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Equation of a sphere in XYZ coordinates

  1. Sep 9, 2016 #1
    1. The problem statement, all variables and given/known data
    Show that the equation represents a sphere, and find its center and radius.

    3x2+3y2+3z2 = 10+ 6y+12z

    2. Relevant equations


    3. The attempt at a solution
    3x2+3y2-6y +3z2 -12z =10

    My equation is how the constants in-front of the squared terms affect the sphere formula? Besides that I should be completing the square for the Y and Z terms right?
     
  2. jcsd
  3. Sep 9, 2016 #2

    Mark44

    Staff: Mentor

    What is the formula for the equation of a sphere? That would have been good to put in the Relevant equations section.
    Yes.
     
  4. Sep 9, 2016 #3
    The formula of a sphere is
    R2=(X-X0)2 +(Y-Y0)2 +(Z-Z0)2

    So I tried working further and eventually got to
    3Y2-6y=10
    (Y-1)2 =13/3

    3Z2-12Z =13/3
    (Z-6)2=37.44
    3X2 +(Y-1)2+(Z-6)2= 37.44
    so R2=37.44 and R=6.12
     
  5. Sep 9, 2016 #4

    Ray Vickson

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    Homework Helper

    Your "equations" ##3Y^2-6y=10##, ##(Y-1)^2 = 13/3##, etc., are meaningless in this problem. You are trying to find constants ##a, b, r## so that the equation
    $$3 x^2+3 y^2 + 3 z^2 -6y - 12z - 10 = 0$$
    becomes
    $$3 x^2 + 3 (y-a)^2 + 3 (z-b)^2 - r^2 = 0$$
    The second equation must hold for ALL values of ##x,y,z## that satisfy the first equation, and vice-versa.
     
  6. Sep 11, 2016 #5

    vela

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    Staff Emeritus
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    You shouldn't drag the constant term from the original equation along. Instead, you can say
    \begin{align*}
    3y^2 - 6y &= 3(y^2-2y)\\
    &= 3[(y^2-2y+1)-1]\\
    &= 3(y-1)^2 - 3.
    \end{align*} Note that ##3y^2-6y=3(y-1)^2-3## holds for all y whereas the equation you started with, ##3y^2-6y=10##, doesn't. (Also, don't use y and Y interchangeably. It's sloppy and incorrect.)

    The original equation can then be rewritten as
    \begin{align*}
    3x^2 + (3y^2 - 6y) + (3z^2-12z) &= 10 \\
    3x^2 + [3(y-1)^2 - 3] + (3z^2-12z) &= 10 \\
    3x^2 + 3(y-1)^2 + (3z^2-12z) &= 13
    \end{align*}

    You should check your work in completing the square. I don't get (z-6)^2.
     
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