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Volume of Curve

  1. Mar 19, 2004 #1

    Zurtex

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    Hi could you please say if this is right or wrong and if it is not where I went wrong.

    I got the question:

    Find the volume of the area between y=x-x^3 from 0 to 1 revolved about the y-axis.

    So here are my workings:

    [tex]V = \pi \int_0^1{x^2dy}[/tex]
    [tex]V = \pi \int_0^1{x^2 \frac{dy}{dx}dx}[/tex]
    [tex]\frac{dy}{dx} = 1 - 3x^2[/tex]
    [tex]V = \pi \int_0^1{x^2 \left(1 - 3x^2 \right)dx} [/tex]
    [tex]V = \pi \int_0^1{x^2 - 3x^4 dx}[/tex]
    [tex]V = \pi \left[ \frac{x^3}{3} - \frac{3x^5}{5} \right]_0^1 [/tex]
    [tex]V = \pi \left \left( \frac{1}{3} - \frac{3}{5} \right) - \pi(0)[/tex]
    [tex]V = \frac{-4\pi}{15} [/tex]
     
    Last edited: Mar 19, 2004
  2. jcsd
  3. Mar 19, 2004 #2
    Well, the volume can't be negative, can it? :smile:

    As far as I remember, the volumne you get by revolving the area below f(x) between points A and B around the X axis is:

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

    Edit: argh, just noticed you're revolving it around the Y axis. Sorry.
     
    Last edited: Mar 19, 2004
  4. Mar 19, 2004 #3

    Zurtex

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    Grrr, I just wrote up a load of code in Latex and it didn't post up.

    I think the problem is that when I am integrating:

    [tex]x^2 - 3x^4[/tex]

    Should I separate it into the parts above and below the x - axis, work out the area of the two parts individually?
     
  5. Mar 19, 2004 #4

    Doc Al

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    I presume you want to find the volume of revolution of the area between that curve f(x) and the x-axis (between x = 0, 1), revolved around the y-axis?

    Wouldn't that just be:
    [tex]V = 2\pi \int_0^1{xf(x)dx}[/tex]
     
  6. Mar 19, 2004 #5

    Zurtex

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    Yes but I've never heard of that equation before, that only briefly reminds me of something I did in statistics.
     
  7. Mar 19, 2004 #6
    If you take a point [tex](x, y)[/tex] on the graph, connect it to the X axis and revolve the line around the Y axis, you would get a very thin but high donut. The perimeter of the donut is [tex]2\pi x[/tex] and its height is [tex]y[/tex], bringing its volum to [tex]2\pi xy[/tex] or [tex]2\pi xf(x)[/tex]. That's how you get to:

    [tex]V = 2\pi \int_0^1{xf(x)dx}[/tex]
     
  8. Mar 19, 2004 #7
    Ahh, I think I found your mistake. Since you integrate by [tex]dy[/tex], you need the take Y limits and not the X limits. You are asked to find the volume between X = 0 and X = 1, those points are (0, 1) and (1, 0). So when you integrate by Y, the definite integral needs to be from 1 to 0, and not from 0 to 1. This is why you got a negative volume, and if you solve it with the equation Zurtex posted you would in fact get 4pi/15.
     
  9. Mar 19, 2004 #8

    Zurtex

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    Thanks loads :D

    Knew it was just some sill mistake. Although I don't understand your volume thingy but then I am only doing A level further maths.
     
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