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Could someone help me with this problem:

The change in spherical coordinates is given by the equations:

$x= \rho \sin \phi \cos \theta \tag{I}$ $y= \rho \sin \phi \sin \theta \tag{II}$ $z = \rho \cos \phi \tag{III}$

The set of points $A = \{(x, y, z)\in\mathbb{R^3} \mid \rho=\text{constant}\}$ is a spherical surface.

The set of points $A = \{(x, y, z)\in \mathbb{R^3} \mid \theta =\text{constant}\}$ is a vertical semiplane passing through the $Oz$ axis.

And the set of points $A = \{(x, y, z)\in \mathbb{R^3} \mid \phi=\text{constant}\}$ is a cone.

Based on this theme, mark the alternative that correctly indicates the equation, in spherical coordinates, that describes the sphere:

$x^2 + y^2 + (z - a) ^2 = b^2 \tag{IV}$

Resolution:

I replaced (I), (II) and (III) in (IV) and arrived in:

$$\rho ^2 + a^2 -2a \rho \cos \phi= b^2 \text{ or } \rho^2 = b^2 -a^2 +2a \rho \cos \phi.$$

But the feedback says that the answer must be $\rho = b^2 -a^2 +2a \rho \cos \phi$

I don't understand where I'm going wrong.

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  • $\begingroup$ Welcome to MathSE. Please use MathJax for to write the problem. $\endgroup$
    – user798113
    Dec 24, 2020 at 18:52
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    $\begingroup$ Just on the grounds of units, it's definitely $\rho^2 = b^2 + \dots$. Presumably there is some typo somewhere. $\endgroup$
    – Ian
    Dec 24, 2020 at 19:07

1 Answer 1

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$$ x^2+y^2+(z-a)^2=b^2 ; $$ $$ \rho^2\sin^2\phi\cos^2\theta+\rho^2\sin^2\phi\sin^2\theta+(\rho\cos\phi-a)^2=b^2 ; $$ $$ \rho^2\sin^2\phi\cos^2\theta+\rho^2\sin^2\phi\sin^2\theta+\rho^2\cos^2\phi+a^2-2a\rho\cos\phi =b^2 ; $$ Use $\sin^2+\cos^2=1$ on the first 2 terms: $$ \rho^2\sin^2\phi+\rho^2\cos^2\phi+a^2-2a\rho\cos\phi =b^2 ; $$ and again $$ \rho^2+a^2-2a\rho\cos\phi =b^2 ; $$ $$ \rho =\sqrt{b^2-a^2+2a\rho\cos\phi} ; $$ In the answer $\rho$ is a distance, so obviously the positive sign in front of the root is the relevant one. The feedback is wrong and missing the square root.

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