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Need help getting formula for volume of any tree log.

  1. Oct 6, 2011 #1
    1. The problem statement, all variables and given/known data

    For calc 2 we have to write a formula using integration that will find the volume of any tree log that is cut. Using the picture that is attacked I came up with this equation. The integral from 0 to h of pi*(((R-r/h)x+r))^2)dx. Any ideas about my formula or anything would be appreciated. Thanks

    2. Relevant equations

    3. The attempt at a solution

    Attached Files:

  2. jcsd
  3. Oct 6, 2011 #2


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    If that formula is the radius at x then it looks correct, if a little vague.
  4. Oct 6, 2011 #3


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    Your equation, y= ((R-r)/h)x+ r, is vague because there is no "x" nor "y" in your picture. I presume you mean y to be the radius of the tree at distance x from the large end. If that is what is meant, yes, when x= 0, that is y= 0+ r= r and when x= h, y= R- r+ r= R. And you are assuming that the radius is a linear function of x.

    Now, at a given x, a cross section will be a disk with radius h. That disk has area [itex]\pi y^2[/itex] and so a disk with thickness "dx" will have volume "area times thickness"= [itex]\pi y^2 dx= \pi ([(R-r)/r]x+ r)^2)dx[/itex].

    The approximate volume would be a sum of such things, a "Rieman sum", and the limit, as the thickness goes to 0 will be the integral
    [tex]\pi \int_{x=0}^h (((R-r)/r)x+ r)^2 dx= \pi \int_{x=0}^h \left[\left(\frac{R-r}{r}\right)^2x^2+ 2(R-r)x+ r^2\right] dx[/tex]
  5. Oct 6, 2011 #4


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    Yes, it looks perfectly fine.
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