To solve the problem, we need to compute the surface integral of $x^2 z$ over the portion of the sphere $x^2 + y^2 + z^2 = 1$ above the plane $z = \frac{1}{2}$. Let us proceed with the steps.

Step 1: Parameterization of the Sphere

The sphere $x^2 + y^2 + z^2 = 1$ can be parameterized as: $x = \sin\phi \cos\theta, \quad y = \sin\phi \sin\theta, \quad z = \cos\phi,$ where $\phi$ is the polar angle ($0 \leq \phi \leq \pi$) and $\theta$ is the azimuthal angle ($0 \leq \theta < 2\pi$).

Since the surface is only the portion above $z = \frac{1}{2}$, we need to determine the range of $\phi$. Since $z = \cos\phi$, the condition $z \geq \frac{1}{2}$ implies: [ \cos\phi \geq \frac{1}{2} \quad \Rightarrow \quad \phi \in [0, \arccos(\frac{1}{2})] = [0, \frac{\pi}{3}]. ]

Thus, the parameter ranges are: [ \phi \in [0, \frac{\pi}{3}], \quad \theta \in [0, 2\pi]. ]

Step 2: Surface Element $dS$

The surface element $dS$ on a sphere is given by: $dS = |\mathbf{r}_\phi \times \mathbf{r}_\theta| d\phi d\theta,$ where $\mathbf{r}_\phi$ and $\mathbf{r}_\theta$ are partial derivatives of the parameterization: $\mathbf{r}(\phi, \theta) = (\sin\phi\cos\theta, \sin\phi\sin\theta, \cos\phi).$

The cross product $|\mathbf{r}_\phi \times \mathbf{r}_\theta|$ simplifies to $\sin\phi$. Thus: $dS = \sin\phi \, d\phi \, d\theta.$

Step 3: Setting Up the Integral

The function to be integrated is $x^2 z$. Substituting the parameterization: $x^2 = (\sin\phi\cos\theta)^2 = \sin^2\phi\cos^2\theta, \quad z = \cos\phi.$ Thus: $x^2 z = \sin^2\phi \cos^2\theta \cos\phi.$

The surface integral becomes: $\iint_\sigma x^2 z \, dS = \int_0^{2\pi} \int_0^{\pi/3} \sin^2\phi \cos^2\theta \cos\phi \sin\phi \, d\phi \, d\theta.$

Step 4: Separating the Integral

Separate the integral into $\phi$- and $\theta$-dependent parts: $\int_0^{2\pi} \cos^2\theta \, d\theta \quad \text{and} \quad \int_0^{\pi/3} \sin^2\phi \cos\phi \sin\phi \, d\phi.$

$\theta$-Integral:

$\int_0^{2\pi} \cos^2\theta \, d\theta = \int_0^{2\pi} \frac{1 + \cos(2\theta)}{2} \, d\theta = \pi.$

$\phi$-Integral:

Simplify $\sin^2\phi \cos\phi \sin\phi$ as $\sin^3\phi \cos\phi$: $\int_0^{\pi/3} \sin^3\phi \cos\phi \, d\phi = \int_0^{\pi/3} \frac{1}{4} \frac{d}{d\phi} (\sin^4\phi) \, d\phi.$ This evaluates to: $\frac{1}{4} [\sin^4\phi]_0^{\pi/3} = \frac{1}{4} \left( \sin^4\frac{\pi}{3} - \sin^4 0 \right) = \frac{1}{4} \left( \left( \frac{\sqrt{3}}{2} \right)^4 - 0 \right) = \frac{1}{4} \cdot \frac{9}{16} = \frac{9}{64}.$

Step 5: Combine Results

Multiply the results: $\iint_\sigma x^2 z \, dS = \pi \cdot \frac{9}{64} = \frac{9\pi}{64}.$

Thus, the final result is: $\boxed{\frac{9\pi}{64}}.$

Let me know if you'd like clarification or additional steps! Here are some follow-up questions for further exploration:

1. Can you explain how to derive the surface element $dS$ for other parameterizations?
2. How would the integral change if the sphere were shifted in space?
3. What happens if the plane $z = \frac{1}{2}$ intersects a different section of the sphere?
4. How do you interpret surface integrals geometrically?
5. Can this problem be solved using spherical coordinates directly instead of parameterization?

Tip: When working with surface integrals, always verify your limits of integration to ensure you cover the correct surface region.