# The change in spherical coordinates

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.

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

$$x^2+y^2+(z-a)^2=b^2 ;$$ $$\rho^2\sin^2\phi\cos^2\theta+\rho^2\sin^2\phi\sin^2\theta+(\rho\cos\phi-a)^2=b^2 ;$$ $$\rho^2\sin^2\phi\cos^2\theta+\rho^2\sin^2\phi\sin^2\theta+\rho^2\cos^2\phi+a^2-2a\rho\cos\phi =b^2 ;$$ 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.