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?

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.