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Some more volume integral questions

  1. Feb 25, 2013 #1
    Formulas:
    Shell Method: dV = 2pi(radius) * (height) * (thickness)
    Disk method: dV = pi(radius)^2 * (thickness)

    Question 1 (26-3-15)
    Statement
    Using Shell method, find the volume generated by revolving the region bounded by the given curve about the x-axis.
    x = 4y - y^2 - 3, x = 0

    Attempt
    integrating:
    4y^2 - y^3 - 3y = x
    4/3y^3 - 1/4y^4 - 3/2y^2

    at this point I would plug a boundary number into the variable, what number should it be?

    Question 2 (26-3-19)
    Statement
    Using disk method, find the volume generated by revolving the region bounded by the given curve about the y axis.
    y = 2(x^1/2), x = 0, y = 3

    Attempt
    y/2 = x^1/2
    (y/2)^2 = x
    y^2/4
    (y^2/4)^2
    y^4/16

    is this the correct path?

    Question 3 (26-3-21)
    Statement
    Using shell method, find the volume generated by revolving the region bounded by the given curve about the y axis.
    x^2 - 4y^2 = 4, x = 3

    Attempt
    2pixy principal formula
    isolate y
    -4y^2 = 4 - x^2
    -y^2 = (4-x^2)/4
    -y^2 = -x^2
    x = y
    2pix^2 = 1/3x^3 = 9

    but this answer is incorrect

    Thank you for your time.
     
  2. jcsd
  3. Feb 25, 2013 #2

    eumyang

    User Avatar
    Homework Helper

    Be precise in your setup. There are a number of errors in your work. I see no integral sign, I see no dy, and I see no 2pi in the front. As to the limits of integration, well the graph is a sideways parabola. Find the y-intercepts.

    This is wrong. It should be
    [itex]-y^2 = 1 - \frac{x^2}{4}[/itex]
     
  4. Feb 27, 2013 #3
    thanks for the tips how do you use latex
     
  5. Feb 27, 2013 #4

    Mark44

    Staff: Mentor

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