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To find volume of a cylindrical shell with holes around it

  1. Jun 28, 2013 #1
    ITS NOT A HOMEWORK PROBLEM
    PROBLEM::- THINK OF A CYLINDRICAL SHELL WITH NUMEROUS SPHERICAL HOLES ALL AROUND IT.
    (FOR EG. http://sell.lulusoso.com/upload/20120317/Underground_Water_Pipe.jpg)
    HOW TO FIND VOLUME & SURFACE AREA OF SUCH A CYLINDRICAL SHELL.
    (A GENERALIZED CASE FOR N HOLES AROUND THE SHELL.)
     
    Last edited by a moderator: Jun 28, 2013
  2. jcsd
  3. Jun 28, 2013 #2
    What's the formula for the volume of a cylinder?
     
  4. Jun 28, 2013 #3
    (Pi).r^2. h

    & i need the volume of the shell, not the cylindrical cavity.
     
  5. Jun 28, 2013 #4
    OK, I'm not totally sure how they differ. Do you mean the amount of material required to make one of the things you describe.

    Alternatively what is the thickness of the "rim" of the cylinder, and how big are the holes wrt to the cylinder?
     
  6. Jun 28, 2013 #5
    .... there's a cylinder with inner dia R and outer dia R+dR..... it has holes all around penetrating the whole thickness of the shell.... assume the rim to be hollow and radius of holes to be "r" ....
    Now, as we knw, the volume of a normal cylindrical shell is (pi)(R+dr-R)^2.(h).. Now what i want to ask is that --what would be the volume of the cylindrical shell with holes all around..... (i have attached a picture of water-pipe, u can see that for reference.)
     
  7. Jun 28, 2013 #6
    How big are the holes compared to the pipe? If they're small, can we make an assumption that they are flat?
     
  8. Jun 28, 2013 #7
    no, we can't .....
     
  9. Jun 28, 2013 #8

    LCKurtz

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    Can you assume the holes were drilled with a bit so the sides of the holes are portions of cylinders and the projection of the holes perpendicular to their axes are circles?
     
  10. Jun 28, 2013 #9

    LCKurtz

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    Let's assume the pipe has inner radius ##a## and outer radius ##b## and orient the pipe so the surfaces are ##x^2+z^2 = a^2## and ##x^2+z^2 = b^2##. Now we drill a vertical hole of radius ##c## centered on the ##z## axis through the upper side of the cylinder. The volume of material removed is$$
    \int_0^{2\pi}\int_0^c\int_{\sqrt{a^2-r^2\cos^2\theta}}^{\sqrt{b^2-r^2\cos^2\theta}}r\, dzdrd\theta$$All you have to do is work that out and multiply by the number of holes. :tongue2:

    [Edit, added] If you have numbers for ##a,b,c## Maple will crank out an answer.
     
    Last edited: Jun 28, 2013
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