Find a parametrization of the intersection curve between two surfaces in $\mathbb{R^3}$ $x^2+y^2+z^2=1$ and $x^2+y^2=x$.

Question

Find a parametrization of the intersection curve between two surfaces in $\mathbb{R}^3$ $$x^2+y^2+z^2=1$$ and $$x^2+y^2=x.$$

I know that $x^2+y^2+z^2=1$ is a sphere and that $x^2+y^2=x$ is a circular cylinder. Any help is greatly appreciated, thank you.

mathcounterexamples.net · Answer 1 · 2015-01-22T22:26:48.437

2

A point of the intersection belongs to the cylinder. So $(x-1/2)^2+y^2=1/4$. Take for parameter the angle $\theta$ such that $x-1/2=1/2 \cos \theta$ and $y=1/2 \sin \theta$. This is just the angle of the cylindrical coordinates.

Then you have $x+z^2=1$ which implies $z^2=1/2-1/2 \cos \theta=\sin^2(\theta/2)$. You then get $z=\sin(\theta/2)$.

edited Jan 22 '15 at 22:26

answered Jan 22 '15 at 22:18

mathcounterexamples.net

70,018

yea, there are two choices for $z$, you can avoid a $\pm$ by going to a double angle and a shift by $\pi$. – orangeskid Jan 22 '15 at 22:28

score 0 · Answer 2 · answered Jan 22 '15 at 22:17

0

+1) The polar equation of $x^2+y^2=x$ is $r=cos\theta$ and so we have $x=rcos\theta$ and $y=rsin\theta$ as parametrics (with $r$ known) Substituting into the first equation, you can solve for $z$. Note that $1-cos^2\theta=sin^2\theta$ Hope this helps, if not or incorrect, I will take it off.

answered Jan 22 '15 at 22:17

imranfat

10,029

Totally nice, you just forgot to add the $r$ factor in $x$ and $y$ – orangeskid Jan 22 '15 at 22:27

Find a parametrization of the intersection curve between two surfaces in $\mathbb{R^3}$ $x^2+y^2+z^2=1$ and $x^2+y^2=x$.

2 Answers2