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Homework Help: Volume of Solids with Known Cross Sections (Calculus AB)

  1. Mar 8, 2010 #1
    Can anyone help me with this calculus problem? I have tried it a couple of different ways, but keep getting various incorrect answers; I think my problem is probably in the setup.

    I am trying to find the volume of the solid created by revolving the area bounded by these two equations
    y = 6 - 2x - x[tex]^{2}[/tex]
    y = x + 6
    around the x-axis, and around y = 3.

    First, I found the intersection points of the two equations, which are at (-3,3) and (0,6)
    For the revolution around the x-axis, I tried doing
    [tex]\pi \int ^{0}_{-3}((6 - 2x - x[^2}[/tex])[tex]^{2}[/tex] - (x+6)[tex]^{2}[/tex] )dx,

    simplified to [tex]\pi [/tex] \int ^{6}_{0}(-32x -13x[tex]^{2}[/tex] +4x[tex]^{3}[/tex] + x[tex]^{4}[/tex] )dx ,
    and got a final answer of -27[tex]\pi[/tex]/5
    The correct answer is 243[tex]\pi[/tex] /5

    For the revolution around y = 3, I got -3-x^{2} from subtracting the lower function, x + 6, from the top function, 6 - 2x - x^{2}, and then tried
    [tex]\pi \int^{6}_{0}(6-(-3-x^{2}))^{2} dx, simplified to \pi^{6}_{0}\int (6+3x+x^{2} ) dx, and ended up with 29376[tex]\pi[/tex]/5.

    The answer for this one is supposed to be 108[tex]\pi[/tex]/5

    Again, I think my real problems are in the setup, probably with the "top minus bottom" part.

    Also, I am new to Physics Forum, and can't seem to get rid of how, in the second integral, my words ended up included in the integral. I would appreciate it if anyone could tell me how to fix/avoid that.

  2. jcsd
  3. Mar 9, 2010 #2


    Staff: Mentor

    Note: Fixed up some of your LaTeX by removing a lot of superfluous [ tex] tags.
    The first integral above was fine, but there are some mistakes in the second one. You switched the limits of integration. They should still be -3 and 0. Also, the coefficients of the x^4 and x^3 terms are right, but those for the x^2 and x terms are incorrect.
    I'll take a look at the 2nd integral in a separate post.
  4. Mar 9, 2010 #3


    Staff: Mentor

    For the second integral,
    [tex]\Delta V = \pi (R^2 - r^2)\Delta x[/tex]
    The limits of integration are the same, namely x = -3 to x = 0. R = the y-value on the parabola - 3, and r = the y-value on the line - 3.
  5. Mar 9, 2010 #4
    Hm, I don't know what I must have been thinking when I decided to start integrating from 0 to 6, especially when I had already started using -3 to 0 as well... I suppose I was particularly jumbled yesterday, or something of that sort.

    Thanks so much for the help; goodness knows I needed it. May I ask though, how can I differentiate between [yes, a very bad choice of words; I mean when to use one as opposed to the other, not take the derivative of] when R(x)^2 - r(x)^2 is equal to (top equation minus line)^2 - (bottom equation - line)^2 and when it is equal to (top minus bottom)^2 - (line)^2?
  6. Mar 10, 2010 #5


    Staff: Mentor

    R is the larger radius and r is the smaller radius. Both radii are as measured from the axis of rotation. If you have drawn a sketch, it should be pretty obvious.
  7. Mar 10, 2010 #6
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