Polar parametrization surface intersection

Question

here is my problem:

I need some help, i need the parametrization of the intersection of this two surfaces:

$\ z^2= x^2+y^2 $

$\ (x-1)^2+y^2=1 $

Well, i can do it with cartesian equations

$\ F(x,y)= (x,y,{(x^2+y^2)}^{1/2}) $

Where:

$\ -2<x<2 $

$\ {-(1-(x-1)^2 )^{1/2}}<y<{(1-(x-1)^2)^{1/2}} $

How can i do a polar parametrization of this intersection? Thanks

Answer 1

The off-center circle is given by $r=2\cos\theta$, $-\pi/2\le\theta\le\pi/2$. Since $x=r\cos\theta$, $y=r\sin\theta$, and $z=r$, you should have it easily.

The intersection is a curve, whereas what you've drawn with Mathematica seems to be the portion of the cone inside the cylinder. You need to decide what you really mean. (Also, your $-2<x<2$ is wrong. You have $|x-1|\le 1$, so $0\le x\le 2$.)