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Spherical coordinates Domain

  1. Nov 14, 2011 #1
    1. The problem statement, all variables and given/known data
    f(x,y,z) = 1
    [itex]x^{2}[/itex] + [itex]y^{2}[/itex] + [itex]z^{2}[/itex] ≤ 4z
    z ≥ [itex]\sqrt{x^2 + y^2}[/itex]

    2. Relevant equations



    3. The attempt at a solution
    How can I know ρ if there is z variable? Do I just square root both 4 and z?

    For θ, since it is sphere it would be 0 ≤ θ ≤ 2[itex]\pi[/itex] right?

    for [itex]\phi[/itex], is it 0 ≤ [itex]\phi[/itex] ≤ [itex]\pi[/itex]?
     
    Last edited: Nov 14, 2011
  2. jcsd
  3. Nov 14, 2011 #2

    LCKurtz

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    Well [itex]\phi > \frac \pi 2[/itex] is always necessary to get below the xy plane regardless of what quadrant in xy you are in. What is its upper limit?
     
  4. Nov 14, 2011 #3
    I changed the question sorry
     
  5. Nov 14, 2011 #4

    LCKurtz

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    Your first step is to draw a picture of your two surfaces. For your first equation, take the z onto the other side and complete the square. Then you should recognize both surfaces and you will hopefully be able to see the correct limits.

    Also, it is not hard to convert your first equation to spherical coordinates. Try it.
     
  6. Nov 14, 2011 #5
    So I must complete the square or I don't have to?
     
  7. Nov 14, 2011 #6

    LCKurtz

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    You need to do whatever it takes to understand the picture. I'm guessing you can't draw the first surface without completing the square. And you will need the spherical coordinates version of the first surface in order to figure out [itex]\rho[/itex] on the surface.
     
  8. Nov 15, 2011 #7
    OHHH ok I went back and reviewed completing the squares.

    So it comes about as

    -[itex]x^{2}[/itex]-[itex]y^{2}[/itex]-[itex](z-2)^{2}[/itex] = 4

    From this I know that

    0 ≤ ρ ≤ 2

    is z ≥ √x^2+y^2 a cone?
     
  9. Nov 15, 2011 #8

    LCKurtz

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    You have the signs wrong. Yes the second surface is a cone. But regarding the first surface, please answer these questions:

    1. Describe the shape and location of the first surface.
    2. What is its spherical equation?
    3. How did you get ρ ≤ 2?
     
  10. Nov 15, 2011 #9
    Since it's there is only Z translation, the sphere would go 2 up from the origin
    spherical equation would be... x^2 + y^2 + (z-2)^2 = -4?
    ρ is from 0 to 2 because it can't be negative and the sphere of the radius is 2
     
  11. Nov 15, 2011 #10

    LCKurtz

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    You can't have positive on one side and negative on the other. Still wrong.
    If you have a sphere of radius two moved up 2, then the spherical coordinate ρ is not the radius of the sphere. Draw a picture (as I suggested earlier).

    And you ignored my second question without which you will never solve this problem.
     
  12. Nov 15, 2011 #11
    That's how completing the square came out as. Can I just take out the negative and make it equal to 4?

    Why does z translation change the radius of the sphere??

    and the spherical equation is [itex]\int\int\int[/itex][itex]ρ^{2}[/itex]sin[itex]\phi[/itex]dρd[itex]\phi[/itex]dθ

    is ρ 4?
     
    Last edited: Nov 15, 2011
  13. Nov 15, 2011 #12

    LCKurtz

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    No, that isn't how completing the square comes out. Check your arithmetic. I can't correct it if you don't show it but you should be able to find the sign mistakes anyway.
    It doesn't. When the sphere is not at the origin ρ is not the radius of the sphere. Have you drawn a picture yet?
    That is the formula for volume once you have the limits, but that is not what I asked you. You need to get the spherical coordinate equation of the sphere or you will never get the limits figured out.
     
  14. Nov 15, 2011 #13
    ρ is a distance from the origin, so it would be 4.

    and I have no idea what spherical coordinate equation you are talking about... I really don't... Would you give me some hint?

    x^2 + y^2 + z^2 = 4z

    x^2 + y^2 = -z^2 + 4z

    -x^2 - y^2 = z^2 - 4z

    -x^2 - y^2 - 4 = z^2 - 4z - 4

    = (z-2)^2
    -x^2 - y^2 - (z - 2)^2 = 4

    x^2 + y^2 + (z - 2)^2 = -4
     
  15. Nov 15, 2011 #14

    LCKurtz

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    ρ is not a constant on this surface.
    That isn't equal to (z-2)2 as you will see if you multiply it out. When completing the square you add, not subtract the necessary term.

    Hint on spherical coordinates: You have x2+y2+z2=4z

    What is the left side in spherical coordinates? What is the equation for z in spherical coordinates? (Both questions are easy.) Solve the result for [itex]\rho[/itex].
     
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