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Spherical surface

  1. Jun 6, 2014 #1
    1. The problem statement, all variables and given/known data

    Sqrt(x^2+y^2)<=z<=x^2+y^2+z^2
    With this problem I run into a few questions

    The first of which arises at the statment 1/2<=cos^2a
    Here I go about writing
    -1/sqrt(2)<=cosa<=1/sqrt(2)
    But when dealing with trigs it doesn't make any sense to write 3pi/4<=a<=pi/4

    So here I suppose we stop thinking with algebra and think logically saying that 3pi/4<=a<=piand 0<=a<=pi/4
    But since z >=0 with take only a between 0 and pi/4
    Which all semms a bit off track as if there's should be one to follow algebreically

    But anyways this is techinaclly correct by me in any case


    Next my question is at the stament cosa<=p
    Here I must assume the the surface is only that of the sphere above the cone because well the equation of the cone is not in these terms and a is less than Pi/4 giving the straight line rather than the curve of the cone
    Only Here p is never going to equal 0 a never goes past pi/4 so its minimum technicHlly is 1/srt(2) I don't know Wtf the book is writting 0<=p<=cosa for








    2. Relevant equations



    3. The attempt at a solution
     
    Last edited: Jun 7, 2014
  2. jcsd
  3. Jun 7, 2014 #2
    Had a typo in the intitial problem corrected now
     
  4. Jun 7, 2014 #3

    ehild

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    What is the question????


    ehild
     
  5. Jun 7, 2014 #4
    Read the post confused about the expression on the right in terms of p
     
  6. Jun 7, 2014 #5
    Corrected mistake or 3pi/4 not -pi/4
     
  7. Jun 7, 2014 #6
    Seems correct to write 0<=cosa<=p
     
  8. Jun 7, 2014 #7

    ehild

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    What are a and p? What does the problem want you to do? Copy the whole problem text, please.

    ehild
     
  9. Jun 7, 2014 #8
    Another typo a between pi and 3pi/4
     
  10. Jun 7, 2014 #9

    SammyS

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    As ehild said:

    Please post the whole problem word for word as it was given to you.
     
  11. Jun 7, 2014 #10
    The surface above the cone underneath the sphere
     
  12. Jun 7, 2014 #11

    HallsofIvy

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    So inside the sphere [tex]x^2+ y^2+ (z- 1/2)^2= 1/4[/tex] which has center at (0, 0, 1/2) and radius 1/2 but above the cone [tex]x^2+ y^2= z^2[/tex], which has vertex at (0, 0, 0), a point on the surface of the sphere? Have you thought about exactly where those do intersect?
    On the unit circle, 3pi/4 is the same as -5pi/4. -1/sqrt{2}<= cos(a)<= 1/sqrt{2} is the same as -5pi/4<= a<= pi/4.

    ?? "Thinking with algebra" is "thinking logically".

    I'm not sure what you mean by this- in particular, what do you mean by "the curve of the cone"?

    By "p" do you mean the Greek letter "rho", [itex]\rho[/itex], for the radius?

     
  13. Jun 7, 2014 #12

    LCKurtz

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    That isn't even a sentence, let alone the statement of the whole problem. What is a "surface" above a cone and underneath a sphere"? What are you asked to calculate? Even though I think I know what you want to do, I refuse to guess. State the exact problem.
     
  14. Jun 9, 2014 #13
    The question reads exactly
    A solid that lies above the cone z=sqrt(x^2+y^2) and below the sphere x^2+y^2+z^2=z write a description in terms of inequalities involving spherical cordinatres
     
  15. Jun 9, 2014 #14

    LCKurtz

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    Your first step should be to draw a picture and your second step should be to write both equations in spherical coordinates. What do you get for the equations?
     
  16. Jun 10, 2014 #15
    Well we have the cone under the sphere
    For 0<=alpha<=pi/4
    Cos(theta)<=p
    What exactley does this describe and is this correctly written
     
  17. Jun 10, 2014 #16

    LCKurtz

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    You haven't done the second step, which is write the equation of the two surfaces in spherical coordinates. You know, equations in terms of ##\rho,~\theta,~\phi##. Your inequality with ##\cos\theta## is incorrect, and until you write the equations in spherical coordinates you aren't going to know where the ##\cos\theta## comes from.
     
  18. Jun 13, 2014 #17
    What do you mean
     
  19. Jun 13, 2014 #18

    LCKurtz

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    What does who mean? About what? Learn to use the quote button.
     
  20. Jun 14, 2014 #19
    Tell me this much
    Does the surface include the inner part of the cone I don't see how that is possible given that the equation is only for the sphere but let's say the straight line from the origin rotates around 2pi it would be that only a straight line not the cone
     
  21. Jun 14, 2014 #20

    haruspex

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    Why do you keep mentioning the surface? The question, as quoted by you in post #15, says nothing about surfaces. All you are asked to do is to rewrite the given inequalities in spherical coordinates. For that, you need three equations
    x = some function of ρ, θ, ϕ,
    y = some function of ρ, θ, ϕ,
    z = some function of ρ, θ, ϕ,
    and use those to replace x, y and z in the given inequalities.
     
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