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could someone remind me how to do this with spherical coordinates?
I need to find the cartesian formula for this spherical surface \(\phi=\frac{\pi}{6}\) I can easily see that this is a cone. no idea how to get the answer \(z^2=3x^2+3y^2\) my textbook solutions don't even show how to do it once!
 one year ago
 one year ago
could someone remind me how to do this with spherical coordinates? I need to find the cartesian formula for this spherical surface \(\phi=\frac{\pi}{6}\) I can easily see that this is a cone. no idea how to get the answer \(z^2=3x^2+3y^2\) my textbook solutions don't even show how to do it once!
 one year ago
 one year ago

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richywBest ResponseYou've already chosen the best response.0
using these definitions
 one year ago

beginnersmindBest ResponseYou've already chosen the best response.1
I'm just guessing but why not try writing \[x =r sin \theta cos\phi\] \[y =r sin \theta sin\phi\] \[z= rcos\phi\] Now look at x^2+y^2 and z^2 substituting \[\phi=\pi/6\] Maybe...
 one year ago

richywBest ResponseYou've already chosen the best response.0
I have tried that. I still get stuck.
 one year ago

beginnersmindBest ResponseYou've already chosen the best response.1
r from my equations is rho in your picture
 one year ago

beginnersmindBest ResponseYou've already chosen the best response.1
Ok, let me see if I can get anything useful.
 one year ago

beginnersmindBest ResponseYou've already chosen the best response.1
Ah, ok, I think I got it. Using the letters from your picture: Starting from \[z^2 = 3x^2 +3 y^2\] we have \[z^2 = 3\rho^2\] or \[(z/\rho)^2 = 3 \] where z/p is the arctangent of phi.
 one year ago

beginnersmindBest ResponseYou've already chosen the best response.1
I mean z^2 = 3r^2, sorry.
 one year ago

richywBest ResponseYou've already chosen the best response.0
thanks but I can't figure out how to go backwards like the question requires!
 one year ago

beginnersmindBest ResponseYou've already chosen the best response.1
Sorry, didn't see your comment. You just need to reverse the argument. From the picture we see that ctg(phi) = z/sqrt[(x^2+y^2)]. Plugging in phi=pi/6 and squaring both sides gives 3=z^2/(x^2+y^2) or 3x^2+3y^2=z^2
 one year ago
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