- #1
StephenDoty
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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
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
Last edited: