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Chapter 13: Q 9 (page 1065)

. To convert from rectangular to spherical coordinates
$p =, 胃 =, 蠒 =$

Short Answer

Expert verified

$p = \sqrt{x^{2} + y^{2} + z^{2}} 胃 = \tan^{- 1} (\frac{y}{x}) 蠒 = \cos^{- 1} (\frac{z}{\sqrt{x^{2} + y^{2} + z^{2}}})$

Step by step solution

01

Step 1:Given information

Given $(x, y, z)$ are rectangular coodinates and $(p, 胃, 蠒)$ are spherical coordinates

02

Simplification

$(x, y, z)$ are rectangular coordinates and $(p, 胃, 蠒)$ are spherical coordinates

The relation is

From the figure

$鈬� \cos 蠒 = \frac{z}{r}, \sin 蠒 = \frac{蚁}{r}$

Now

$x = 蚁 \cos 蠒$

$= r \sin 胃 \cos 蠒$ where $(蚁 = r \sin 胃)$

similarly

$y = 蚁 \sin 蠒$

$= r \sin 胃 \sin 蠒$

And $z = r \cos 蠒$

now

$x^{2} + y^{2} + z^{2} = r^{2} [\sin^{2} 蠒 (\sin^{2} 胃 + \cos^{2} 胃) + \cos^{2} 蠒] 鈬� x^{2} + y^{2} + z^{2} = r^{2} [\sin^{2} 蠒 + \cos^{2} 蠒] 鈬� x^{2} + y^{2} + z^{2} = r^{2} 鈬� r = \sqrt{x^{2} + y^{2} + z^{2}}$

Similarly

$鈬� 胃 = \tan^{- 1} \frac{y}{x}$

and we have $\cos 蠒 = \frac{z}{r}$

$鈬� 蠒 = \cos^{- 1} \frac{z}{r} 鈬� 蠒 = \cos^{- 1} (\frac{z}{\sqrt{x^{2} + y^{2} + z^{2}}})$

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In Exercises 45鈥�52, rewrite the indicated integral with the specified order of integration.

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