1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Describe the cylinder

  1. Nov 4, 2012 #1
    1. The problem statement, all variables and given/known data

    Describe the cylinder x2+y2=9
    in terms of parameters theta and phi, where (p,theta,phi) are spherical coordinates of a point on the surface.

    I'm a bit confused I know that spherical coords are



    z= Pcos(phi)=3cos(phi)

    But the answer is way off it's
    x = 3 cos(theta) ; y = 3sin(theta) ; z = 3 cot(phi) 

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Nov 4, 2012 #2


    Staff: Mentor

    In your third equation you're thinking that P is the radius of the circle.

    Slice the the cylinder with a plane containing X and Z axes and you'll see two vertical lines so a point on the cylinder is a vector from the origin to the point which makes a triangle with the P being the bottom side and the z value being the vertical side .

    Now describe Z in terms of spherical coordinates phi angle.
  4. Nov 4, 2012 #3
    Not really sure what you mean?
  5. Nov 4, 2012 #4


    Staff: Mentor

    your eqn is a cylinder along the z axis, right?

    So if you slice the cylinder with a plane containing the X and Z axes you'll see two vertical lines at x=3 and x=-3, right?
  6. Nov 4, 2012 #5
  7. Nov 4, 2012 #6


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    jedishrfu is pointing out that there is some confusion regarding p (or P).
    In the problem statement, p is the distance from the origin to a point on the cylinder. You have used P as the radius of the cylinder (=3). A point on the cylinder may be further than 3 from the XY plane, so z cannot be given by 3 cos(phi).
  8. Nov 5, 2012 #7


    User Avatar
    Science Advisor

    [itex]\rho[/itex], what you are calling "P", is the straight line distance from the origin to the point, (x, y, z). It is NOT [itex]\sqrt{x^2+ y^2}[/itex], the distance, in the xy-plane, to the projection (x, y, 0).

    From [itex]x= \rho cos(\theta)sin(\phi)[/itex] and [itex]y= \rho sin(\theta)sin(\phi)[/itex], you can get [itex]x^2+ y^2= \rho^2 cos^2(\theta) sin^2(\phi)+ \rho^2sin^2(\theta)sin^2(\phi)= \rho^2sin^2(\phi)[/itex].
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook