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    take a diagonal slice across a cylinder, starting from the top 'rim', and finishing somewhere on the base. take the smaller of the two slices. given we know most of the dimensions (radius, height, angle of cut...) can we find the volume? i would guess so. but how?
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    lol i had a silly idea this cant be right

    treating it as a parralelogram

    I took the larger piece but you could minus it from the original
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    No. Just did a triple integral on mathematica. It's very very very not pretty!
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    (Original post by Infinity_Kev)
    lol i had a silly idea this cant be right

    treating it as a parralelogram

    I took the larger piece but you could minus it from the original
    i'm interested in the other kind of cut, which goes down through the base, not the side.
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    (Original post by SsEe)
    No. Just did a triple integral on mathematica. It's very very very not pretty!
    oh. how depressing.
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    can you creat a picture chewwy??
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    If you meant the cylinder was cut by a plane passing through A and E like in the image below, then here's the answer for it.
    You said we knew r, h, angle of cutting, say x (x = EAC)

    Then EC = a = h.tanx
    V(cylinder ACEF) = ¼πa².h
    Then you only need to divide that volume by 2, you'll get the volume of the small piece

    edit: forgot 1/4 lol
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    (Original post by BCHL85)
    If you meant the cylinder was cut by a plane passing through A and E like in the image below, then here's the answer for it.
    You said we knew r, h, angle of cutting, say x (x = EAC)

    Then EC = a = h.tanx
    V(cylinder ACEF) = ¼πa².h
    Then you only need to divide that volume by 2, you'll get the volume of the small piece

    edit: forgot 1/4 lol
    ACEF is not a cylinder.
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    (Original post by chewwy)
    ACEF is not a cylinder.
    why not then???
    edit: this is the picture in 1 face ... plz imagine your head... I am not good at drawing
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    (Original post by BCHL85)
    why not then???
    edit: this is the picture in 1 face ... plz imagine your head... I am not good at drawing
    get a glass of water, fill it less than 50%, and tilt. you'll see.
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    (Original post by chewwy)
    get a glass of water, fill it less than 50%, and tilt. you'll see.
    actually u cant see that ACEF isnt a cylinder using the glass of water... although it's a nice idea
    attachment shows why it's not a cylinder... (it's a top view)
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    (Original post by yazan_l)
    actually u cant see that ACEF isnt a cylinder using the glass of water... although it's a nice idea
    attachment shows why it's not a cylinder... (it's a top view)
    oh .. I made a mistake ... lol
    very stupid of me. Let me try again, then
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    I've got an idea that it is the
    (final cross-sectional area / initial cross sectional area) h
    but thats probably quite stupid.
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    I don't think you could do it with an integral (volume of revolution), could you? I mean any curve on a graph that I can think of would not revolve 360 degrees around the axis and give the desired shape. Hmmm....
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    the volume would depend on the angle of the cut, the height and the radius. I'm not sure about integrals on mathematica, but this is definately a very complicated problem. It's calculus of 4 variables ie. f(x,y,z,w). pfft, ask a cambridge mathmo lol.
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    partial derivatives?
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    Has anybody else tried doing it on mathematica?
 
 
 
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