Problem 7 - Thomas Calculus

Question

Find a parametrization of the surface. (There are many correct ways to do these, so your answers may not be the same as those in the back of the book.) Spherical band The portion of the sphere \(x^{2}+y^{2}+z^{2}=3\) between the planes \(z=\sqrt{3} / 2\) and \(z=-\sqrt{3} / 2\)

Step-by-Step Solution

Answer

Parametrization is given by \(x = \sqrt{3}\sin\theta\cos\phi\), \(y = \sqrt{3}\sin\theta\sin\phi\), \(z = \sqrt{3}\cos\theta\) with \(\pi/3 \leq \theta \leq 2\pi/3\).

Step 1: Understanding the Problem

We're given a sphere defined by the equation \(x^2 + y^2 + z^2 = 3\) and we need to find a parametrization for the surface that lies between the planes \(z = \sqrt{3}/2\) and \(z = -\sqrt{3}/2\). This surface is a spherical band.

Step 2: Spherical Coordinates

To parameterize the surface, we'll use spherical coordinates, where \(x = \rho \sin \theta \cos \phi\), \(y = \rho \sin \theta \sin \phi\), and \(z = \rho \cos \theta\). In our sphere, \(\rho = \sqrt{3}\) because \(x^2 + y^2 + z^2 = \rho^2 = 3\).

Step 3: Determine Limits for \(\theta\) and \(\phi\)

The variable \(\theta\) varies from 0 to \(\pi\), but it needs to be restricted depending on the z-values. Using \(z = \rho \cos \theta\) gives \(\theta\) when \(z = \sqrt{3}/2\) is \(\cos^{-1}(1/2)\) which equals \(\pi/3\). Similarly, for \(z = -\sqrt{3}/2\), \(\theta = \cos^{-1}(-1/2)\), which equals \(2\pi/3\). Thus, \(\theta\) varies from \(\pi/3\) to \(2\pi/3\). \(\phi\) can vary from 0 to \(2\pi\) to cover the whole band.

Step 4: Writing the Parametrization

Using the spherical coordinates and limits we found, the parametrization can be expressed as: \(x(\theta, \phi) = \sqrt{3} \sin(\theta) \cos(\phi)\), \(y(\theta, \phi) = \sqrt{3} \sin(\theta) \sin(\phi)\), \(z(\theta) = \sqrt{3} \cos(\theta)\) where \(\pi/3 \leq \theta \leq 2\pi/3\), and \(0 \leq \phi \leq 2\pi\).

Key Concepts

Spherical Coordinates

Spherical coordinates offer a unique way to describe points in three-dimensional space. Instead of using Cartesian coordinates (x, y, z), spherical coordinates use a radius and two angles to define a point. This system is particularly useful for dealing with spherical surfaces, or any shapes with circular symmetry.

• The radius \(\rho\) is the distance from the origin to the point in space. For a sphere of radius \(\sqrt{3}\), \(\rho\) remains constant.
• The angle \(\phi\) measures the rotation around the vertical z-axis; it ranges from 0 to \(2\pi\), allowing full traversal of a horizontal circle.
• The angle \(\theta\) measures the inclination from the positive z-axis, ranging from 0 to \(\pi\). This allows for specifying how "high" or "low" a point is over or under the xy-plane.

For our exercise, the sphere is defined by the equation \(x^2 + y^2 + z^2 = 3\), translating to \(\rho = \sqrt{3}\). This means we're mainly working with adjusting the angles \(\theta\) and \(\phi\) to describe different points on the spherical band.

Spherical Band

A spherical band is a specific portion of a sphere, bounded by two parallel planes. In the given exercise, the spherical band is carved out of the sphere \(x^2 + y^2 + z^2 = 3\) by the planes \(z = \sqrt{3}/2\) and \(z = -\sqrt{3}/2\). Such a band looks like a ring or belt circumscribing the sphere.

To find the band, we need to ensure that our parametrization takes into account the limitations imposed by these planes. This translates to restricting the angle \(\theta\) instead of \(z\) directly, due to its dependency on the angle:

• For \(z = \sqrt{3}/2\): we have \(\theta = \cos^{-1}(1/2)\), which is \(\pi/3\).
• For \(z = -\sqrt{3}/2\): similarly, \(\theta = \cos^{-1}(-1/2)\), yielding \(2\pi/3\).

Thus, \(\theta\) ranges from \(\pi/3\) to \(2\pi/3\), forming the spherical band between the specified planes on the sphere.

Parametrization Techniques

Parametrization is the method of representing a geometric surface through variables, here \(\theta\) and \(\phi\), which makes calculations involving the surface more straightforward. With spherical coordinates, we can elegantly describe the surface as well as its bounds without algebraic complexity.

To parametrize a piece of a sphere, follow these steps:

• Express each Cartesian coordinate in terms of spherical coordinates \(\rho\), \(\theta\), and \(\phi\):
• \(x = \rho \sin \theta \cos \phi\)
• \(y = \rho \sin \theta \sin \phi\)
• \(z = \rho \cos \theta\)

For the exercise's spherical band, the specific parametrization would be:
- \(x(\theta, \phi) = \sqrt{3} \sin(\theta) \cos(\phi)\)
- \(y(\theta, \phi) = \sqrt{3} \sin(\theta) \sin(\phi)\)
- \(z(\theta) = \sqrt{3} \cos(\theta)\)
where \(\pi/3 \leq \theta \leq 2\pi/3\) and \(0 \leq \phi \leq 2\pi\). This efficiently describes every point on the spherical band's surface, making it clear how each aspect of the surface is mathematically constructed.