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Turn spherical coordinates into rectangular coordinates

  1. Nov 6, 2015 #1
    1. The problem statement, all variables and given/known data
    Find the volume of the solid region that lies inside the cone φ= pi/6 and inside the sphere ρ=4. Use rectangular coordinates.

    2. Relevant equations
    x=ρ sinφ cos θ
    y=ρsinφ sin θ
    z=ρ cos φ
    ρ^2=x^2+y^2+z^2
    x= r cos θ
    y= r sin θ
    r^2=x^2+y^2


    3. The attempt at a solution

    at first, I tired to use rectangular coordinates, but I dont even know how to express this cone in rectangular coordinates. so I i divided them into 2 parts and used polar coordinates, where the lower part is just a straightforward volume of a cone, the upper part is the small dome where the sphere and the cone intersects each other.

    since ρ=4, z=4cosφ = 4cos (π/6) = 2√3 which is the height when the sphere and the cone intersects at.
    16=x^2+y^2+z^2
    z=√(16-x^2-y^2)
    set them equal to each other : √(16-x^2-y^2)=2√3 , by simplify, i get 4=x^2+y^2
    which means radius =2 for the R where the sphere and the cone meet.
    the volume of the cone in the lower part: 1/3*π*(2)^2*(2√3)= 14.51 [using volume formula for a cone]
    the upper dome volume= ∫∫ [(√(16-r^2)- 2√3 ] r dr dθ where r goes from 0 to 2, and θ goes from 0 to 2π.
    I get V=3.449, adding it with the volume of the cone, i get 17.958.

    Snapshot.jpg
    I am confused on how to do this problem in x.y,z coordinates( without dividing it into 2 parts). more precisely, I dont know how to get the expression of the cone in x,y,z coordinates from the given φ=pi/6 .
     
  2. jcsd
  3. Nov 6, 2015 #2

    Simon Bridge

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    Hint: the equation of a cone, in rectangular coords, pointing along the z axis, is the equation of a circle whose radius depends linearly on z.
     
  4. Nov 6, 2015 #3
    thank you for the hint

    i found the equation of a general cone from wolfram:
    NumberedEquation2.gif

    where

    NumberedEquation3.gif
    in my case, the equation of the cone would be z= sqrt ( (x^2)/3 + (y^2)/3 )
    since, r=2, h = 2√3 , c^2= 1/3 , correct?
     
  5. Nov 6, 2015 #4

    Simon Bridge

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    Do you not understand the equation wolfram gives you?

    The equation of a circle whose radius depends on z is: ##x^2 + y^2 = f^2(z)##
    It varies linearly with z so f has the form of a straight line. ##f(z) = az + b##
    Now its a matter of finding a and b using the information you have.
    Is this consistent with what you got by rote from wolfram?
     
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