1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook