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More volume of a washer cross section(integration)

  1. Aug 2, 2012 #1
    1. The problem statement, all variables and given/known data

    Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis.


    2. Relevant equations

    [itex]y = x^2, x = y^2[/itex] about [itex] y = -1 [/itex]


    3. The attempt at a solution

    I tried using both cylindrical and washer methods - but for cylindrical I couldn't figure out what the shell radius would be, so I switched to washer.

    [itex]\pi\int_0^1(x^2+1)^2 - (√x+1)^2 dx[/itex]
    [itex]\pi\int_0^1 x^4 + 2x^2 - x - 2√x dx[/itex]
    [itex]\pi(x^5/5 + (2x^3)/3 - x^2/2 - (4x[/itex]^3/2[itex]))/3)[/itex]
    [itex]\pi(1/5 + 2/3 - 1/2 - 4/3)[/itex]
    [itex]-29\pi/30[/itex]

    And I get a negative volume. Any ideas?
    EDIT: Ah. I just noticed I subtracted in the wrong order - would that fix it?
    Edit2: Also, would you mind helping me with using the cylindrical method for this? Because I wasn't able to narrow down the shell radius - there was always a little unknown portion.
     
    Last edited: Aug 2, 2012
  2. jcsd
  3. Aug 2, 2012 #2

    SammyS

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    For the cylindrical method, the radius is y - (-1), which is y + 1 .
     
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