(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Hi there. Well, I have the next exercise, which I've solved, but I don't know if the solution I got is the right one.

It says Given the next region on spheric coordinates find the expression for it in rectangular coordinates, and plot.

[tex]\phi\in{(0,\displaystyle\frac{\pi}{6}}),\rho\in{[1,2],\theta\in{(0,\displaystyle\frac{\pi}{2}})}[/tex]

Well, In think it is a part of a cone, without a sphere of radius 1, with a sphere of radius 2 on the top.

2. Relevant equations

[tex]\begin{Bmatrix}{ x=\rho\cos\theta\sin\phi}\\y=\rho^2\sin\theta\sin\phi\\z=\rho\cos\phi \end{matrix}[/tex]

3. The attempt at a solution

[tex]x^2+y^2=\rho^2\sin^2\phi(\cos^2\theta+\sin^2\theta)\Rightarrow{x^2+y^2=\rho\sin^2\phi}[/tex]

[tex]z^2=\rho^2\cos^2\phi\Rightarrow{\rho^2=\displaystyle\frac{z^2}{\cos^\phi}}[/tex]

[tex]x^2+y^2=z^2\displaystyle\frac{\sin^2\phi}{\cos^2\phi}\theta}\Rightarrow{x^2+y^2=z^2\tan^2\phi}\Rightarrow{x^2+y^2\leq{\displaystyle\frac{z^2}{3}}}[/tex]

[tex]f(x,y)=\begin{Bmatrix}{ x^2+y^2\leq{\frac{z^2}{3}}\forall{}z\in{[\frac{\sqrt[ ]{3}}{2}},\sqrt[ ]{3}]}\\x^2+y^2+z^2\leq{4}\forall{z\in{(\sqrt[ ]{3},2]}}\\x^2+y^2+z^2\geq{1} \end{matrix},x>0,y>0[/tex]

Is this right?

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# Homework Help: Espheric to cartesian coordinates

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