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Volumes of solid revolution

  1. Sep 1, 2006 #1
    Find the volume of the solid generated by revolving the region bounded by [tex]y = 2x - x^2[/tex] and y = x about the line x = 1

    Ok, so I can solve this if it were revolved around the y-axis, it would be: [tex]\pi \int_0^1 (x)^2 - (2x - x^2)^2 dy[/tex]

    If I have to revolve it around x = 1, what do I need to change with the equation?
     
  2. jcsd
  3. Sep 1, 2006 #2

    Galileo

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    If you cut up the region into horizontal slabs, you need the horizontal distance of either curve to the axis of rotation as a function of y.

    If you rotate around the y-axis, you are right to integrate wrt y (though you could also do it wrt x, using cylindrical shells instead of horizontal slabs), but then you need to write the distance as a function of y. (2x-x^2) is the vertical distance from the left curve to the x-axis, as a function of x, that's not what you're looking for.
     
  4. Sep 1, 2006 #3

    J77

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    Shift the axis - ie. replace x with x-1: remembering to do (x-1)^2 etc.
     
  5. Sep 1, 2006 #4
    Oh, sorry about that typo. If revolving around the y-axis it would be:
    [tex]\pi \int_0^1 (y)^2 - (\sqrt{1 - y} +1)^2 dy[/tex].

    So the answer when integrating with respect to y and revolved around x = 1 would be the same if I integrated with respect to x and replaced x with x-1?
     
  6. Sep 1, 2006 #5

    Galileo

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    If you can visualize dividing the area in slabs (make a picture), it's clear that the distance of a point on the left curve to the line x=1 is 1-f(y), where f(y) is simply the x coordinate of the point at height y. You should thus solve y=2x-x^2 for x in terms of y (you'll get 2 answers, take a look at the picture to see which one you should take). That'll be the outer radius, the inner radius from the line x=y is easy.
     
  7. Sep 1, 2006 #6

    J77

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    Yeah - sketch it out.

    Am I having a completely Friday brain-drain moment here, or should the line y=x around x=1 become the consideration of the line y=x+1 (+ not -) etc.

    (You can also work the volume by cylindrical shells which allows differentiation w.r.t. x and an easy rotation about given x.)
     
  8. Sep 1, 2006 #7
    Ok, using cylindrical shells method, will my equation be:
    [tex]2\pi \int_0^1 (x - 1) (x - x^2) dx[/tex]
     
  9. Sep 1, 2006 #8

    J77

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    Looks good to me.
     
  10. Sep 1, 2006 #9
    Alright, I did that and I get [tex]V = 2\pi \left[\frac{2}{3}x^3 - \frac{1}{4}x^4 - \frac{1}{2}x^2\right]_{0}^{1}[/tex]
    Then when I put in 1 and 0, my final volume is a negative value...
     
  11. Sep 1, 2006 #10
    [tex]y = (x)^2 - (2x - x^2)^2[/tex]

    Right?

    Then shouldn't it be,

    [tex]-2\pi \int_0^1 (1 - x)(x^2 - (2x - x^2)^2) dx[/tex]

    The negative at the front is due to the fact that when sketched, this function goes in the negative direction. Since you can't have a negative height, the height has to be the opposite of y.
     
    Last edited: Sep 1, 2006
  12. Sep 4, 2006 #11

    J77

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    Sorry for being a bit vague on Friday, end of week and all...

    Hopefully, this will be better.

    Firstly, from post above, using cylindrical shells, you don't square the y's.

    In the OP example, it looks like the volume of a shell will be:

    [tex]2\pi(1-x)y\delta x=2\pi(1-x)((2x-x^2)-x)\delta x[/tex]

    I graphed the curves and I think this is the correct order.

    Integrate over this for result...
     
    Last edited: Sep 4, 2006
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