Problem 8 - Thomas Calculus

Question

In Exercises \(1-16,\) 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.) $$ \begin{array}{l}{\text { Spherical cap The upper portion cut from the sphere }} \\ {x^{2}+y^{2}+z^{2}=8 \text { by the plane } z=-2}\end{array} $$

Step-by-Step Solution

Answer

Parametrize using \((u, v)\) with \(u\) for \(\phi\) and \(v\) for \(\theta\).

Step 1: Understand the Geometry of the Sphere and Plane

The given sphere's equation is \(x^2 + y^2 + z^2 = 8\), which is a sphere centered at the origin with radius \(\sqrt{8}\). The plane \(z = -2\) cuts the sphere into a spherical cap. We need to find the surface that represents the cap above this plane.

Step 2: Convert to Spherical Coordinates

To parametrize the sphere, we use spherical coordinates: \(x = \rho \sin \phi \cos \theta\), \(y = \rho \sin \phi \sin \theta\), \(z = \rho \cos \phi\), where \(\rho = \sqrt{8}\). Here, \(\phi\) is the polar angle measured from the positive z-axis, and \(\theta\) is the azimuthal angle.

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

The plane \(z = -2\) intersects the sphere where \(\rho \cos \phi = -2\). Since \(\rho = \sqrt{8}\), we find \(\phi\) by solving \(\cos \phi = -\frac{2}{\sqrt{8}} = -\frac{1}{\sqrt{2}}\), thus \(\phi = \arccos(-\frac{1}{\sqrt{2}})\). The azimuthal angle \(\theta\) varies from 0 to \(2\pi\).

Step 4: Establish the Parametrization

The parametrization of the spherical cap can be given by:\[\begin{align*}x(u, v) &= \sqrt{8} \sin u \cos v, \y(u, v) &= \sqrt{8} \sin u \sin v, \z(u, v) &= \sqrt{8} \cos u,\end{align*}\]where the parameter \(u\) ranges from 0 to \(\arccos(-\frac{1}{\sqrt{2}})\) and \(v\) ranges from 0 to \(2\pi\).

Key Concepts

Spherical Coordinates

Spherical coordinates provide a way to describe points in three-dimensional space using three values:

• the radial distance (\(\rho\)) from the origin to the point,
• the polar angle (\(\phi\)) measured from the positive z-axis,
• and the azimuthal angle (\(\theta\)) measured in the xy-plane from the positive x-axis.

In essence, rather than using rectangular coordinates (\(x, y, z\)), which consist of perpendicular distances along the x, y, and z axes, spherical coordinates describe a sphere of radius \(\rho\).
For instance, a point given by coordinates \(x = \rho \sin \phi \cos \theta\), \(y = \rho \sin \phi \sin \theta\), and \(z = \rho \cos \phi\) adapts well to objects having radial symmetry, like spheres.

Spherical Cap

A spherical cap is a portion of a sphere cut off by a plane. Imagine slicing a basketball with a flat plane to get a dome-shaped piece—that's a spherical cap. The size and position of this cap depends on how high or low the plane cuts through the sphere.
In the exercise you're tackling, the plane cuts the sphere at \(z = -2\), resulting in a spherical cap above this plane. This particular surface can be expressed parametrically, meaning with parameters that vary independently. The formula for the sphere is vital because it helps define where this cap sits mathematically.

Sphere

A sphere is a perfectly symmetrical three-dimensional shape, where every point on its surface is equidistant from its center. In mathematical terms, a sphere centered at the origin with radius \(r\) is defined by the equation \(x^2 + y^2 + z^2 = r^2\).
The exercise above refers to a sphere with the equation \(x^2 + y^2 + z^2 = 8\), meaning it has a radius equal to \(\sqrt{8}\).

• This equation tells us the distance of all points on the surface of the sphere from the origin is the same: \(\sqrt{8}\).
• A sphere can be parameterized using spherical coordinates, providing an efficient manner to find surfaces like caps.

Plane Intersection

A plane intersection with a sphere can carve out various shapes. Specifically, when the plane intersects a sphere, it often forms a shape similar to a disk on the sphere's surface. However, when a plane does not pass through the center of the sphere, it creates more complex surfaces such as caps or segments.
In this exercise, the plane given by \(z = -2\) intersects the sphere \(x^2 + y^2 + z^2 = 8\), forming a spherical cap above \(z = -2\).

• To understand where the plane cuts the sphere, we solve for the polar angle \(\phi\) at the intersection.
• This is done by recognizing \(\cos \phi = \frac{-2}{\sqrt{8}}\), leading to the polar angle \(\phi = \arccos\left(\frac{-1}{\sqrt{2}}\right)\).

The specific placement of the cap is determined by this intersection, emphasizing the importance of both spherical and planar geometry in describing the surface.