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Multiple Integral

  1. Feb 19, 2012 #1
    1. The problem statement, all variables and given/known data

    Ok, so I am going to French University, I have to translate in English.

    There is a hemisphere with the radius of 2. Inside of it, there is an empty space shaped as a cylinder with the radius (a< 2) which is perpendicular to the base of the hemisphere. The density of each point is different. Note that the farthest point from the base is 4.

    a) Find the mass using cylinder coordinates (do not evaluate the integral)

    b) Find the mass using sphere coordinates (do not evaluate the integral)



    2. Relevant equations



    3. The attempt at a solution



    The equation of the sphere is x^2+y^2+z^2=2 where z >= 0

    For a) we have to find the density equation with the given situation.

    p(x,y,z) = 2(x^2 + y^2 +z^2-2)

    now we have to find the interval. Before that we should change
    x to rcos(theta)
    y to rsing(theta)
    z = z

    r must be a <= r <= 2
    theta must be 0 <= theta <= 2*pi
    and the interval of z must be

    (2-a^2)^(1/2) <= z <= 2 coz x^2+y^2=a^2


    Am I right?? I feel dumb,,,


    and I have no idea how to do b)


    Please can you help me out? Thank you http://img338.imageshack.us/img338/5822/32344458.jpg [Broken]
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Feb 19, 2012 #2

    vela

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    The equation of a sphere of radius r centered at the origin is ##x^2+y^2+z^2 = r^2##, so the righthand side should be equal to 4.

    At the base, z=0, so for a fixed r and θ, z is going to go from 0 until it hits the sphere. So what should the upper limit for z be?
     
  4. Feb 19, 2012 #3


    We have to know the height of the cylinder.
    The cylinder equation is x^2+y^2 = a^2 where a is the radius.

    z must be from 0 to z=(2^2-a^2)^(1/2)


    Am I right?
     
  5. Feb 19, 2012 #4

    vela

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    No, that's not correct. Which surface defines the top of solid?
     
  6. Feb 19, 2012 #5

    Hemisphere surface without the base surface of the cylinder?
     
  7. Feb 19, 2012 #6

    vela

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    The upper limit of z is the hemisphere, so solve the equation of the sphere for z. That's your upper limit.
     
  8. Feb 19, 2012 #7
    I don;t think so
    the maximum height (the value of z) depends on the height of cylinder.

    z=2-( √(2-x^2-y^2) - a^2 )



    ****,,,this is confusing
     
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