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Chapter 15: Problem 4

\(3-4\) Change from rectangular to spherical coordinates. $$\text { (a) }(0, \sqrt{3}, 1) \quad \text { (b) }(-1,1, \sqrt{6})$$

Short Answer

Expert verified
(a) (2, \(\frac{\pi}{2}, \frac{\pi}{3}\)); (b) (2\sqrt{2}, \(\frac{3\pi}{4}, \frac{\pi}{6}\))

Step by step solution

01

Identify the Rectangular Coordinates

We begin with the given rectangular coordinates. In part (a), the coordinates are \( (0, \sqrt{3}, 1) \) and in part (b), the coordinates are \( (-1, 1, \sqrt{6}) \). These correspond to the \( x, y, \) and \( z \) values.
02

Calculate \( \rho \)

\( \rho \) is the distance from the origin to the point and can be calculated using the formula \( \rho = \sqrt{x^2 + y^2 + z^2} \). For part (a), we have \( \rho = \sqrt{0^2 + (\sqrt{3})^2 + 1^2} = \sqrt{4} = 2 \). For part (b), \( \rho = \sqrt{(-1)^2 + 1^2 + (\sqrt{6})^2} = \sqrt{8} = 2\sqrt{2} \).
03

Calculate \( \theta \)

\( \theta \) is the angle in the \( xy \)-plane from the positive \( x \)-axis and can be calculated using \( \theta = \tan^{-1}\left(\frac{y}{x}\right) \). For part (a), since \( x = 0 \), we define \( \theta = \frac{\pi}{2} \) if \( y > 0 \). Therefore, \( \theta = \frac{\pi}{2} \). For part (b), \( \theta = \tan^{-1}\left(\frac{1}{-1}\right) = \tan^{-1}(-1) = \frac{3\pi}{4} \) (in the second quadrant).
04

Calculate \( \phi \)

\( \phi \) is the angle from the positive \( z \)-axis and is found using \( \phi = \cos^{-1}\left(\frac{z}{\rho}\right) \). For part (a), \( \phi = \cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3} \). For part (b), \( \phi = \cos^{-1}\left(\frac{\sqrt{6}}{2\sqrt{2}}\right) = \cos^{-1}(\sqrt{\frac{3}{4}}) = \frac{\pi}{6} \).
05

Write the Spherical Coordinates

Using the values from the calculations above, write the spherical coordinates. For part (a), they are \( (2, \frac{\pi}{2}, \frac{\pi}{3}) \). For part (b), they are \( (2\sqrt{2}, \frac{3\pi}{4}, \frac{\pi}{6}) \).

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Key Concepts

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

Rectangular Coordinates
Rectangular coordinates are like the grid on a sheet of graph paper. Each point on this grid can be described by three values:
  • The x-coordinate (how far along the horizontal axis)
  • The y-coordinate (how far along the vertical axis)
  • The z-coordinate (how high or low, adding a 3D element)
In this exercise, we convert these rectangular coordinates to spherical coordinates. Visualizing each rectangular coordinate as a position in space can help.
Understanding where a point lies based on its x, y, and z values is foundational to switching to a new coordinate system.
Coordinate Conversion
Switching between rectangular and spherical coordinates is known as coordinate conversion. This is crucial in many applications like physics and engineering where analyzing objects or systems in different coordinate systems provides better clarity or simplicity.
The core process involves three steps:
  • Finding \( \rho \), the distance from the origin.
  • Determining the angle \( \theta \) in the xy-plane.
  • Calculating the angle \( \phi \) from the vertical axis.
These conversions make it easier to solve problems by allowing us to choose the most convenient method of representation.
Trigonometric Functions
Trigonometric functions are deeply tied to angles and relationships in geometry. They are essential for calculating \( \theta \) and \( \phi \) in spherical coordinates. For instance, the arctan or \( \tan^{-1} \) function helps in finding \( \theta \), given by:
  • \( \theta = \tan^{-1}\left( \frac{y}{x} \right) \)
Similarly, the arccos or \( \cos^{-1} \) function calculates \( \phi \):
  • \( \phi = \cos^{-1}\left( \frac{z}{\rho} \right) \)
By using these functions, angles can be accurately measured, ensuring precise spherical coordinates.
Distance Formula
The distance formula, \( \rho = \sqrt{x^2 + y^2 + z^2} \), plays a critical role in converting coordinates. It calculates the distance from the origin to any given point in 3D space, giving us the value of \( \rho \) for spherical coordinates.
This formula extends the Pythagorean theorem into three dimensions:
  • The square root of the sum of squares of each coordinate.
  • It defines the spherical radius in this setting.
This foundational formula emphasizes the relationship between distance and space, crucial for any coordinate conversion.

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Most popular questions from this chapter

Evaluate the integral by making an appropriate change of variables. \(\iint_{R}(x+y) e^{x^{2}-y^{2}} d A,\) where \(R\) is the rectangle enclosed by the lines \(x-y=0, x-y=2, x+y=0,\) and \(x+y=3\)

\(21-22\) Use a computer algebra system to find the mass, center of mass, and moments of inertia of the lamina that occupies the region \(D\) and has the given density function. \(D\) is enclosed by the cardioid \(r=1+\cos \theta\) \(\rho(x, y)=\sqrt{x^{2}+y^{2}}\)

Evaluate the integral by reversing the order of integration. $$\int_{0}^{\sqrt{\pi}} \int_{y}^{\sqrt{\pi}} \cos \left(x^{2}\right) d x d y$$

When studying the formation of mountain ranges, geologists estimate the amount of work required to lift a mountain from sea level. Consider a mountain that is essentially in the shape of a right circular cone. Suppose that the weight density of the material in the vicinity of a point \(P\) is \(g(P)\) and the height is \(h(P) .\) (a) Find a definite integral that represents the total work done in forming the mountain. (b) Assume that Mount Fuji in Japan is in the shape of a right circular cone with radius \(62,000\) ft, height \(12,400 \mathrm{ft}\) , and density a constant 200 \(\mathrm{lb} / \mathrm{ft}^{3} .\) How much work was done in forming Mount Fuji if the land was initially at sea level?

The latitude and longitude of a point \(P\) in the Northern Hemisphere are related to spherical coordinates \(\rho, \theta, \phi\) as follows. We take the origin to be the center of the earth and the positive \(z\) -axis to pass through the North Pole. The positive \(x\) -axis passes through the point where the prime meridian (the meridian through Greenwich, England) intersects the equator. Then the latitude of \(P\) is \(\alpha=90^{\circ}-\phi^{\circ}\) and the longitude is \(\beta=360^{\circ}-\theta^{\circ} .\) Find the great-circle distance from Los Angeles (lat. \(34.06^{\circ} \mathrm{N},\) long. \(118.25^{\circ} \mathrm{W}\) ) to Montreal (lat. \(45.50^{\circ} \mathrm{N},\) long. \(73.60^{\circ} \mathrm{W} ) .\) Take the radius of the earth to be 3960 \(\mathrm{mi} .\) (A great circle is the circle of intersection of a sphere and a plane through the center of the sphere.)
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