I have the following... $$z=\sqrt{x^2+y^2}$$
I need to write this as an equation in spherical coordinates.
I know that $p^2 = x^2+y^2+z^2$
and that...
$$x = p\sin\phi \cos\theta$$ $$y=p\sin\phi \sin\theta$$ $$z = p\cos\theta$$
The answer is $\phi = \pi/4$
How do you get t0 this answer? I tried doing
$$p^2-z^2 = x^2+y^2$$ $$z^2 = p^2-z^2$$ $$z = p/2$$
But I really do not know what I am doing. How do I solve this?
$\endgroup$2 Answers
$\begingroup$The idea is to plug in the values of $x$, $y$ and $z$ in $$z = \sqrt{x^2+y^2}.$$ Specifically, by using the given expressions, we get $$p \cos \phi = \sqrt{p^2\sin^2\phi \cos^2 \theta + p^2\sin^2\theta \sin^2 \phi}$$ $$p \cos\phi = \sqrt{p^2\sin^2 \phi \ (\sin^2 \theta + \cos^2 \theta)} $$ $$p \cos\phi = p \sin \phi$$ $$\cos \phi = \sin \phi$$ $$\phi = \pi/4.$$
$\endgroup$ 1 $\begingroup$When latitude is not $\pi/4 $ the following can be said in addition:
$ \phi $ is latitude,$ \,\pi/2-\phi= \alpha $ complementatry or co-latitude, $ r$ radius in polar ( or in cylindrical coordinates), $\rho$ is in spherical coordinates with
$$ r= \rho \sin \alpha = \rho \cos \phi $$
