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!

Changing from rectangular coordinate to sperical

  1. Nov 14, 2009 #1
    1. The problem statement, all variables and given/known data

    change from rectangle to spherical coordinate :

    z^2 = x^2 + y^2


    I know that :

    z = pcos(phi)

    x = psin(phi)cos(theta)

    y = psin(phi)sin(theta)

    there fore

    z^2 = x^2 + y^2 in spherical coordinate is

    p^2cos(phi)^2 = (psin(phi)cos(theta))^2 + (psin(phi)sin(theta))^2

    =

    cos^2(phi) = sin^2(phi)*cos^2(theta) + sin^2(phi)*sin^2(theta)

    =

    cos^2(phi) = sin^2(phi) + sin^2(phi)

    =

    cos^2(phi) = 2*sin^2(phi)


    But the book has it as : cos^2(phi) = sin^2(phi) ?? what gives?
     
  2. jcsd
  3. Nov 14, 2009 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    How did you conclude that sin^2(phi)*cos^2(theta) + sin^2(phi)*sin^2(theta) equals sin^2(phi) + sin^2(phi)??
     
  4. Nov 14, 2009 #3
    Look at the third to the last line again:

    cos^2(phi) = sin^2(phi)[cos^2(theta) + sin^2(theta)]

    hope that helps.
     
  5. Nov 14, 2009 #4

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Once you correct your mistake that Dick has pointed out, you might notice that your book's answer:

    [tex] \cos^2(\phi) = \sin^2(\phi)[/tex]

    while correct, isn't completely simplified. Can you see that the final answer should be of the form [itex]\phi = C[/itex] for the appropriate value of the constant C, and that the surface is actually one of the parameter surfaces for spherical coordinates?
     
  6. Nov 14, 2009 #5
    Yes I see what I did wrong.

    It is cos^2(phi) = sin^2(phi)

    =

    cos(phi) = sin(phi)

    >>
    . Can you see that the final answer should be of the form phi = C.
    Yes I guess pi/2 would be correct.

    So the equation z^2 = x^2 + y^2 is phi = pi/2 in spherical coord?
     
  7. Nov 14, 2009 #6

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You mean [itex]|\cos(\phi)| = |\sin(\phi)|[/itex]
    Nope. Guess again. Think about [itex]|\tan(\phi)| = 1[/itex].
     
  8. Nov 14, 2009 #7
    I mean pi/4.
     
  9. Nov 14, 2009 #8

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Remember [itex]\tan(\phi)[/itex] can be ± 1. [itex]\pi/4[/itex] will get you the top half of the cone. What value of [itex]\phi[/itex] will get you the bottom half?
     
  10. Nov 14, 2009 #9
    -pi/4 ?

    would this be ok :


    plus/minus : n*pi/4n : where n is a integer. OR no because phi is limited to 0 <= phi <= pi
     
  11. Nov 14, 2009 #10

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    So if phi is between 0 and pi, what value gives the bottom half of the cone? Visualize the surface.
     
  12. Nov 14, 2009 #11
    Is there a bottom half? I don't think it will be within bounds because
    only pi/4 would solve | tan(phi) | = 1 where 0<= phi <= pi.

    | tan( -pi/4) |= 1 but thats no within bounds and

    |tan( 5pi/4 ) |= 1 but thats also not within bounds, and those 2 points are the
    next points which satisfies the equation.
     
  13. Nov 14, 2009 #12

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Yes, there is a bottom half. In your original xyz equation z can be negative. Draw a sketch of the cone, top and bottom. Just look at it. There obviously is a [itex]\phi[/itex] that gives the bottom half.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook