Chapter 11: Problem 21
Make the required change in the given equation. \(2 x^{2}+2 y^{2}-4 z^{2}=0\) to spherical coordinates
Short Answer
Expert verified
In spherical coordinates, the surface is \(\phi = \arccos\left(\frac{1}{\sqrt{3}}\right)\).
Step by step solution
01
Understand Spherical Coordinates
Spherical coordinates express a point in terms of a radius and two angles. The conversion formula relates Cartesian coordinates \((x,y,z)\) to spherical coordinates \((\rho, \theta, \phi)\) with the following equations: \(x = \rho \sin \phi \cos \theta\), \(y = \rho \sin \phi \sin \theta\), \(z = \rho \cos \phi\). Here, \(\rho\) is the radial distance, \(\phi\) is the polar angle, and \(\theta\) is the azimuthal angle.
02
Substitute Cartesian with Spherical
Substitute \(x\), \(y\), and \(z\) from spherical coordinates into the given equation. The equation \( 2x^2 + 2y^2 - 4z^2 = 0 \) becomes: \(2(\rho \sin \phi \cos \theta)^2 + 2(\rho \sin \phi \sin \theta)^2 - 4(\rho \cos \phi)^2 = 0\).
03
Simplify
Factor out \(\rho^2\) from each term: \(\rho^2[2(\sin^2 \phi \cos^2 \theta + \sin^2 \phi \sin^2 \theta) - 4\cos^2 \phi] = 0\). Simplify the terms inside the bracket using \(\cos^2 \theta + \sin^2 \theta = 1\): \[2 \sin^2 \phi (\cos^2 \theta + \sin^2 \theta) - 4\cos^2 \phi = 0\] to get \(2\sin^2 \phi - 4\cos^2 \phi = 0\).
04
Solve for Angles
Further simplify by dividing the equation by 2: \(\sin^2 \phi - 2\cos^2 \phi = 0\). Reorganize the equation: \(\sin^2 \phi = 2\cos^2 \phi\). Use the identity \(\sin^2 \phi = 1 - \cos^2 \phi\), substitute it into the equation, resulting in \(1 - \cos^2 \phi = 2\cos^2 \phi\).This simplifies to \(1 = 3\cos^2 \phi\) or \(\cos \phi = \frac{1}{\sqrt{3}}\).
05
Final Spherical Equation
The solution shows that \(\phi\) is constrained such that \(\cos \phi = \frac{1}{\sqrt{3}}\). In spherical coordinates, the equation becomes a condition on \(\phi\), while \(\theta\) and \(\rho\) can take any values. Therefore, the surface described in spherical coordinates is defined by \(\phi = \arccos(\frac{1}{\sqrt{3}})\), which is a constant angle.
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 Vaia!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Coordinate Transformation
In mathematics, coordinate transformation refers to the conversion of a set of coordinates for a point from one coordinate system to another. This is an essential tool in many fields such as physics, engineering, and computer graphics.
In our exercise, we change the coordinates of a geometrical shape from the Cartesian system to the spherical system.
Spherical coordinates offer an intuitive way to describe points in three-dimensional space using:
In our exercise, we change the coordinates of a geometrical shape from the Cartesian system to the spherical system.
Spherical coordinates offer an intuitive way to describe points in three-dimensional space using:
- A radial distance ( \( \rho \) )
- A polar angle ( \( \phi \) )
- An azimuthal angle ( \( \theta \) )
Polar Angles
Polar angles are crucial in spherical coordinates because they help in understanding the position of a point relative to the poles of the coordinate system. Specifically:
In the context of the problem, simplifying the equation involves understanding how \( \phi \) relates to the other variables.
We arrived at a condition: \( \phi = \arccos(\frac{1}{\sqrt{3}}) \) .
This mean, the point's radial and azimuthal components are largely determined by this fixed polar angle.
- \( \phi \) represents the angle made with the positive z-axis
- This angle is also referred to as the polar angle
- It ranges from \( 0 \) to \( \pi \)
In the context of the problem, simplifying the equation involves understanding how \( \phi \) relates to the other variables.
We arrived at a condition: \( \phi = \arccos(\frac{1}{\sqrt{3}}) \) .
This mean, the point's radial and azimuthal components are largely determined by this fixed polar angle.
Cartesian Coordinates
Cartesian coordinates are a coordinate system that specifies each point uniquely in a plane by a pair of numerical coordinates. These are essential for representations in two or three-dimensional space, like our problem:
This set of coordinates can be converted to spherical coordinates using the formulas: \( x = \rho \sin \phi \cos \theta \), \( y = \rho \sin \phi \sin \theta \), \( z = \rho \cos \phi \).
Understanding this system is key to performing transformations between different coordinate types.
- X-coordinate ( \( x \) ): Describes the point's distance along the x-axis.
- Y-coordinate ( \( y \) ): Represents the point's distance along the y-axis.
- Z-coordinate ( \( z \) ): Describes distance along the z-axis in a 3D context.
This set of coordinates can be converted to spherical coordinates using the formulas: \( x = \rho \sin \phi \cos \theta \), \( y = \rho \sin \phi \sin \theta \), \( z = \rho \cos \phi \).
Understanding this system is key to performing transformations between different coordinate types.
Mathematical Equations
Mathematical equations are used to express relationships between different variables. In this problem, we took an equation in terms of Cartesian coordinates and transformed it into the spherical coordinate form.
By factoring and simplifying, we arrived at a condition: \( \phi = \arccos(\frac{1}{\sqrt{3}}) \), illustrating how equations transform variables between coordinate systems.
Such transformations are central to simplifying complex surfaces or understanding geometrical properties in different perspectives.
- The original problem was given as: \( 2 x^2 + 2 y^2 - 4 z^2 = 0 \)
- Substitution with spherical coordinates changed it to: \( 2(\rho \sin \phi \cos \theta)^2 + 2(\rho \sin \phi \sin \theta)^2 - 4(\rho \cos \phi)^2 = 0 \)
By factoring and simplifying, we arrived at a condition: \( \phi = \arccos(\frac{1}{\sqrt{3}}) \), illustrating how equations transform variables between coordinate systems.
Such transformations are central to simplifying complex surfaces or understanding geometrical properties in different perspectives.