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Chapter 9: Problem 61

Convert the point given in spherical coordinates to (a) rectangular coordinates and (b) cylindrical coordinates. $$ \left(8, \frac{\pi}{4}, \frac{3 \pi}{4}\right) $$

Short Answer

Expert verified
Rectangular coordinates: (4, 4, -4鈭�2); Cylindrical coordinates: (4鈭�2, 蟺/4, -4鈭�2).

Step by step solution

01

Understanding Spherical Coordinates

Spherical coordinates are given as \(( ho, heta, ext{ and } heta)\), where \(\rho\) is the radial distance from the origin, \(\phi\) is the polar angle measured from the positive z-axis, and \(\theta\) is the azimuthal angle in the xy-plane from the positive x-axis.
02

Converting to Rectangular Coordinates

Rectangular (Cartesian) coordinates \((x, y, z)\) are determined by the following formulas:\[ x = \rho \sin \phi \cos \theta \]\[ y = \rho \sin \phi \sin \theta \]\[ z = \rho \cos \phi \]Substitute the given values \(\rho = 8\), \(\phi = \frac{3\pi}{4}\), and \(\theta = \frac{\pi}{4}\) into these formulas:\[ x = 8 \sin\left(\frac{3\pi}{4}\right) \cos\left(\frac{\pi}{4}\right) \]\[ y = 8 \sin\left(\frac{3\pi}{4}\right) \sin\left(\frac{\pi}{4}\right) \]\[ z = 8 \cos\left(\frac{3\pi}{4}\right) \]
03

Calculating x, y, and z

First, calculate each sine and cosine value:- \(\sin\left(\frac{3\pi}{4}\right) = \frac{\sqrt{2}}{2}\)- \(\cos\left(\frac{3\pi}{4}\right) = -\frac{\sqrt{2}}{2}\)- \(\cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}\)- \(\sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}\)Now, substitute these values into the rectangular coordinate equations:\[ x = 8 \times \frac{\sqrt{2}}{2} \times \frac{\sqrt{2}}{2} = 8 \times \frac{1}{2} = 4 \]\[ y = 8 \times \frac{\sqrt{2}}{2} \times \frac{\sqrt{2}}{2} = 8 \times \frac{1}{2} = 4 \]\[ z = 8 \times -\frac{\sqrt{2}}{2} = -4\sqrt{2} \]So, the rectangular coordinates are \((4, 4, -4\sqrt{2})\).
04

Converting to Cylindrical Coordinates

Cylindrical coordinates \((r, \theta, z)\) consist of the radial distance \(r\) from the z-axis, angle \(\theta\), and the same z as in rectangular coordinates. The conversions to cylindrical are:\[ r = \rho \sin \phi \]\[ z = \rho \cos \phi \]Given \(\rho = 8\) and \(\phi = \frac{3\pi}{4}\), calculate:\[ r = 8 \times \sin\left(\frac{3\pi}{4}\right) = 8 \times \frac{\sqrt{2}}{2} = 4\sqrt{2} \]The angle \(\theta\) remains the same and \(z\) is calculated from previous:So, the cylindrical coordinates are \((4\sqrt{2}, \frac{\pi}{4}, -4\sqrt{2})\).

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Key Concepts

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

Spherical Coordinates
Spherical coordinates are a way of representing points in space using a set of three different values. Unlike the regular x, y, and z from rectangular coordinates, spherical coordinates involve the following elements:
  • \(\rho\) (rho), which is the radial distance from the origin to the point.
  • \(\phi\) (phi), the polar angle from the positive z-axis.
  • \(\theta\) (theta), the azimuthal angle in the xy-plane measured from the positive x-axis.
The point is expressed as \((\rho, \theta, \phi)\). This system is especially useful in fields such as physics and engineering for representing points on the surface of a sphere or when dealing with rotational symmetries. It can often simplify the mathematics involved with three-dimensional problems, especially when symmetry about a point is relevant.
Rectangular Coordinates
Rectangular coordinates, also known as Cartesian coordinates, represent positions in space using three perpendicular axes. Each point in this system is defined by three values: \((x, y, z)\). This is the most common coordinate system and underpins much of the geometry and algebra we learn.
To convert from spherical to rectangular coordinates, we use the following equations:
  • \[ x = \rho \sin \phi \cos \theta \]
  • \[ y = \rho \sin \phi \sin \theta \]
  • \[ z = \rho \cos \phi \]
These transformations allow us to find where a spherical point lies in regular 3D space, described in terms of x, y, and z, bridging the spherical perspective with the more intuitive rectangular approach.
Cylindrical Coordinates
Cylindrical coordinates mix elements of both spherical and rectangular systems to describe a point's position in a three-dimensional space. They are structured as \((r, \theta, z)\) where:
  • \(r\) is the radial distance from the z-axis in the xy-plane.
  • \(\theta\) is the same azimuthal angle as in spherical coordinates.
  • \(z\) represents the height above the xy-plane, equivalent to the z in rectangular coordinates.
The conversion from spherical coordinates to cylindrical involves using:
  • \[ r = \rho \sin \phi \]
  • The \(\theta\) value remains the same.
  • \[ z = \rho \cos \phi \]
Cylindrical coordinates are incredibly useful in scenarios involving circular symmetry, like those in mechanical engineering or when navigating objects around a central axis, providing a middle ground approach between the two other systems.

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Most popular questions from this chapter

In Problems \(75-78\), find the volume of the solid that is bounded by the graphs of the given equations. $$ z^{2}=3 x^{2}+3 y^{2}, x=0, \quad y=0, z=2, \text { first octant } $$

Evaluate the given integral by means of the indicated change of variables. $$ \begin{aligned} &\iint_{R} \frac{\cos \frac{1}{2}(x-y)}{3 x+y} d A, \text { where } R \text { is the region bounded by the }\\\ &\text { graphs of } y=x, y=x-\pi, y=-3 x+3, y=-3 x+6 \text { ; }\\\ &u=x-y, v=3 x+y \end{aligned} $$

In Problems \(63-66\), convert the points given in rectangular coordinates to spherical coordinates. $$ (1,-\sqrt{3}, 1) $$

In Problems \(35-38\), convert the point given in cylindrical coardinates to rectangular cocrdinates. $$ \left(\sqrt{3}, \frac{\pi}{3},-4\right) $$

Evaluate the given integral by means of the indicated change of variables. $$ \begin{aligned} &\iint_{R}\left(x^{2}+y^{2}\right)^{-3} d A \text { , where } R \text { is the region bounded by the circles }\\\ &x^{2}+y^{2}=2 x, x^{2}+y^{2}=4 x, x^{2}+y^{2}=2 y, x^{2}+y^{2}=6 y\\\ &u=\frac{2 x}{x^{2}+y^{2}}, v=\frac{2 y}{x^{2}+y^{2}}\left[\text { Hint: } \text { Form } u^{2}+v^{2}\right. \text { .] } \end{aligned} $$
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