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I Finding the volume under the curve of a rotated function

  1. Apr 27, 2016 #1
    why can't it be like this?

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  2. jcsd
  3. Apr 27, 2016 #2

    Simon Bridge

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    Your strategy is to find the area under the curve and multiply it by ##2\pi##?
    The reason it cannot be like that is because the result is not the volume... it doesn't even have the right units.

    Try it for some simple volumes ... i.e. the volume of a cylinder height h and radius b with a hole radius a punched in it. Then f(x) = h when a<x<b and 0 elsewhere.
  4. Apr 28, 2016 #3

    Simon Bridge

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    You may be thinking "well OK, I can see it's not correct, but how come? It looks like the formulae I been taught!"
    This is probably because you have been taught calculus "by rule" ... so you get a bunch of formulas and you don't know where they come from.

    The idea behind integrating for a volume is to divide the shape into lots of small shapes, find the volume of each smaller shape, and add them up.
    The clever part is to choose how you divide up the shape to make the maths easier.

    In your example ... a convenient shape to use would be a cylindrical shell of thickness ##dx##.

    A shell with radius between x and x+dx inside the limits has volume ##dV = 2\pi x f(x)\; dx## ... see how that's a volume?
    It's a length (##2\pi x##) multiplied by a height (##f(x)##) and a width (##dx##).
    Find the overall volume by adding up all the shells with radii between a and b ... ##V=2\pi \int_a^b xf(x)\; dx##

    In your pic you have written, in effect, ##dV = f(x)\;dx## but that's just a height times a width: it's an area.

    You could do it by the basic method you illustrated ... but you have to divide the overall f(x)-donut shape into small wedges instead of shells.
    That gets you a different equation from what you got and it can take harder maths, but there are situations where it works well.

    eg. the cylinder with a hole in it example in the post#2
    ... the volume of a wedge with apex angle ##d\theta## is ##dV=\frac{1}{2}d\theta (b^2-a^2)h## (how did I get that?)
    Now we integrate around the whole circle: $$V = \frac{1}{2}(b^2-a^2)h\int_0^{2\pi}\; d\theta$$

    Compare - method of cylinders for the same problem: ##V = 2\pi h\int_a^b x\; dx##

    Look up "cylindrical-polar coordinates".
  5. Apr 28, 2016 #4
    thank you for your input sir. yes i do agree and understand the intuition of the correct method. 2pi*x is simply the circumference(to add a "dimension" to it), the f(x) stacks up the circumferences as we move along the horizontal axis and deltaX is simply the width because circumference has no width dimension to it, so we have to have that to calculate the volume. my erroneous formula just popped out of my intuition which apparently failed me this time from trying to find other ways to find the volume of it.
  6. Apr 28, 2016 #5

    Simon Bridge

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    Your intuition comes from the human faculty for pattern recognition. It works well as long as you understand where the patterns come from.
    You got the formula you did by following the general pattern ... but the setup was generally fine: you can rotate a function like that, so long at the volume integrated is wedge shaped slices - like cutting a cake. It comes in handy if the shape being cut varies in distance from the symmetry axis ... i.e. what if the cylinder in the example above had an oval hole in it instead of a circular one?

    If you explore this area you'll run ahead of your course ;)
  7. Apr 28, 2016 #6


    Staff: Mentor

    Actually, this isn't too far off. See http://mathworld.wolfram.com/PappussCentroidTheorem.html, where they talk about the 2nd Theorem of Pappus. If you determine the centroid of the region that's being revolved, and then determine the distance the centroid moves in rotation, the volume of the solid of revolution is the product of the distance the centroid moves, and the area of the region being revolved.
  8. Apr 28, 2016 #7
    AHHH! that's it. wedge shape like cutting cakes. i now see why my formula can't work haha! yes it has to vary from the axis of symmetry! thank you for adding more insight into it :D
  9. Apr 28, 2016 #8
    i've always wondered if this was how it's supposed to be in geometry class. thank you for bringing that theorem up sir!
  10. Apr 28, 2016 #9

    Simon Bridge

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    Your formula would have been spot on if the centroid of the area was 1 unit from the rotation axis :)
    Finding the centroid can be a pain - but sometimes it is easy... like maybe f(x) is a circle so the rotated shape is a donut.
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