I need convert this spherical expression, to a rectangular form (specific surface). $$\rho^2\cos(2\phi)-1=0$$ Thanks for a while.
Identifying a surface $\rho^2\cos(2\phi)-1=0$
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calculus
spherical-coordinates
2 Answers
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If $\phi$ is the polar and $\theta$ the azimuthal coordinate, using double angle trigonometric identity your surface is described by $$\rho^2\cos(2\phi)=\rho^2\cos^2(\phi)-\rho^2\sin^2(\phi)=1,$$i.e. $$\rho^2\cos^2(\phi)-\rho^2\sin^2(\phi)(\sin^2\theta+\cos^2\theta)=1$$ which transforming to Cartesian coordinates yields the hyperboloid $$z^2-(y^2+x^2)=1$$.
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0Can you explain a bit more your answer? Im working with $$x=sin\phi cos\theta$$ $$y=sin\phi sin\theta $$ $$ z=cos\phi$$ – 2012-12-24
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0you mean, what you wrote multiplied by $\rho$ on the right-hand side, right? or is $\rho=\sqrt{x^2+y^2}$ ? (my confusion arises 'cause one usually reserves $\rho$ for cylindric coordinates.) – 2012-12-24
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0@Rakisbro I updated... – 2012-12-24
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0Now It is very clear. Thanks for your time Jorge Campos. – 2012-12-24
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$(x^2+y^2)\cos(2\arctan(y/x))-1 = (x^2+y^2).\frac{x^2-y^2}{x^2+y^2} -1 = x^2-y^2-1$