Problem 7
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.) $$ \begin{array}{l}{\text { Spherical band The portion of the sphere } x^{2}+y^{2}+z^{2}=3} \\ {\text { between the planes } z=\sqrt{3} / 2 \text { and } z=-\sqrt{3} / 2}\end{array} $$
Step-by-Step Solution
Verified Answer
Parametrization: \(x(u,v) = \sqrt{3}\sin(u)\cos(v)\), \(y(u,v) = \sqrt{3}\sin(u)\sin(v)\), \(z(u,v) = \sqrt{3}\cos(u)\); with \(u \in [\pi/3, 2\pi/3]\) and \(v \in [0, 2\pi]\).
1Step 1: Understand the Problem
We're tasked with parametrizing the spherical band on the sphere defined by the equation \(x^2 + y^2 + z^2 = 3\) that lies between the planes \(z = \sqrt{3}/2\) and \(z = -\sqrt{3}/2\). By parametrization, we need to express the surface in terms of two parameters, usually \(u\) and \(v\). This allows us to describe every point on the surface using these parameters.
2Step 2: Define the Sphere Parametrization
For the sphere \(x^2+y^2+z^2 = 3\), a common parametrization using spherical coordinates is: \[x = \sqrt{3} \sin\theta \cos\phi\] \[y = \sqrt{3} \sin\theta \sin\phi\] \[z = \sqrt{3} \cos\theta\] where \(\theta\) is the polar angle ranging between \(0\) and \(\pi\), and \(\phi\) is the azimuthal angle ranging between \(0\) and \(2\pi\). However, the values of \(\theta\) are restricted by the planes given in the problem.
3Step 3: Determine the Range of Parameters
The planes \(z = \sqrt{3}/2\) and \(z = -\sqrt{3}/2\) impose restrictions on \(\theta\). Using \(z = \sqrt{3} \cos\theta\), solve for \(\theta\):1. For \(z = \sqrt{3}/2\), we have \(\sqrt{3} \cos\theta = \sqrt{3}/2\) \(\Rightarrow \cos\theta = 1/2\). So \(\theta\) can be \(\pi/3\) (due to positive value). 2. For \(z = -\sqrt{3}/2\), \(\sqrt{3} \cos\theta = -\sqrt{3}/2\) \(\Rightarrow \cos\theta = -1/2\). Therefore, \(\theta\) can be \(2\pi/3\) (as we look for values between 0 and \(\pi\)). Therefore, \(\theta\) ranges from \(\pi/3\) to \(2\pi/3\), while \(\phi\) varies from 0 to \(2\pi\).
4Step 4: Write the Parametrization
Using the determined ranges of \(\theta\) and \(\phi\), the parametrization for the spherical band is:\[ \begin{align*} x(u, v) &= \sqrt{3} \sin(u) \cos(v), \ y(u, v) &= \sqrt{3} \sin(u) \sin(v), \ z(u, v) &= \sqrt{3} \cos(u), \end{align*} \]where \(u\) ranges from \(\pi/3\) to \(2\pi/3\) and \(v\) from \(0\) to \(2\pi\).
Key Concepts
Spherical CoordinatesSpherical SurfacePolar Angle RangeAzimuthal AngleRestriction by Planes
Spherical Coordinates
Spherical coordinates are a system used to specify points in three-dimensional space. Unlike Cartesian coordinates which are represented by \((x, y, z)\), spherical coordinates use three different parameters: \(r\), \(\theta\), and \(\phi\). Each of these parameters describes a different aspect of a point's position:
- The radius \(r\) is the distance from the origin to the point.
- The polar angle \(\theta\) is the angle between the positive z-axis and the line connecting the origin to the point.
- The azimuthal angle \(\phi\) is the angle in the xy-plane from the positive x-axis.
Spherical Surface
A spherical surface is a set of points in three-dimensional space that are all equidistant from a single point, known as the center. This distance is called the radius. The equation of a sphere in Cartesian coordinates is \(x^2 + y^2 + z^2 = r^2\), where \(r\) is the radius of the sphere.
The given exercise involves a spherical surface, specifically a sphere with its equation given as \(x^2 + y^2 + z^2 = 3\). Here, the radius is the square root of 3. Being a perfect three-dimensional shape, the sphere is symmetrical in all directions, making it a suitable candidate for spherical coordinates. By using spherical coordinates, we can easily describe precise sections or bands of the sphere, like the one given in the problem statement.
The given exercise involves a spherical surface, specifically a sphere with its equation given as \(x^2 + y^2 + z^2 = 3\). Here, the radius is the square root of 3. Being a perfect three-dimensional shape, the sphere is symmetrical in all directions, making it a suitable candidate for spherical coordinates. By using spherical coordinates, we can easily describe precise sections or bands of the sphere, like the one given in the problem statement.
Polar Angle Range
The polar angle \(\theta\) is crucial in defining the position of a point on a sphere. It ranges from 0 to \(\pi\) and determines how far a point is from the z-axis:
- When \(\theta = 0\), the point is at the top of the sphere on the positive z-axis.
- When \(\theta = \pi\), the point is at the bottom of the sphere on the negative z-axis.
Azimuthal Angle
The azimuthal angle, denoted as \(\phi\), is the angle on the xy-plane from the positive x-axis. It gives the point's direction from the z-axis once the polar angle sets the height.
In a complete spherical coordinate system, it ranges from 0 to \(2\pi\), which essentially covers all directions in the horizontal plane. This provides a full 360-degree rotation around the z-axis.
In the example given, \(\phi\) is unrestricted, allowing for a complete rotation as the section of the sphere is like a band encircling it completely. This forms the circular pattern needed for visualizing spherical bands or cylindrical sections.
In a complete spherical coordinate system, it ranges from 0 to \(2\pi\), which essentially covers all directions in the horizontal plane. This provides a full 360-degree rotation around the z-axis.
In the example given, \(\phi\) is unrestricted, allowing for a complete rotation as the section of the sphere is like a band encircling it completely. This forms the circular pattern needed for visualizing spherical bands or cylindrical sections.
Restriction by Planes
When portions of a spherical surface are defined by certain bounds, such as those created by intersecting planes, parameters like \(\theta\) and \(\phi\) become crucial. In our exercise, the sphere is bounded by the planes \(z = \sqrt{3}/2\) and \(z = -\sqrt{3}/2\). This means the sphere is constrained to the narrow region of the spherical band:
- The upper plane \(z = \sqrt{3}/2\) corresponds to \(\theta = \pi/3\). This cuts the sphere near its top.
- The lower plane \(z = -\sqrt{3}/2\) aligns with \(\theta = 2\pi/3\), effectively restricting the region near the bottom.
Other exercises in this chapter
Problem 6
Give a formula \(\mathbf{F}=M(x, y) \mathbf{i}+N(x, y) \mathbf{j}\) for the vector field in the plane that has the properties that \(\mathbf{F}=0\) at \((0,0)\)
View solution Problem 7
Verify the conclusion of Green's Theorem by evaluating both sides of Equations \((3)\) and \((4)\) for the field \(\mathbf{F}=M \mathbf{i}+N \mathbf{j}\) . Take
View solution Problem 7
In Exercises \(7-12,\) find a potential function \(f\) for the field \(\mathbf{F}\). $$\mathbf{F}=2 x \mathbf{i}+3 y \mathbf{j}+4 z \mathbf{k}$$
View solution Problem 7
In Exercises \(1-8,\) integrate the given function over the given surface. Parabolic dome \(H(x, y, z)=x^{2} \sqrt{5-4 z},\) over the parabolic dome \(z=1-x^{2}
View solution