Tricky Multiple integral word problem

  1. Nov 24, 2007 #1
    So Ive been talking with some people about this problem, but I cant seem to find the answer, or even set it up!

    "A region in space, when viewed from 3 different views, looks like a circle, looks like a square and looks like a triangle. Describe this object. And then use multiple integration to determine its volume"

    So if you think of any two shapes its pretty easy. Such as square and triangle = pyramid, circle and square = cylinder and so on. But I couldnt figure out a shape for when you add the third shape. And then Figuring out the integral would be even more difficult.
    What do you guys think about it?
    =D
     
  2. jcsd
  3. Nov 24, 2007 #2
    you almost had it with the circle+square=cylinder, what if you were to have only a square from the "top/bottom" and the sides a triangle?
     
  4. Nov 24, 2007 #3
    Im still not getting it.. Pictures would help. >.<
    So Square base and top. Sides triangle? Where would the circle go?
     
  5. Nov 24, 2007 #4
    it's originally a cylinder, 2r=height, so from 2 sides you'd have a square. and now cut off some part in order to get a triangle from the other 2 sides. So it looks like a wedge with a circular base and square top/bottom.
     
  6. Nov 24, 2007 #5
  7. Nov 24, 2007 #6
    hm..depends how you place the axis's on it, but I think you'll need to add 2 integrals due to the circle you'll have the top and bottom part of it.

    and it might be better to cut the cylinder by 2 lines so looking at the triangle side you can cut it at the center of the circle and get 2 symmetric sides so you can just do 2xsome integral to find the volume.
     
  8. Nov 24, 2007 #7
    Well if I cut the cylinder in half diagonally, couldn't I just take the volume of the cylinder and then divide it by two? When you say cut the cylinder with two lines what do you mean?

    And I have no idea how to set up the integral. >.<
    could you get me started please?
     
  9. Nov 24, 2007 #8
    o yea you could just do that.

    Use cylindrical coordinates to setup your integral and then just multiply it by .5 to find the volume.
     
  10. Nov 24, 2007 #9
    could you help me set it up? thats the part im really bad at. Many thanks for discussing this problem with me. =)
     
  11. Nov 24, 2007 #10
    well you have [tex]r,\theta,z[/tex] in cylindrical coordinates.

    r-?
    theta-?
    z-?

    z being the height and from what we've written you know what it is.
    so just find the r and the z and you can setup the integral.
     
  12. Nov 24, 2007 #11
    theta would be 0 - 2pi?
    z would be the height soo lets say 0 - a
    what would r be?
     
  13. Nov 24, 2007 #12
    r would be 0 to the circle's radius, say R, and the height of the cylinder can be described in terms of R. So you can have the square when looking at it from 2 sides.
     
  14. Nov 24, 2007 #13
    what would the actual integrand be? [tex]rdrdz\theta[/tex]
    r - 0-R
    theta - 0-2pi
    z - 0-R

    ?
     
  15. Nov 24, 2007 #14
    no the z=2R because you need the diameter of the circle to = the height in order to get the square when looking at it from that side.

    and yes the integrand will just be rdrdzd(theta).
     
  16. Nov 24, 2007 #15
    [tex] 2\pi (\int \int rdrdz) [/tex]
    0-2R for first integral and 0-R for second one. how would I go from there? the rdr is confusing me a little.
     
  17. Nov 24, 2007 #16
    [tex]\int_0^{2\pi}d\theta\int_0^R rdr\int_0^{2R}dz[/tex]

    the rdr just means integrate r w.r.t. r just like: [tex]\int x dx[/tex]
     
  18. Nov 24, 2007 #17
    I got 2pi r^3 for final answer
    after dividing by two. How does that sound?
     
  19. Nov 24, 2007 #18
    yep that's what I got.

    the volume for a cylinder is pi r^2 *h. h=2r ->2pir^3 and then *.5 = pi r^3
     
  20. Nov 24, 2007 #19
    cool! Thanks a lot for all your help. ^_^
     
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