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Homework Help: Solid of Revolution

  1. Feb 16, 2009 #1
    1. The problem statement, all variables and given/known data
    I have to go around and find the volume of a silo-shaped trash can using solid of revolution
    height 91cm
    Circumference 119.3cm
    Diameter 15cm
    http://common.csnstores.com/United-Receptacle-European-Designer-15-Gal.-Round-Top-Receptacle~img~UR~UR1180_l.jpg [Broken] is what the trash can looks like.

    I need list of the equations(s) of the curve(s) which are going to be revolved around an axis. I also need to state the upper and lower limits for the equation(s). Then intergrate the curve(s)

    2. Relevant equations
    I want to use [tex]\int[/tex] pi r^2 dx and from 0 to r
    then do the volume of a sphere and divide by half.
    But I believe this is wrong.


    3. The attempt at a solution
    I tried that equation I wrote and was not right. I think I have to use the shells intergration but dont know the set up.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Feb 17, 2009 #2

    HallsofIvy

    User Avatar
    Science Advisor

    It's hard to understand what you are doing. What I make of this is that your trash can would be represented, on cordinate system, as a vertical straight line, at x= 15, from (15, 0) to (15, 91-15)= (15, 76) and then a quarter circle with center at (0, 76), radius 15, so that its ends are (15, 76) and (0, 91). Rotated around the y-axis, the solid generated is a cylinder topped by a hemisphere.

    But you start talking about doing this as a solid of revolution but then say "then do the volume of a sphere and divide by half." Certainly the easiest way to do this would be to use the formulas for volume of a cylinder and sphere, but that is not doing it "as a solid of revolution". I would not use "shells" (and integrating with respect to x), "disks" (and integrating with respect to y) is much simpler. At each y, the radius of a "disk" is the x value at on the curve and the area is [tex]\pi x^2[/tex] so the volume is [tex]\pi r^2 dy[/itex]. That's what you want to integrate as y goes from the bottom to the top of the figure. Because the formula changes at y= 76, it is probably best to do this at as two separate integrals. For the first x= 15, a constant, while y goes from 0 to 76. For the second, the equation of a circle with center at (0, 76), radius 15, is [tex]x^2+ (y- 76)^2= 225[/tex] so [tex]x= \sqrt{225- (y- 76)^2}[/itex] and y goes from 76 to 91.
     
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