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Homework Help: Integration Problems

  1. Feb 9, 2005 #1
    if I want to find the volume of y= (x + 2) ^(1/2) about the x-axis with the boundaries of -1 and 1, then would it be :

    V = pi times integral of (x+2) dx?? Its confusing trying to set it up. The volume of a section would be v = pi r^2 (delta x)

    Also I'm having trouble finding the volume of the region enclosed by y = (2-x)(2+x), y = 3x, and the y- axis when rotated around the x - axis. It's hard to even picture what the figure will be like, much harder to find the volume. Please point me in the right direction! Thanks!!
  2. jcsd
  3. Feb 9, 2005 #2


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    You seem to have the first part, but don't forget the limits of integration.
    [tex]\pi\int_{-1}^{1}(x+2) dx[/tex]

    For the second part, one thing that jumped out at me was:
    is a parabola that opens downward, so it will hit the line
    [tex]y=3x[/tex] in two (or zero) places.

    So, let's look for intersections by solving:
    to get
    so we can use the quadratic formula
    [tex]x=\frac{3\pm\sqrt{9-4(1)(-4)}}{2}=\frac{3 \pm 5}{2}[/tex]

    That tells us that the functions intersect where [itex]x=-1[/itex] and where [itex]x=4[/itex]. Those will be the limits of integration. A quick check, say at [itex]x=0[/itex] confirms that [itex]4-x^2 > 3x[/itex] in the region.

    So, to calculate the volume, you can (equivalently) calculate the volume from [tex]y=4-x^2[/tex]
    on and then subtract the volume from
    on the region.
    Or you can integrate using annuli (annuluses?) instead of disks.
  4. Feb 9, 2005 #3
    So, for the first one, the answer would be 4pi, correct?

    For the second one, I thought the 2 functions interesected at (1,3), so I said the volume was pi times integral (with boundaries of 0 and 1) of [(4-x^2) -(3x)] dx since it has to be subracted. But I think I need to find the integral of r^2 times delta x, so would i have to square (4-x^2) and make it (16-x^4) then subtract 3x (or would it be 9x^2)??? Thanks
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