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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?
  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
    OH I think I get it now. So bascially, its 1/2 a cylinder right?

    http://img77.imageshack.us/img77/8178/shapehi9.gif [Broken]

    So side view is square, bottom is circle
    and other sides are triangles. Thanks!!

    But how would I find the volume of that shape? :bugeye:
    Last edited by a moderator: May 3, 2017
  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.


    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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