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Why isn't the volume of a pyramid Bh/2?

  1. Jun 26, 2010 #1
    Hey, so I guess my question is as follows. I've got a problem picturing a pyramid only being Bh/3 and not Bh/2. I've looked at the proof why it is so, and it seems reasonable when looking at it, BUT if I imagine a cone vs. a cylinder sliced with vertical rather than horizontal cuts I'm lost. See, you get a cylinder if you rotate a rectangle, but you get a cone with the same height if you rotate a triangle whose area is half of that of a rectangle. And if I then imagine the cone and cylinder sliced vertically to slices of virtually zero width (z component, as opposed to height and radius of the underlying circle) I get an indefinite number of them shaped as either rectangles with area x or triangles with an area of x/2.

    So I guess what I wanted to ask is what am I missing. Because if my thinking as presented above was true, the volume of the cone vs. the cylinder would have a ratio of a half and not a third. Where is the rest of the cylinder and why is it not covered in the said image of an indefinite number of rectangles vs. triangles?

    Thanks in advance.
  2. jcsd
  3. Jun 27, 2010 #2


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    Here's the point you are missing. Like you said, if you rotate a rectangle around an axis, you get a cylinder. Now imagine cutting that rectangle into two triangles of equal area. As you said, if you rotate the lower(green) triangle around the same axis, you get a cone. But look at the upper(blue) triangle - it's area is further away from the axis of rotation than the lower triangle that makes up the cone. So the volume of the solid created when you rotate the upper triangle around the axis is greater than the volume of the cone. Integration shows that the volume of the lower solid (the cone) is 1/3 the volume of the cylinder, and the volume of the upper solid (which is a cylinder with a cone removed) is 2/3 the volume of the cylinder. Does this make sense?

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  4. Jun 27, 2010 #3
    Yeah, thanks, I get it now. I was stuck in the mindset that the volume of the slice at width x is the same regardless of whether the blue or green triangle was rotated. I now see that the ring, although still of the same width as compared to the green triangle's one packs more volume because it is on the outer layer of the circle as opposed to being directly adjacent to the centre.

    And am I then right in thinking that even an indefinite number of vertical slices I've described in my original post does not cover the whole of the cylinder?
  5. Jun 27, 2010 #4


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    I'm not sure I uderstand what you mean, but you can do the integrals with either horizontal slices (adding up a bunch of disks stacked on top of each other), or with vertical slices (adding up a bunch of cylindrical shells), and if you do it right you will get the same thing. For the green solid, using horizontal slices gives an integral:
    [tex]\int_0^h\pi R^2\frac{(h-z)^2}{h^2}dz = \frac{\pi R^2 h}{3}[/tex]
    While using vertical slices gives an integral:
    [tex]\int_0^R2\pi r h \frac{R-r}{R}dr = \frac{\pi R^2 h}{3}[/tex]

    Can you write the corresponding integrals for the blue solid?
  6. Jun 28, 2010 #5
    Well, what I mean is that if you sliced the cone into vertical slices with no width dimension, each slice representing a rectangle of two identical triangles, you would have to conclude that the volume of the cone is only half of the one of the cylinder. But since this isn't the case, then not every green triangle has a corresponding blue one, but rather there are more blue ones in the cylinder than green ones. Do you see what I'm trying to say now and where I got into trouble? :)

    As for the corresponding integrals for the blue solid, no, I can't write them yet. Basically, I'm revising high school Maths to prepare for a second degree in Physics and I haven't yet come to the part with integrals. I guess I'll come back to this when I'm there, because I'd rather not jump ahead. But thanks for your help!
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