To solve the given surface integral problem, let’s carefully break it down step by step.

Problem

We need to evaluate the surface integral: $\iint_S f(x, y, z) \, d\sigma$ where $f(x, y, z) = z^2$, and $S$ is the part of the sphere $x^2 + y^2 + z^2 = 4$ that lies above the plane $z = 1$.

Solution

1. Surface Description

The surface $S$ is a spherical cap of the sphere $x^2 + y^2 + z^2 = 4$, limited to $z \geq 1$. We rewrite the equation of the sphere in spherical coordinates: $x = 2\sin\phi\cos\theta, \quad y = 2\sin\phi\sin\theta, \quad z = 2\cos\phi.$ Here:

• $\phi$ is the polar angle (measured from the positive $z$-axis),
• $\theta$ is the azimuthal angle (measured in the $xy$-plane).

The cap lies above $z = 1$, so: $2\cos\phi \geq 1 \quad \implies \quad \cos\phi \geq \frac{1}{2} \quad \implies \quad \phi \in [0, \pi/3].$ The azimuthal angle $\theta$ spans $[0, 2\pi]$.

2. Surface Element

The surface element $d\sigma$ in spherical coordinates is given by: $d\sigma = R^2 \sin\phi \, d\phi \, d\theta,$ where $R = 2$ (radius of the sphere).

So, for this sphere: $d\sigma = 4\sin\phi \, d\phi \, d\theta.$

3. Integral Setup

The function $f(x, y, z) = z^2$ in spherical coordinates becomes: $f(x, y, z) = z^2 = (2\cos\phi)^2 = 4\cos^2\phi.$

Thus, the surface integral becomes: $\iint_S f(x, y, z) \, d\sigma = \int_0^{2\pi} \int_0^{\pi/3} 4\cos^2\phi \cdot 4\sin\phi \, d\phi \, d\theta.$

Simplify: $\iint_S f(x, y, z) \, d\sigma = 16 \int_0^{2\pi} d\theta \int_0^{\pi/3} \cos^2\phi \sin\phi \, d\phi.$

4. Solve the Integral

• First, evaluate the $\theta$-integral: $\int_0^{2\pi} d\theta = 2\pi.$

• Next, evaluate the $\phi$-integral. Use the substitution $u = \cos\phi$, so $du = -\sin\phi \, d\phi$. The limits change as follows: $\phi = 0 \implies u = 1, \quad \phi = \pi/3 \implies u = \frac{1}{2}.$

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

Evaluate: $\int_{1/2}^1 u^2 \, du = \left[\frac{u^3}{3}\right]_{1/2}^1 = \frac{1^3}{3} - \frac{\left(\frac{1}{2}\right)^3}{3} = \frac{1}{3} - \frac{1}{24} = \frac{8}{24} - \frac{1}{24} = \frac{7}{24}.$

5. Final Result

Multiply the results: $\iint_S f(x, y, z) \, d\sigma = 16 \cdot 2\pi \cdot \frac{7}{24} = \frac{224\pi}{24} = \frac{28\pi}{3}.$

Final Answer:

$\boxed{\frac{28\pi}{3}}$

Let me know if you need a breakdown of any step!

Related Questions:

1. How would the limits change if the plane was $z = 0$ instead of $z = 1$?
2. What is the role of spherical coordinates in simplifying such surface integrals?
3. How do you compute surface integrals over parametric surfaces in general?
4. How would the problem change if the scalar function was $f(x, y, z) = z$ instead of $z^2$?
5. How can you use symmetry to simplify similar problems?

Tip:

Always visualize the surface and identify boundaries clearly before setting up integrals in spherical or other coordinate systems!