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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!

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- richyw

- katieb

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- richyw

using these definitions

- beginnersmind

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...

- richyw

I have tried that. I still get stuck.

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- beginnersmind

r from my equations is rho in your picture

- beginnersmind

Ok, let me see if I can get anything useful.

- beginnersmind

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.

- beginnersmind

I mean z^2 = 3r^2, sorry.

- richyw

thanks but I can't figure out how to go backwards like the question requires!

- beginnersmind

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

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