Spherical Coordinates and Mathematica

[StephenDoty]

Use spherical coordinates to draw the cone z=Sqrt[x^2+y^2].
Hint: You will need to determine \[Phi].

How would I go about finding phi?
Below are the x y and z components, but I cannot figure out how to find the range of phi:

z^2=x^2 + y^2
(Rho)=sqrt(2x^2+2y^2)

x = sqrt(2x^2+2y^2) sin [Phi] cos [Theta]
y = sqrt(2x^2+2y^2) sin [Phi] sin [Theta]
z = sqrt(2x^2+2y^2) cos [Phi]

Mathematica Code:
ParametricPlot3D[{sqrt(2x^2+2y^2) Sin[[Phi]]*Cos[[Theta]],
sqrt(2x^2+2y^2)Sin[[Phi]] Sin[[Theta]], sqrt(2x^2+2y^2) Cos[[Phi]]}, {[Phi], 0,
Pi}, {[Theta], 0, 2*Pi}, ViewPoint -> {6, 3, 2}]

Any help would be appreciated.
Thanks
Stephen

 

[HallsofIvy]

Use spherical coordinates to draw the cone z=Sqrt[x^2+y^2].
Hint: You will need to determine \[Phi].

How would I go about finding phi?
Below are the x y and z components, but I cannot figure out how to find the range of phi:

z^2=x^2 + y^2
(Rho)=sqrt(2x^2+2y^2)

x = sqrt(2x^2+2y^2) sin [Phi] cos [Theta]
y = sqrt(2x^2+2y^2) sin [Phi] sin [Theta]
z = sqrt(2x^2+2y^2) cos [Phi]

Mathematica Code:
ParametricPlot3D[{sqrt(2x^2+2y^2) Sin[[Phi]]*Cos[[Theta]],
sqrt(2x^2+2y^2)Sin[[Phi]] Sin[[Theta]], sqrt(2x^2+2y^2) Cos[[Phi]]}, {[Phi], 0,
Pi}, {[Theta], 0, 2*Pi}, ViewPoint -> {6, 3, 2}]

Any help would be appreciated.
Thanks
Stephen

Oh, please don't think of Mathematica as a Panacea. Like any computer program, GIGO! You are mixing Cartesian and Spherical coordinates there: you don't want "x" and "y" in your spherical coordinates formula.

x= \rho cos(\theta)sin(\phi)
and
y= \rho sin(\theta) sin(\phi)
so
x^2+ y^2= \rho^2 sin^2(\phi)(cos^2(\theta)+ sin^2(\theta))= \rho^2 sin^2(\phi)
and so the equation becomes
z= \rho cos(\phi)= \sqrt{\rho^2sin(\phi)}= \rho sin\phi
or simply
cos(\phi)= sin(\phi)
tan(\phi)= 1[/itex]<br /> <br /> Of course, there is a much easier way to find \phi! Since the equation is symmetric in x and y, just look in the xz-plane. If y= 0, the equation becomes z= x, a line which makes an angle of \pi/4 with the x-axis and so the entire cone makes an angle of \pi/2- \pi/4= \pi/4 with the z-axis. That is, \phi= \pi/4. <br /> (And, of course, tan(\pi/4)= 1.)

 

[StephenDoty]

so to graph it would I use
x= (Rho)sin(phi)cos(theta)
y=(Rho)sin(phi)sin(theta)
z=(Rho)cos(phi)

with theta going from 0 to 2pi
phi = pi/4
Rho= some range like (0,3)

or is there a way to solve for Rho?
Thanks
Stephen