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I have the following sphere and cylinder, respectively:

$$\begin{align} x^2+y^2+z^2&=(2R)^2,\\ (x-R)^2+y^2&=R^2,\qquad R>0. \end{align}$$

How can I parametrize the space curve formed by their intersection?

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up vote 0 down vote accepted

Equation (2) gives $y^2=R^2-(x-R)^2$. Substitute it in Equation (1): $$ z^2=4R^2-(R^2-(x-R)^2)-x^2=4R^2-2Rx.$$ So $$x=(4R^2-z^2)/2R.$$ Going back to (2) you get the equation $$y^2=z^2(1-\frac{z^2}{4R^2}).$$

If you replace $x, y, z$ by their quotients by $2R$, then the equations have better look $$x=1-z^2, \quad y^2=z^2(1-z^2).$$

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