Mathematica Stack Exchange is a question and answer site for users of Mathematica. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|improve this question
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.

share|improve this answer
    
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

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.