find the equation of the cone $z=\sqrt{x^2+y^2}$ in spherical coordinates 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?

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
$$p \cos \phi = \sqrt{p^2\sin^2\phi \cos^2 \theta + p^2\sin^2\theta \sin^2 \phi}$$ $$p \cos\phi = \sqrt{p^2\sin^2 \phi \ (\sin^2 \theta + \cos^2 \theta)} $$
$$p \cos\phi = p \sin \phi$$
$$\cos \phi = \sin \phi$$
$$\phi = \pi/4.$$
