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I need to represent the cone $z=\sqrt{3x^2+3y^2}$ parametrically in terms of $\rho$ and $\theta$ where $(\rho,\theta,\phi)$ are spherical coordinates.

Attempt. I tried using: $$x=\rho\sin\phi\cos\theta \\y=\rho\sin\phi\sin\theta\\z=\rho\cos\phi$$

and $$\rho^2=x^2+y^2+z^2\\\cos\phi=\frac{z}{\sqrt{x^2+y^2+z^2}}\\\rho^2\sin^2\phi=x^2+y^2$$

I cannot find a way to get rid of $\phi$. Hints please.

This is the graph of it. It is a cone.

enter image description here

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If you substitute the equation of the cone in the formula giving $\cos\phi,$ what do you get? –  Giuseppe Tortorella Jun 9 '12 at 10:08

1 Answer 1

up vote 1 down vote accepted

If you have $z=\sqrt{3x^2+3y^2}$ and $z^2=3x^2+3y^2$, making the appropriate replacements in $\cos\phi=\dfrac{z}{\sqrt{x^2+y^2+z^2}}$ gives

$$\cos\phi=\dfrac{z}{\sqrt{\frac{z^2}{3}+z^2}}$$

Can you see what to do next? (Hint: your Cartesian equation guarantees that $z$ is always positive, so $\sqrt{z^2}=?$)

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