Parametralizing curves in 3 dimensions

[kaoken]

I need help on how to approach this problem:

you have a sphere centered at the origin with radius 2 and a cylinder, (x-1)^2 +y^2 = 1

How do i parametrize their intersection?

[OfficeShredder]

The first question would be how do you parametrize either of the other curves?

[HallsofIvy]

What other curves?

The sphere can be written as x^2+y^2+ z^2= 4. The cylinder is (x-1)^2+ y^2= 4. Their intersection is the set of points (x,y,z) that solve both.
One way to parametrize that, since the second equation involves only x and y would be to solve for y as a function of x, then put that into the first equation, solve it for z as a function of x. Use x as the parameter. That suffers from the problem that it involves square roots. To get both parts of the figures, you would have to have "piecewise" functions, one piece + the other -.

Another way would be to parameterize one of the solids (which is what OfficeShredder meant). One way to parameterize the surface of the sphere is to use spherical coordinates with rho= 2: x= 2sin(phi)cos(theta), y= 2sin(phi)sin(theta), z= 2cos(phi). Now put those into the second equation:
(x-1)^2+ y^2= (2sin(phi)cos(theta)-1)^2+ 4sin^2(phi)cos^2(theta)= 4. Solve for one of the parameters as a function of the other and substitute that back into the equations for x, y, and z.

It might be easier to parameterize the cylinder. if theta is the angle a radius makes with the x axis, we have x= 2cos(theta)+ 1, y= 2sin(theta). Since z does not appear in the equation of the circle, we have z as the second parameter. Now put those into the equation of the sphere:
(3cos(theta)+1)^2+ 4sin^2(theta)+ z^2= 4. Solve that for z so that you have also z as a function of theta.

[OfficeShredder]

Actually, it should be surfaces. So we're both wrong

This just goes to show that it's better to agree to disagree than to learn both sides are wrong :(