2 + z^2 = 1 \) above the plane \( z = \frac{1}{2} \). We will use the standard parameterization for the sphere.

### Step 1: Parameterization of the Sphere
The parameterization of the sphere \( x^2 + y^2 + z^2 = 1 \) can be written in spherical coordinates as:
\[
x = \sin\phi \cos\theta, \quad y = \sin\phi \sin\theta, \quad z = \cos\phi,
\]
where:
- \( \phi \) is the polar angle (\( 0 \leq \phi \leq \pi \)),
- \( \theta \) is the azimuthal angle (\( 0 \leq \theta \leq 2\pi \)).

Since the surface is above \( z = \frac{1}{2} \), the polar angle \( \phi \) is restricted to \( 0 \leq \phi \leq \arccos\left(\frac{1}{2}\right) = \frac{\pi}{3} \).

### Step 2: Surface Area Element
The surface area element \( dS \) in spherical coordinates is given by:
\[
dS = \left| \frac{\partial \vec{r}}{\partial \phi} \times \frac{\partial \vec{r}}{\partial \theta} \right| d\phi \, d\theta = \sin\phi \, d\phi \, d\theta.
\]

### Step 3: Expression for the Integrand
From the parameterization:
\[
x^2 = \sin^2\phi \cos^2\theta, \quad z = \cos\phi.
\]
The integrand \( x^2 z \) becomes:
\[
x^2 z = \sin^2\phi \cos^2\theta \cos\phi.
\]

### Step 4: Setting Up the Integral
The limits for \( \phi \) are \( 0 \) to \( \frac{\pi}{3} \), and the limits for \( \theta \) are \( 0 \) to \( 2\pi \). The integral is:
\[
\int_\sigma x^2 z \, dS = \int_0^{2\pi} \int_0^{\frac{\pi}{3}} \sin^2\phi \cos^2\theta \cos\phi \sin\phi \, d\phi \, d\theta.
\]

### Step 5: Simplify and Separate the Integral
Separate the angular and polar components:
\[
\int_\sigma x^2 z \, dS = \int_0^{2\pi} \cos^2\theta \, d\theta \int_0^{\frac{\pi}{3}} \sin^3\phi \cos\phi \, d\phi.
\]

#### Solve the Angular Integral:
\[
\int_0^{2\pi} \cos^2\theta \, d\theta = \pi.
\]

#### Solve the Polar Integral:
Let \( u = \cos\phi \), so \( du = -\sin\phi \, d\phi \). The limits change as:
- When \( \phi = 0 \), \( u = 1 \).
- When \( \phi = \frac{\pi}{3} \), \( u = \frac{1}{2} \).

The integral becomes:
\[
\int_0^{\frac{\pi}{3}} \sin^3\phi \cos\phi \, d\phi = \int_1^{\frac{1}{2}} (1 - u^2) u \, (-du) = \int_{\frac{1}{2}}^1 (u^3 - u) \, du.
\]

Evaluate:
\[
\int_{\frac{1}{2}}^1 u^3 \, du = \left[ \frac{u^4}{4} \right]_{\frac{1}{2}}^1 = \frac{1}{4} - \frac{1}{64} = \frac{16}{64} - \frac{1}{64} = \frac{15}{64}.
\]
\[
\int_{\frac{1}{2}}^1 u \, du = \left[ \frac{u^2}{2} \right]_{\frac{1}{2}}^1 = \frac{1}{2} - \frac{1}{8} = \frac{4}{8} - \frac{1}{8} = \frac{3}{8}.
\]

Thus:
\[
\int_0^{\frac{\pi}{3}} \sin^3\phi \cos\phi \, d\phi = \frac{15}{64} - \frac{24}{64} = -\frac{9}{64}.
\]

### Step 6: Combine Results
\[
\int_\sigma x^2 z \, dS = \pi \cdot -\frac{9}{64} = -\frac{9\pi}{64}.
\]

The final answer is:
\[
\boxed{-\frac{9\pi}{64}}
\]

Would you like me to explain any step in detail or go further with alternative approaches?

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#### Related Questions:
1. What is the geometric interpretation of the integral \( \int_\sigma x^2 z \, dS \)?
2. How does the plane \( z = \frac{1}{2} \) affect the limits of integration for \( \phi \)?
3. Can this integral be solved using Cartesian coordinates instead of spherical coordinates?
4. What is the significance of \( dS = \sin\phi \, d\phi \, d\theta \) in spherical parametrization?
5. How would the answer change if the sphere were larger (e.g., \( x^2 + y^2 + z^2 = 4 \))?

#### Tip:
When integrating over surfaces, always carefully analyze symmetry or boundaries to simplify computations.

Let σ be a portion of the sphere x² + y² + z² = 1 above the plane z = 1/2. Compute the integral ∫∫σ x²z dS.

Learn how to compute the surface integral of x²z over the portion of the unit sphere above z = 1/2. This problem demonstrates surface integrals, spherical parameterization, and multivariable calculus techniques. Step-by-step solution included.

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