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Homework Help: Finding the volume with a triple integral

  1. Mar 25, 2010 #1
    1. The problem statement, all variables and given/known data
    Use spherical coordinates to find the volume of the solid that lies above the cone z2 = x2 + y2 and below the sphere x2 + y2 + z2 = z



    2. Relevant equations
    I'm going to use { as an integral sign.

    Volume = {{{ P2 Sin[Φ] dP dΦ dΘ


    3. The attempt at a solution

    P2 = x2+y2+z2

    P2 = z
    P = Sqrt[z]

    (P Cos[Φ])2 = P2 Sin2[Φ]
    Cos2[Φ] = Sin2[Φ]
    Tan2[Φ] = 1
    Φ = Pi/4

    And for the limits of integration i got:

    from 0 to 2 Pi on the first integral bracket
    from 0 to Pi/4 on the second integral bracket
    from 0 to Sqrt[z] on the thrid integral bracket

    I was hoping someone could please confirm that these are the right limits of integration before i evaluate it, thanks in advanced!
     
  2. jcsd
  3. Mar 25, 2010 #2

    vela

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    The angular limits are okay, but your upper limit for ρ isn't. Try completing the square in the equation for the sphere to get it into standard form, so you have an idea of where the sphere is centered and what its radius is.

    Spherical coordinates might not be the best choice here.

    Edit: I see the problem says to use spherical coordinates. Oh well.
     
  4. Mar 25, 2010 #3
    how do i get the equation of the sphere into standard form since the Right hand side of the equation equals z?
     
  5. Mar 25, 2010 #4

    vela

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    Move the z to the LHS and then complete the square.
     
  6. Mar 25, 2010 #5

    gabbagabbahey

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    Does your original problem statement explicitly tell you to use spherical coordinates centered at the origin? If not, you may find it easier to use spherical coordinates centered at (0,0,1/2)
     
  7. Mar 25, 2010 #6
    when i move it to the LHS i get: x2+y2+z2-z =0

    I'm not sure how to complete the square on that. and once i have the completed square formula how do i use that to get the limits of integration?

    gabbagabbahey << yes sorry it has to be at the origin.
     
  8. Mar 25, 2010 #7

    gabbagabbahey

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    Just complete the square on [itex]z^2-z[/itex]...
     
  9. Mar 25, 2010 #8
    this completing the square is really screwing with me ha,

    z^2 - z = 0

    z^2 - z - 1/2 = -1/2

    (z-1/2)^2 = -1/2

    is that right?
     
  10. Mar 25, 2010 #9

    gabbagabbahey

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    [tex]\left(z-\frac{1}{2}\right)^2=z^2-z+\frac{1}{4}[/tex]

    Studying university level mathematics is no excuse for forgetting basic high school algebra.
     
  11. Mar 25, 2010 #10

    vela

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    Nope. If you multiply the LHS out, you get z^2-z+1/4, not what you have on the second line. Take the coefficient of z, divide by two, and add the square of that number.
     
  12. Mar 25, 2010 #11
    I honestly have no idea how to complete the square then...

    what i'm trying to complete the square on is z^2 -z ....right?

    so to do that from the example: (the example i'm folowing is ax^2 + bx + c = 0)

    step 1 says to move c to the other side: well there's no c so: z^2 - z = 0

    step 2) if a does not = 1 divide by a, well a in this case = 1 so: no change

    step 3) divide the z term coefficient by 2 then square it so that value is: (1/2)^2

    step 4) add the term in step 3 to both sides: z^2 - z + (1/2)^2 = (1/2)^2

    step 5) this is where i get confused... it just says to re-write it as a perfect square but i clearly don't know how to do it because what i tried to do came out as (z-1/2)^2 which is wrong.

    I was hoping by writing this all out you can help point out which part i'm doing wrong.
     
  13. Mar 26, 2010 #12

    gabbagabbahey

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    There is nothing wrong here. The problem is that in your previous attempt you subtracted 1/2 rather than adding (1/2)^2.

    The x^2+y^2 on the LHS changes nothing from this process:

    [tex]x^2+y^2+z^2-z=0 \implies x^2+y^2+\left(z-\frac{1}{2}\right)^2=\frac{1}{4}[/tex]
     
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