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Calculus 3 : 3-Dimensional Space

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Example Question #2182 : Calculus 3

Express the three-dimensional (x,y,z) Cartesian coordinates as spherical coordinates (ρ, θ, φ):

$\left ( 2\sqrt{3},6,-4\right )$

Possible Answers:

$\left ( 64,\tan^{-1}\frac{1}{3}\pi, \cos^{-1}\frac{2}{3}\pi\right )$

$\left ( 64,\frac{1}{3}\pi, \frac{2}{3}\pi\right )$

$\left ( 8,\frac{2}{3}\pi,\frac{1}{3}\pi\right )$

$\left ( 8,\tan^{-1}\frac{1}{3}\pi, \cos^{-1}\frac{2}{3}\pi\right )$

$\left ( 8,\frac{1}{3}\pi, \frac{2}{3}\pi\right )$

Correct answer:

$\left ( 8,\frac{1}{3}\pi, \frac{2}{3}\pi\right )$

Explanation:

The coordinates (2√3, 6, -4) corresponds to: x = 2√3, y = 6, z = -4, and are to be converted to the spherical coordinates in form of (ρ, θ, φ), where:

$\\ \rho = \sqrt{x^{2}+y^{2}+z^{2}} \\ \tan \theta = \frac{y}{x} \\ \cos \phi = \frac{z}{\rho}$

So, filling in for x, y, z:

$\\ \rho = \sqrt{\left ( 2\sqrt{3}\right )^{2}+6^{2}+ \left ( -4\right )^{2}} = \sqrt{4\cdot3 + 36 + 16} = \sqrt{64}= 8 \\ \tan \theta = \frac{6}{2\sqrt{3}} \Rightarrow \theta = \tan^{-1}\frac{6}{2\sqrt{3}} \Rightarrow \theta = \tan^{-1}\frac{\sqrt{3}}{1} \Rightarrow \theta = \frac{1}{3}\pi \\ \cos \phi = \frac{-4}{8} \Rightarrow \phi = \cos^{-1}\frac{-1}{2} \Rightarrow \phi = \pi - \frac{1}{3}\pi = \frac{2}{3}\pi$

Then the spherical coordinates are represented as:

$\left ( 8,\frac{1}{3}\pi, \frac{2}{3}\pi\right )$

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Calculus 3 : 3-Dimensional Space

Study concepts, example questions & explanations for Calculus 3

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Example Question #2182 : Calculus 3

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