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I want to show that the smooth surface $$(x^2+y^2+z^2+a^2-b^2)^2=4a^2(x^2+y^2) , \text{ where } a>b>0$$ is the torus $$\sigma (\theta , \phi )=((a+b\cos \theta )\cos \phi , (a+b\cos \theta )\sin \phi , b\sin \theta )$$

For the one direction we set $$x=(a+b\cos \theta )\cos \phi \\ y=(a+b\cos \theta )\sin \phi \\ z=b\sin \theta $$ and show that it satisfies the equation $(x^2+y^2+z^2+a^2-b^2)^2=4a^2(x^2+y^2)$ right?

For the other direction:

We set $$x=(a+b\cos \theta )\cos \phi \\ y=(a+b\cos \theta )\sin \phi $$ and we suppose that the equality $$(x^2+y^2+z^2+a^2-b^2)^2=4a^2(x^2+y^2)$$ stands, then we get the following:

$$(((a+b\cos \theta )\cos \phi )^2+((a+b\cos \theta )\sin \phi)^2+z^2+a^2-b^2)^2=4a^2(((a+b\cos \theta )\cos \phi )^2+((a+b\cos \theta )\sin \phi)^2) \\ \Rightarrow ((a+b\cos \theta )^2 \cos^2 \phi +(a+b\cos \theta )^2 \sin^2 \phi+z^2+a^2-b^2)^2=4a^2((a+b\cos \theta )^2 \cos^2 \phi +(a+b\cos \theta )^2 \sin^2 \phi )\\ \Rightarrow ((a+b\cos \theta )^2 +z^2+a^2-b^2)^2=4a^2(a+b\cos \theta )^2 \\ \Rightarrow (a+b\cos \theta )^2 +z^2+a^2-b^2=\pm 2a(a+b\cos \theta ) $$

From $z^2=\pm 2a(a+b\cos \theta ) -(a+b\cos \theta )^2-a^2+b^2$ we get

$ {\tag 7}$

$$z^2= 2a(a+b\cos \theta ) -(a+b\cos \theta )^2-a^2+b^2 \\ \Rightarrow z^2=2a^2+2ab\cos \theta -a^2-2ab\cos \theta -b^2\cos^2 \theta -a^2+b^2 \\ \Rightarrow z^2= -b^2\cos^2 \theta +b^2 \\ \Rightarrow z^2= -b^2(1-\sin^2 \theta ) +b^2 \\ \Rightarrow z^2=b^2\sin^2 \theta \Rightarrow z=\pm b\sin \theta $$

or

$ {\tag 8}$

$$z^2= -2a(a+b\cos \theta ) -(a+b\cos \theta )^2-a^2+b^2 \\ \Rightarrow z^2=-2a^2-2ab\cos \theta -a^2-2ab\cos \theta -b^2\cos^2 \theta -a^2+b^2 \\ \Rightarrow z^2= -4a^2-4ab\cos \theta -b^2\cos^2 \theta +b^2 \\ \Rightarrow z^2= -(2a+b\cos \theta)^2 +b^2$$

right?

But how do we get only $z=b\sin \theta$ ?

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Squaring and square-rooting is a common pitfall.

$$ (a+b\cos \theta )^2 + z^2 +( a^2- b^2) >0 $$

It must be always positive because you had tacitly/implicitly assumed major radius of torus bigger than tube radius $ a>b. $

Taking negative sign would attempt to validate a physically impossible toroid, the negative sign at $(8)$ should be discarded whenever a spurious solution got in by the squaring process.

Else all ok.

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  • $\begingroup$ I got it!! Thanks a lot! :-) $\endgroup$ – Mary Star Jan 7 '16 at 19:23
  • $\begingroup$ You are welcome.. $\endgroup$ – Narasimham Jan 7 '16 at 19:28
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For the other direction you cannot set $$x=(a+b\cos \theta )\cos \phi \\ y=(a+b\cos \theta )\sin \phi $$ because that already assumes the wanted representation for all points $(x, y, z)$ on the surface. But if you do so then in your calculation at the step

$$ (a+b\cos \theta )^2 +z^2+a^2-b^2=\pm 2a(a+b\cos \theta) $$

you cannot have the minus-sign because the left-hand side is non-negative. That excludes the second case in your proof.


I would proceed as follows. With polar coordinates $$ x = r \cos \phi \, , \quad y = r \sin \phi \label 1\tag 1 $$ the equation becomes $$ (r^2 + z^2 + a^2- b^2)^2 = 4 a^2 r^2 \\ \Longleftrightarrow r^2 + z^2 + a^2- b^2 = \pm 2 a r $$

From $a > b$ it follows that we must have "$+$" on the rhs, and then the equation can be simplified to $$ (r - a)^2 + z^2 = b^2 $$ That is the equation of a circle and can be parameterized as $$ r - a = b \cos \theta \, , \quad z = b \sin \theta \label 2\tag 2 $$

Combining $(\ref{1})$ and $(\ref{2})$ gives the desired representation for $(x, y, z)$.

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  • $\begingroup$ I understand!! Thanks for your answer!! :-) $\endgroup$ – Mary Star Jan 7 '16 at 19:23
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    $\begingroup$ @MaryStar: You are welcome! $\endgroup$ – Martin R Jan 7 '16 at 19:24

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