/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 5 Describe the graph of the equati... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Describe the graph of the equation in three dimensions. (a) \(\rho=3\) (b) \(\phi=\pi / 6\) (c) \(\theta=\pi / 3\)

Short Answer

Expert verified
(a) Sphere with radius 3, (b) Cone with \( \phi = \pi/6 \), (c) Half-plane at \( \theta = \pi/3 \).

Step by step solution

01

Understanding the Equation (a)

The equation \( \rho = 3 \) represents a sphere in spherical coordinates. In three-dimensional space, \( \rho \) indicates the radius from the origin to any point on the sphere. Hence, for \( \rho = 3 \), we have a sphere centered at the origin with a radius of 3.
02

Plotting the Sphere

To plot \( \rho = 3 \) in 3D space, imagine all points that are a constant distance of 3 units from the origin. This creates a spherical surface surrounding the origin at a radius of 3.
03

Understanding the Equation (b)

The equation \( \phi = \pi / 6 \) in spherical coordinates represents a conical surface. The angle \( \phi \) is the angle from the positive z-axis to the radial line. This equation describes a cone with its vertex at the origin and opening symmetrically about the z-axis with an angle of \( \pi / 6 \).
04

Plotting the Cone

To visualize \( \phi = \pi / 6 \), plot a cone extending from the origin and encompassing all points where the angle with the positive z-axis is \( \pi / 6 \). The angle sweeps radially, similar to a traditional 3D cone shape.
05

Understanding the Equation (c)

The equation \( \theta = \pi / 3 \) represents a vertical half-plane in spherical coordinates. The angle \( \theta \) is the azimuthal angle in the xy-plane, measured from the positive x-axis. The result is a plane slicing through the z-axis at the specified angle in the xy-plane.
06

Plotting the Half-Plane

For \( \theta = \pi / 3 \), imagine a plane originating from the z-axis and extending outward. It is specifically at an angle of \( \pi / 3 \) from the positive x-axis, sweeping vertically throughout the 3D space.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

3D graph plotting
Three-dimensional (3D) graph plotting is a powerful tool that allows us to visually understand complex mathematical equations by representing them in a space that can be manipulated and examined from various angles. When dealing with spherical coordinates, 3D graph plotting becomes particularly useful because it provides a means to visualize surfaces that are not easily represented in a two-dimensional framework.

For example, consider plotting the equation \( \rho = 3 \). In three-dimensional space, this describes a sphere centered at the origin with a radius of 3 units. Imagine a balloon whose surface is at a constant distance from a central point—now, plot all those points in 3D. Similarly, the equation \( \phi = \pi / 6 \) creates a conical surface. It's like drawing a party hat where the tip is touching the center of the origin and the angle with the vertical axis is \( \pi / 6 \). For \( \theta = \pi / 3 \), plot a slice or a plane cutting through space from the z-axis, making an angle with the x-axis in the xy-plane. Each of these equations provides unique surfaces or shapes in a 3D setting.
  • 3D plotting renders mathematical surfaces visually accessible.
  • It charts points and lines that define the structure in space.
Using software tools like graphing calculators or computer applications that support 3D graphing, students can manipulate these view angles to see how different components interact spatially.
spherical equations
Spherical equations use a coordinate system alternative to the Cartesian system. In spherical coordinates, any point in 3D space is defined by three values: \( \rho \), \( \phi \), and \( \theta \). These correspond to the radial distance, the polar angle, and the azimuthal angle, respectively.

The equation \( \rho = 3 \) represents a sphere because \( \rho \) specifies how far the points are from the origin. Regardless of angles \( \phi \) and \( \theta \), all points are constant at a radius of 3. Thus, it forms a perfect spherical surface around the center.

Next, \( \phi = \pi / 6 \) describes a cone. Here, \( \phi \) affects the points by controlling the angle to the z-axis. When \( \phi \) is held constant, it forces all radial paths to form a conical shape, symmetrical around the vertical axis.

The equation \( \theta = \pi / 3 \) might seem different, as \( \theta \) determines orientation in the horizontal plane. It's like plotting a specific slice through a cake at \( \pi / 3 \) radian from the x-axis—it extends vertically through space.
  • Spherical coordinates simplify the description of spherical shapes.
  • They focus on radius and angles, providing intuitive visual cues.
Understanding these equations unlocks new ways to describe and plot complex geometries.
mathematical visualization
Mathematical visualization involves the representation and interpretation of mathematical concepts and problems graphically. When dealing with spherical equations, visualization helps to comprehend and analyze the shapes dictated by these equations in a 3D coordinate system.

Let's take the example of the sphere defined by \( \rho = 3 \). Visualizing it allows one to see how every point on this spherical surface is exactly 3 units away from the origin, no matter the angle. Such visualization assists in understanding intrinsic properties like symmetry and surface area that are not readily apparent otherwise.

For the conical representation \( \phi = \pi / 6 \), mathematical visualization displays how the cone narrows symmetrically as it extends from its vertex at the origin. This shape is easier to understand when you see its cross-section lined symmetrically around the z-axis. Similarly, for \( \theta = \pi / 3 \), visualization paints a clear picture of how this equation carves out a vertical plane, revealing how it slices through space and interacts with other forms.

Through mathematical visualization:
  • Abstract concepts become concrete and graspable.
  • Complex shapes are broken down into understandable, interactive parts.
  • It improves comprehension by connecting algebraic expressions to their geometric counterparts.
By using mathematical visualization, students can enhance their spatial reasoning and understanding of how geometric principles apply in higher dimensions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A lamina having area mass density \(\delta(x, y)\) has the shape of the region bounded by the graphs of the equations. Set up an iterated double integral that can be used to find the mass of the lamina. $$ \delta(x, y)=y^{2}: \quad y=e^{x}, \quad x=0, \quad x=1, \quad y=0 $$

Sketch the region of integration for the iterated integral. $$ \int_{-2}^{-1} \int_{3 y}^{2 y} f(x, y) d x d y $$

We can sometimes evaluate a double integral by changing it to a single integral. |a) Use polar coordinates to change the given double integral to a single integral involving only the variable \(r .\) (b) Use Simpson's rule, with \(n=4,\) to approximate the single integral. \(\iint_{R} \sqrt{1+\left(x^{2}+y^{2}\right)^{2}} d A\) \(R\) is the region in the first quadrant bounded by the circle \(x^{2}+y^{2}=4\) and the coordinate axes.

Use spherical coordinates. If the earth is assumed to be spherical with a radius of 6370 kilometers, the density \(\delta\) (in \(\mathrm{kg} \mathrm{m}^{3}\) ) of the atmosphere a distance \(\rho\) meters from the center of the earth can be approximated by $$ \begin{aligned} \delta &=619.09-\left(9.7 \times 10^{-5}\right) \rho \\ \text { for } 6,370,000 \leq & \rho \leq 6,373,000 \end{aligned} $$ (a) Estimate the mass of the atmosphere between ground level and an altitude of 3 kilometers. (b) The atmosphere extends beyond an altitude of 100 kilometers and has a total mass of approximately \(5.1 \times 10^{18}\) kilograms. What percentage of the mass is in the lowest 3 kilometers of the atmosphere?

Reverse the order of integration, and evaluate the resulting integral. $$ \int_{1}^{e} \int_{0}^{\ln x} y d y d x $$

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.