The problem involves calculating the surface integral $\iint_\sigma x^2 z \, dS$ over the portion of the sphere $x^2 + y^2 + z^2 = 1$ that lies above the plane $z = \frac{1}{2}$. Here is the solution:

Step 1: Parametrize the sphere

The sphere $x^2 + y^2 + z^2 = 1$ can be parametrized using spherical coordinates: $x = \sin\phi \cos\theta, \quad y = \sin\phi \sin\theta, \quad z = \cos\phi,$ where:

• $0 \leq \theta \leq 2\pi$,
• $0 \leq \phi \leq \pi$.

Step 2: Restrict to the upper portion of the sphere

The condition $z = \cos\phi \geq \frac{1}{2}$ restricts $\phi$ to: $\cos\phi \geq \frac{1}{2} \implies \phi \leq \cos^{-1}\left(\frac{1}{2}\right).$ Thus, $\phi \in [0, \pi/3]$ (since $\cos^{-1}(1/2) = \pi/3$).

Step 3: Compute the differential surface area $dS$

For a sphere of radius $a = 1$, the surface element is: $dS = \|\mathbf{r}_\phi \times \mathbf{r}_\theta\| \, d\phi \, d\theta = \sin\phi \, d\phi \, d\theta.$

Step 4: Express the integrand in terms of $\phi$ and $\theta$

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

Thus, the surface integral is: $\iint_\sigma x^2 z \, dS = \int_0^{2\pi} \int_0^{\pi/3} (\sin^2\phi \cos^2\theta)(\cos\phi)(\sin\phi) \, d\phi \, d\theta.$

Step 5: Simplify the integral

Separate the integral into $\phi$- and $\theta$-dependent parts: $\iint_\sigma x^2 z \, dS = \int_0^{2\pi} \cos^2\theta \, d\theta \cdot \int_0^{\pi/3} \sin^2\phi \cos\phi \sin\phi \, d\phi.$

$\theta$-integral:

$\int_0^{2\pi} \cos^2\theta \, d\theta = \pi.$

$\phi$-integral:

Simplify $\sin^2\phi \cos\phi \sin\phi = \sin^3\phi \cos\phi$: $\int_0^{\pi/3} \sin^3\phi \cos\phi \, d\phi.$ Use the substitution $u = \sin\phi$, $du = \cos\phi \, d\phi$: $\int_0^{\pi/3} \sin^3\phi \cos\phi \, d\phi = \int_0^{\sin(\pi/3)} u^3 \, du = \int_0^{\sqrt{3}/2} u^3 \, du.$ Evaluate: $\int_0^{\sqrt{3}/2} u^3 \, du = \left[\frac{u^4}{4}\right]_0^{\sqrt{3}/2} = \frac{1}{4}\left(\frac{3\sqrt{3}}{4}\right) = \frac{3\sqrt{3}}{16}.$

Step 6: Combine the results

Multiply the results of the two integrals: $\iint_\sigma x^2 z \, dS = \pi \cdot \frac{3\sqrt{3}}{16} = \frac{3\pi\sqrt{3}}{16}.$

Final Answer:

$\boxed{\frac{3\pi\sqrt{3}}{16}}$

Would you like a more detailed explanation or help with similar problems?

Relative Questions:

1. How can we compute surface integrals for general functions over a sphere?
2. What are common parametrizations for spheres in various coordinate systems?
3. How does the restriction $z > \frac{1}{2}$ affect the integration bounds?
4. What substitution techniques simplify trigonometric integrals?
5. Can we use symmetry to simplify integrals on symmetric domains?

Tip:

Always verify the domain of integration carefully, especially when applying bounds like $z > \frac{1}{2}$, as this directly affects the parametrization limits.