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Volume in a cone, using a double integral.

  1. Mar 29, 2010 #1
    1. The problem statement, all variables and given/known data
    Evaluate the volume under z^2 = x^2 + y^2
    and the disc x^2 + y^2 < 4.

    Just wondering what I should write to constitute a proper solution. Would this do?:

    V=(int)(int) z dA
    R is {x²+y² < 4} [context: R in other problems was the region over which integrals were performed]

    (int)(int) z dA
    (int)(int) sqrt( r² ) r drdT [T for theta]
    (using x²+y²=r² and dA->r dr dT)

    The integral to actually be computed is:
    (int)(int) r² dr dT
    with r in [0,2]
    T in [0, 2pi]
    = 2pi/3 whatever the hell it is.

    For a 1st year student in not-mathematics (which isn't me), is that too concise?
  2. jcsd
  3. Mar 29, 2010 #2


    User Avatar
    Science Advisor
    Homework Helper

    Looks to me like it's got all the essential steps, given the abbreviations you are using like (int). If you aren't the 1st year student in not-mathematics, why are you asking?
  4. Mar 30, 2010 #3
    I need to know how to explain it to 1st year not-mathematics students.
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