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So in my homework problem I'm being asked to convert from a cylindrical equation to spherical coordinates and to sketch the graph, The original function is $$z^2+8z=-3r^2-6$$ I'm using the conversions of $r=\rho\sin\phi \quad z=\rho\cos\phi \quad r^2+z^2=\rho^2$ I get stuck and cannot simplify it enough anymore to get $p$ alone. This is what I have so far.$$z^2+3r^2=-8z-6\\\rho^2\cos^2\phi+3\rho^2\sin^2\phi=-8\rho\cos\phi-6\\\rho^2(1-\sin\phi^2)\\\rho^2-\rho^2\sin^2\phi+3\rho^2\sin^2=-8\rho\cos\phi-6\\\rho^2+2\rho^2\sin^2\phi=-8\rho\cos\phi-6$$ I'm not sure where to move on from this if I could get some advice on how to move forward I'd greatly appreciate it.

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  • $\begingroup$ I think you got confused and changed $x$ to $z$ when you went from $z^2 + 8x = -3r^2 - 6$ to $z^2 + 3r^2 = -8z -6$. Let us know if you still have problems after correcting that! :) $\endgroup$ – 2012ssohn Nov 2 '16 at 0:32
  • $\begingroup$ Sorry it was a typo that I made while writing it out, but the questions still stands. $\endgroup$ – Carlos V Nov 2 '16 at 0:36
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Observe \begin{align} z^2+8z= -3r^2-6 \end{align} in rectangular coordinates is \begin{align} z^2+8z+16+3x^2+3y^2= 10 \ \ \Rightarrow\ \ \ \frac{x^2}{\left( \sqrt{\frac{10}{3}}\right)^2}+\frac{y^2}{\left( \sqrt{\frac{10}{3}}\right)^2}+\frac{(z+4)^2}{(\sqrt{10})^2}= 1 \end{align} which is a translated ellipisoid. In spherical coordinates, we have \begin{align} &\rho^2 \cos^2\phi+8\rho \cos\phi +3\rho^2\sin^2\phi= -6\\ \Rightarrow\ & \rho^2+2\rho^2(1-\cos^2\phi)+8\rho\cos\phi = -6 \\ \Rightarrow\ &\ 3\rho^2-2\rho^2\cos^2\phi +8\rho\cos\phi = -6 \\ \Rightarrow\ &\ 3\rho^2-2(\rho\cos\phi-2)^2=-14 \\ \Rightarrow\ &\ 3\rho^2=2(\rho\cos\phi-2)^2-14. \end{align}

In short, you should treat the problem as two separate problems.

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