To compute the surface integral:

$\iint_\sigma x^2 z \, dS$

for the given portion of the sphere $x^2 + y^2 + z^2 = 1$ above the plane $z = \frac{1}{2}$, we proceed as follows:

Step 1: Parameterization of the sphere

The standard parameterization of the sphere $x^2 + y^2 + z^2 = 1$ is: $x = \sin\phi \cos\theta, \quad y = \sin\phi \sin\theta, \quad z = \cos\phi,$ where:

• $\phi$ is the polar angle ($0 \leq \phi \leq \pi$),
• $\theta$ is the azimuthal angle ($0 \leq \theta < 2\pi$).

For the region above the plane $z = \frac{1}{2}$, we restrict $\phi$ such that $\cos\phi \geq \frac{1}{2}$, or equivalently: $\phi \in [0, \arccos\frac{1}{2}].$

Step 2: Surface area element

The surface area element $dS$ for the sphere is given by: $dS = \|\vec{r}_\phi \times \vec{r}_\theta\| \, d\phi \, d\theta,$ where $\vec{r}_\phi$ and $\vec{r}_\theta$ are partial derivatives of the position vector $\vec{r}(\phi, \theta) = (\sin\phi \cos\theta, \sin\phi \sin\theta, \cos\phi)$.

The magnitude of the cross product is: $\|\vec{r}_\phi \times \vec{r}_\theta\| = \sin\phi.$ Thus: $dS = \sin\phi \, d\phi \, d\theta.$

Step 3: Expressing the integrand

The integrand $x^2 z$ becomes: $x^2 z = (\sin\phi \cos\theta)^2 \cos\phi.$

Substituting everything into the integral: $\iint_\sigma x^2 z \, dS = \int_0^{2\pi} \int_0^{\arccos\frac{1}{2}} (\sin\phi \cos\theta)^2 \cos\phi \sin\phi \, d\phi \, d\theta.$

Step 4: Simplify and evaluate

1. Factorize terms involving $\theta$ and $\phi$: $\int_0^{2\pi} \cos^2\theta \, d\theta \quad \text{and} \quad \int_0^{\arccos\frac{1}{2}} \sin^2\phi \cos\phi \sin\phi \, d\phi.$

2. Solve $\int_0^{2\pi} \cos^2\theta \, d\theta$ using the identity $\cos^2\theta = \frac{1 + \cos2\theta}{2}$: $\int_0^{2\pi} \cos^2\theta \, d\theta = \pi.$

3. Rewrite $\int_0^{\arccos\frac{1}{2}} \sin^2\phi \cos\phi \sin\phi \, d\phi$ as: $\int_0^{\arccos\frac{1}{2}} \sin^3\phi \cos\phi \, d\phi.$

   Use the substitution $u = \sin\phi$, $du = \cos\phi \, d\phi$, with limits $u = 0$ to $u = \sqrt{3}/2$ (since $\sin\phi = \sqrt{1 - \cos^2\phi}$ and $\cos\phi = \frac{1}{2}$): $\int_0^{\arccos\frac{1}{2}} \sin^3\phi \cos\phi \, d\phi = \int_0^{\sqrt{3}/2} u^3 \, du = \left[\frac{u^4}{4}\right]_0^{\sqrt{3}/2} = \frac{(\sqrt{3}/2)^4}{4} = \frac{3}{64}.$

4. Combine results: $\iint_\sigma x^2 z \, dS = \pi \cdot \frac{3}{64} = \frac{3\pi}{64}.$

Final Answer

$\boxed{\frac{3\pi}{64}}$

Let me know if you'd like additional clarification or have more questions!

Related Questions

1. How is the parameterization derived for different surfaces?
2. Why is the surface area element $dS$ dependent on the cross product?
3. What happens if the integration limits for $z$ are extended to the entire sphere?
4. Can this integral be computed in cylindrical or spherical coordinates directly?
5. What other methods exist to solve such surface integrals?

Tip: Always ensure your parameterization is consistent with the given surface and integration bounds!