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- Thread starter Terrell
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- #2

Simon Bridge

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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.

- #3

Simon Bridge

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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".

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- #5

Simon Bridge

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You got the formula you did by following the general pattern ... but the setup was generally fine: you

If you explore this area you'll run ahead of your course ;)

- #6

Mark44

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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.why can't it be like this?

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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

You got the formula you did by following the general pattern ... but the setup was generally fine: youcanrotate 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 ;)

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i've always wondered if this was how it's supposed to be in geometry class. thank you for bringing that theorem up sir!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.

- #9

Simon Bridge

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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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