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A Few Small Integration Problems

  1. May 24, 2010 #1
    I just have a few small questions.

    For this first question I need to setup the integral to find the volume defined by.

    [tex]y=tan^{3}(x)[/tex], [tex]y=1[/tex], and [tex]x=0[/tex] by rotating it about y=1.

    My result was this:
    [tex]\pi\int_{0}^{\frac{\pi}{4}} {(tan^{3}(x) - 1)^{2}dx}[/tex]

    From my understanding all you need to do is offset the function by the axis in order to make it work. But the actual answer is:
    [tex]\pi\int_{0}^{\frac{\pi}{4}} {(1 - tan^{3}(x))^{2}dx}[/tex]

    Which looks like it would do something else entirely. I had a similar issue when doing the same thing for sin(x) about y=1. What's going on here?

    My second issue is a conceptual question involving work. Just to make sure I have the order of everything correct, how does say, the work done to move a spring relate to it's potential energy? Is it the integral of work = potential energy?
     
    Last edited: May 24, 2010
  2. jcsd
  3. May 24, 2010 #2

    Mark44

    Staff: Mentor

    The tangent function (and its cube) are not continuous in the interval [0, pi], which you show below. The region being rotated is bounded by the y-axis, the line y = 1, and the graph of y = tan^3(x). For another, you are apparently using disks, but haven't set up the integral correctly. Using disks, the typical volume element [itex]\Delta V[/itex] is pi * R2[itex]\Delta x[/itex], where R is the radius from the line y = 1 to the curve.
    Yes, that doesn't look right at all, based on the information you have given.
     
  4. May 24, 2010 #3
    Sorry, I forgot to square the function when I entered it in...and I meant to put pi/4. I've corrected it now.

    From what I was told in class though, all I need to do to get the volume using the disk method is this:

    [tex]V = \int_{b}^{a} [f(x)]^{2} dx[/tex]

    and if I was revolving around a differnet axis, say y=-3 then I would just do this:

    [tex]V = \int_{b}^{a} [f(x)+3]^{2} dx[/tex]
     
    Last edited: May 24, 2010
  5. May 24, 2010 #4

    Mark44

    Staff: Mentor

    I think you have it now. To recap, your integral should look like this:
    [tex]V = \pi \int_{0}^{\pi/4} [tan^3(x) - 1]^{2} dx[/tex]
     
  6. May 24, 2010 #5
    That's what I got, but the book has the 1 and the cubic tangent reversed.
     
  7. May 24, 2010 #6

    Mark44

    Staff: Mentor

    When you calculate the overall radius of the shell you should have R - r, with R being the larger y value and r being the smaller y value. In this case, the overall radius is 1 - tan^3(x). Doesn't matter in this case, because it's squared.
     
  8. May 24, 2010 #7

    Mark44

    Staff: Mentor

    I had to look it up in my Halliday & Resnick, which says
    [tex]W = \Delta K = - \Delta U[/tex]
    where W = work, [itex]\Delta K[/itex] = change in kinetic energy, and [itex]\Delta U[/itex] = change in potential energy

    If a force F is applied in a straight line, work W is defined as

    [tex]W = \int_{x_0}^x F(x) dx[/tex]
     
  9. May 24, 2010 #8
    Right, I see now.

    Great. Thanks for the help!
     
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