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Given the following sphere and cylinder,

$\begin{align} x^2+y^2+z^2&=4R^2,\\ (x-R)^2+y^2&=R^2, \end{align}$

find a parametric equation of their intersection.

I know that their intersection is called a hippopede and that on the $x$-$y$ plane, its parametrization is $r(t)=R(\cos t+1)\,\hat i+R\sin t\,\hat j$. However, I have no idea how to find its $\hat k$ component.

Any hint would be appreciated!

Edit: The $\hat k$ component is supposed to be

$ 2R\sin\left(\frac t2\right), $

but I have no idea how that was obtained.

1 Answers 1

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I have no idea how they chose it, but I can show you it works. First let's do some algebra on our second equation.

$x^2-2Rx+R^2+y^2=R^2,2Rx=x^2+y^2$

Substituting into the first equation gives

$2Rx+z^2=4R^2$

$z=2R\sin u$ seems a reasonable enough substitution, yielding $x=2R\cos^2u$. Now plugging that into the second equation gives

$(2R\cos^2u-R)^2+y^2=R^2$

$R^2(2\cos^2u-1)^2+y^2=R^2$

$R^2\cos^22u+y^2=R^2$

$y^2=R^2\sin^22u$

Now to show the functions of x are equivalent, use the power reduction formula.

$2R\cos^2u=R(1+\cos2u)$

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    Thank you. Your approach led me to realize why their pick works. I am still wondering how the substitution $z=2R\sin u$ can be found, though; I would have never come up with it.2012-10-10