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richyw
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!
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...
I have tried that. I still get stuck.
r from my equations is rho in your picture
Ok, let me see if I can get anything useful.
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.
I mean z^2 = 3r^2, sorry.
thanks but I can't figure out how to go backwards like the question requires!
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