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Finding limits on spherical coordinates

  1. Dec 23, 2012 #1
    The problem statement, all variables and given/known data

    find the limits on spherical coordinates. where ε is the region between z²=y²+x² and z = 2(x²+y²)

    no matter what i try i cant seem to find the limits, especially for "ρ",
    so far i got 0<θ<2Pi and 0<φ<Pi.
     
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  3. Dec 23, 2012 #2

    mfb

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    ... limits of what?

    Edit: Ah, the coordinates itself, I see.
     
    Last edited: Dec 23, 2012
  4. Dec 23, 2012 #3

    pasmith

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    Let
    [tex]
    x = \rho\cos\theta\sin\phi \\
    y = \rho\sin\theta\sin\phi \\
    z = \rho\cos\phi
    [/tex]

    The region in question is the region
    [tex]\frac{z}{2} \leq x^2 + y^2 \leq z^2.[/tex]

    There is no limit on [itex]r[/itex]: the region extends arbitrarily far away from the origin. Instead there are limits on [itex]\phi[/itex] so that
    [tex]0 < f(r) \leq \cos\phi \leq g(r) < 1[/tex]
    where [itex]\cos\phi = f(r)[/itex] is the surface [itex]z^2 = (x^2 + y^2)[/itex] and [itex]\cos\phi = g(r)[/itex] is the surface [itex]z = 2(x^2 + y^2)[/itex].

    Substitute [itex]z = \rho\cos\phi[/itex] and [itex]x^2 + y^2 = \rho^2\sin^2\phi[/itex] and solve for [itex]\cos\phi[/itex].
     
  5. Dec 23, 2012 #4
    i still couldn't find the range for ρ, i think i might be missing something and i don't have the answer to check wheter i am right, could you elaborate a little more on how to find ρ,θ and ϕ?

    i tried using sqrt(x²+y²) <ρcosϕ< 2*(x²+y²)

    then, ρsinϕ<ρcosϕ<2*(ρsinϕ)²
    but i dont know how to proceed.

    i did found the ranges for 0<ϕ<Pi and 0<θ<2Pi, tho i dont know if it's correct.
     
    Last edited: Dec 23, 2012
  6. Dec 23, 2012 #5

    vela

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    You have cylindrical symmetry because z for both surfaces depends only on ##x^2+y^2## — z doesn't depend on ##\theta##, so look at the intersection of the solid with a vertical plane, say, y=0. The solid will be that cross-sectional area rotated about the z-axis.

    I've attached a plot of z vs. x for both surfaces for x>0. Can you deduce the limits for ρ and ɸ from that?

    There should actually be a z=-x piece from the cone, but I'm pretty sure you're meant to ignore that.
     

    Attached Files:

  7. Dec 23, 2012 #6
    i'll take a look at the plot z vs x, and i had already tought about cylindrical coordinates but the question asks it to be in spherical coordinates and not in cylindrical.
     
  8. Dec 23, 2012 #7

    vela

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    Who said anything about cylindrical coordinates?
     
  9. Dec 23, 2012 #8
    sorry, i miss read that, are the ranges Pi/4<ϕ<Pi/2 and 0<ρ<0.7?
     
    Last edited: Dec 23, 2012
  10. Dec 23, 2012 #9

    haruspex

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    As in vela's, graph, it's the region [tex]\frac{z}{2} \geq x^2 + y^2 \geq z^2.[/tex]
     
  11. Dec 23, 2012 #10

    Dick

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    The ρ limit won't be constant. Your ρ limit will be a function of ϕ, won't it?
     
  12. Dec 23, 2012 #11
    i'll give it another try tomorrow,it's 4am here and im sleepy(abeit stubborn),and i'm starting to get frustated because im not getting anywhere and the lack of an actual answer in this book makes things worse.
    i'll see if things start to get better in the morning.
    but you are right, that range for ρ looks pretty wrong.
     
  13. Dec 23, 2012 #12

    Dick

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    Sure. But look at vela's picture and put in a straight line making an angle ϕ with the vertical axis. You should be able to see what the appropriate distance is on the graph. Then you just have to calculate it as a function of ϕ.
     
  14. Dec 24, 2012 #13
    do you have an example of a problem that looks like this one?, i haven't found anything similar in my book or on the internet, so it makes things a little bit difficult to see.
     
  15. Dec 24, 2012 #14

    Dick

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    Did you figure out what you want from the picture? You want ρ to be the distance from a point on the parabola to the origin. In terms of r=sqrt(x^2+y^2) the equations you need are z=2r^2 (the 'parabola'), z=ρcos(ϕ) (the 'line') and r^2+z^2=ρ^2. You want to eliminate r and z and just get a relation between ρ and ϕ.
     
    Last edited: Dec 24, 2012
  16. Dec 24, 2012 #15
    so i got that z/2 < x²+y² < z² becomes z/2 < r² < z², i added z² in the equation and got
    z/2 + z² < P² < 2z², any thoughts of what should i do now?
    or isn't this the right way to proceed?
     
  17. Dec 24, 2012 #16

    Dick

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    You are heading in the right direction. But you still aren't paying enough attention to the graph. The cone r^2=z^2 defines the ϕ=pi/4 lower limit. If ϕ>pi/4 then the upper limit for ρ is on the parabola. And yes, it satisfies z/2+z^2=ρ^2. Now you want to find the relation with ϕ. Put z=ρcos(ϕ) into that.
     
  18. Dec 24, 2012 #17

    Dick

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    Also, you have your inequality backwards. z/2>=z^2 over your region.
     
  19. Dec 24, 2012 #18
    so now i have 2ρ²cos²ϕ < ρ² < ρcosϕ/2 + ρ²cos²ϕ , and the next step should be isolating ρ to the center of the inequality, right?
    so i tried dividing for ρ, getting, 2ρcos²ϕ < ρ < cosϕ/2 + ρcos²ϕ
    but i can't seem to isolate ρ.
    how should i proceed?
     
  20. Dec 24, 2012 #19

    Dick

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    If you look at the graph, your lower limit for ρ is 0 when ϕ>=pi/4. Look at the graph! What you care about is the upper limit so change it to equality. ρ² = ρcosϕ/2 + ρ²cos²ϕ. And yes, divide by ρ so ρ = cosϕ/2 + ρcos²ϕ. Now solve for ρ.
     
  21. Dec 24, 2012 #20
    ok, i think i finally understood (thanks to you), the lower limit for ρ is 0, and the upper limit is now ρ = cosϕ/2 + ρcos²ϕ, solving for ρ gives me ρ = cosϕ/(2-cos²ϕ) and the ranges for the spherical coordinates are:
    0 < θ < 2Pi
    pi/4 < ϕ < pi/2
    0 < ρ < cosϕ/(2-cos²ϕ)
    is that it?
    thanks for all the patience in advance, i'll start paying more attention to graphs now.
     
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