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Volume between truncated cone and an inclined plane

  1. Nov 5, 2011 #1

    Simon Bridge

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    This is something I remember as a standard problem given to college math and physics students ... I've been hunting for a model answer online but no luck: everyone is happy to do the cylinder on it's side or a truncated cone or the intersection of two objects with a lot of symmetry in common but not this one for some reason.

    I can brute-force it but was wondering if there is an elegant method.

    Basic Description
    Imagine a glass whose inner surface can be modeled as a truncated cone open at the wide end (eg. a martini glass or a conical beer glass). Fill it with water. Tilt the glass so some water spills ... keep going, until the level of the water reaches the top limb of the bottom surface of the glass.

    Q. what is the volume of water remaining?

    My reasoning
    - this is a truncated cone intersected by an inclined plane:

    cone: [itex]x^2+y^2=mz: b<z<t[/itex]

    so b is the bottom of the glass and t is the top, and m is the radial gradient of the sides.

    If I tilt the glass towards +y about x, but rotate the axis with the tilt, then gravity rotates the other way.
    The plane would be: [itex](x,y,ay+c): a=m(t-b)/(t+b), c=t(1-a/m)[/itex];


    A. Take a volume integral between the cone and the plane.

    temptation is to slice the volume along z, thicknesses dz.
    The cross-section at each z will be a circle radius R(z) cut by a chord at y=r(z):|r|≤R.
    This area can be evaluated with a trig substitution so the volume integral becomes:

    [tex]\int_b^t \pi R^2 dz - R^2 \cos^{-1} \left ( \frac{r}{R} \right ) -r \sqrt{R^2-r^2}dz[/tex]

    Where [itex]R^2=mz[/itex] and [itex]r=\frac{b}{m}-\frac{z-c}{a}[/itex]
    (The first term is just the volume of the cone.)

    Which looks ... cosy.

    So you see why I feel I may have missed a simplification?

    .... hmmm....
    B: take the volume under the cone and subtract it from the volume under the plane inside the ellipse (the intersection of the plane and the cone is an ellipse - so it's projection onto a horizontal plane at, say, z=b, would also be something eggy... this doesn't look any easier).

    the real challenge
    I've been asked for a model answer...
    I need to be able to show this to someone who is not very comfy with calculus.
    Probably I really want the solution in terms of the depth of the glass and the radii or the top and bottom ... I figured I could plug that in later.
    Last edited: Nov 5, 2011
  2. jcsd
  3. Nov 6, 2011 #2

    I like Serena

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    Hi Simon Bridge! :smile:

    Cosy... but the integral is quite doable.
    No weird functions that can't be integrated!
    So I suspect you can't do it much simpler with your approach.

    Btw, your equation for the cone is not quite right.
    It should have [itex]m^2 z^2[/itex] at the RHS.

    I can give you an alternative method.
    I haven't worked out the details though.

    You can tilt the cone and simply integrate without any chords or other special stuff.
    The equation for a tilted cone (rotated along the x-axis with angle phi) is:
    [tex]x^2 + (c^2 - m^2 s^2)y^2 + (s^2 - m^2 c^2) z^2 - 2 y z (c s + m^2 c s)=0[/tex]
    where [itex]c=\cos \phi[/itex] and [itex]s=\sin \phi[/itex].

    Fixate z, and solve y as a function of x to find an expression you can integrate.

    To find this equation, take a look at: http://en.wikipedia.org/wiki/Quadric

    Your cone can be written as: [itex]x^T Q x = 0[/itex], where Q is a diagonal matrix.
    Replace it by [itex]x^T R^T Q R x = 0[/itex], where R is a rotation matrix, and you have a tilted cone.

    To explain this to someone not comfy with calculus though... o:)
  4. Nov 6, 2011 #3

    Simon Bridge

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    Argh! I knew that! <mutter mumble grumble head-slap!>
    I was even thinking that when I derived it (I don't memorize this stuff.)
    "Its a whole bunch of circles," I said to myself said I, "with radius mz." said I.

    OK - I'm over it now :)

    Interesting - my cone is truncated though ... I suppose one could work out the equivalent scenario for a non-truncated cone and just subtract the volume where it was truncated rather than work out rotation (below).

    I know - right?!

    The sliced method is probably the easiest to visualize. :/
    When I outlined the strategy the person just went all blank-face: this is a college grad (IT though). Told him it was like slicing bread and he brightened up - now he's chugging through it ... I can't wait.

    To think I've had 16yo Secondary students discover calculus and there's college grads who run for cover at the sight of a medial s! <mutter> Oh well - we persevere.

    Still open to tips - though I'll probably just have to build the concepts instead.

    Hmmmm ... if someone with a nice modelling package could womp up a couple of pics for me - one of the situation and one illustrating the slices, exploded diagram style? That would be cool.
    (I don't feel like arm-wrestling Blender - the other one I have is gnu-octave. I'd probably just edit a pic of a beer glass - hmmm... beeer...)
  5. Nov 6, 2011 #4

    I like Serena

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    Really! I expected better of you!
    Sloppy work! I'm taking points off!! :grumpy: :rolleyes: :wink:


    Here you go!
    I used a different program to make it.


    And here's another one:


    It illustrates that you're trying to bend the laws of nature, just to be able to find out what you can directly measure if you poor it into a measuring beaker (without bending gravity to you own purposes)!
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