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I am attempting to revolve $f(x)=-x(x-3)(x-2)^2$ about $x=0$. What I have come up with is that I need to redefine each interval between the relative extrema as a separate function, and then treat it as a washer problem. However, I don't have the requisite skill set to redefine the function in this way. Is there a simpler approach to this problem? A delineation of the steps would be useful.

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up vote 7 down vote accepted

RevolutionPlot3D[] supports a RevolutionAxis option that controls how you want to revolve the curve you are interested in. Since it treats functions in its first argument as $z=f(x)$, your requirement here corresponds to rotating about the $z$-axis. Thus,

RevolutionPlot3D[-x (x - 3) (x - 2)^2, {x, 0, 4}, RevolutionAxis -> {0, 0, 1}]

some weird hat

For comparison, here is what RevolutionAxis -> {1, 0, 0} does:

I think I saw pottery like this once

As m_goldberg notes, one can use the shorter forms RevolutionAxis -> "Z" or RevolutionAxis -> "X" instead of the explicit axis settings in the examples given above.

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i think your answer is clearly the better answer rather than my paltry effort. – ubpdqn Feb 19 at 5:09

You can use the built-in function RevolutionPlot3D:

f[x_] := -x (x - 3) (x - 2)^2
RevolutionPlot3D[f[x], {x, 0, 3}, Mesh -> None, Background -> Black, 
 PlotStyle -> Opacity[0.5]]

enter image description here

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Thank you, I realize that I should have posted this in Mathematics, not Mathematica. Either way, I hope your answer helps someone. I will accept your answer when I can. – fmi11 Feb 19 at 3:40
Sorry...ok, good luck :) – ubpdqn Feb 19 at 3:41
@fmi11 i think J.M.'s answer is clearly the better answer rather than my paltry effort. – ubpdqn Feb 19 at 5:09

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