Change from rectangular to spherical coordinates. (Let p 2 0, 0 < 0 < 2n, and 0 < ¢ < n.)(a)(2, 2/3, 4V3)(р, в, ф) %3D(b)(0, -5, -5)(p, 0, $) =

Answered: Change from rectangular to spherical…

Change from rectangular to spherical coordinates. (Let p 2 0, 0 < 0 < 21, and 0 s¢ s n.) (a) (2, 2/3,4V3) (p, 6, ф) - (b) (0, -5, -5) (p, 6, ф) 3 %3D

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter11: Topics From Analytic Geometry

Section11.2: Ellipses

Problem 23E

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Question

Change from rectangular to spherical coordinates. (Let p 2 0, 0 < 0 < 2n, and 0 < ¢ < n.)
(a)
(2, 2/3, 4V3)
(р, в, ф) %3D
(b)
(0, -5, -5)
(p, 0, $) =

Expert Solution

Step 1

To change the coordinate from rectangular to spherical we can use these formulae:

$ρ = \sqrt{x^{2} + y^{2} + z^{2}} θ = \tan^{- 1} (\frac{y}{x}) ϕ = \cos^{- 1} (\frac{z}{\sqrt{x^{2} + y^{2} + z^{2}}})$

Step 2

Part (a):

Given that $(x, y, z) = (2, 2 \sqrt{3}, 4 \sqrt{3})$

$\begin{array}{rcl} x & = & 2, y = 2 \sqrt{3}, z = 4 \sqrt{3} \\ ρ & = & \sqrt{2^{2} + {(2 \sqrt{3})}^{2} + {(4 \sqrt{3})}^{2}} \\ = & \sqrt{4 + 12 + 48} \\ = & \sqrt{64} \\ = & 8 \end{array}$

$\begin{array}{rcl} θ & = & \tan^{- 1} (\frac{y}{x}) \\ = & \tan^{- 1} (\frac{2 \sqrt{3}}{2}) \\ = & \tan^{- 1} (\sqrt{3}) \\ = & \frac{π}{3} \end{array}$

$\begin{array}{rcl} ϕ & = & \cos^{- 1} (\frac{z}{\sqrt{x^{2} + y^{2} + z^{2}}}) \\ = & \cos^{- 1} (\frac{4 \sqrt{3}}{8}) \\ = & \cos^{- 1} (\frac{\sqrt{3}}{2}) \\ = & \frac{π}{6} \end{array}$

Thus the spherical co-ordinates are $(ρ, θ, ϕ) = (8, \frac{π}{3}, \frac{π}{6})$ .

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