I'm struggling to figure out how the format this in Mathematica and could really use some help. My problem is this: Plot the solid of revolution obtained by rotating the region enclosed by the graphs of

    {y == 16 − x^4, y == 0, x == 2, x == 3}

about the y-axis and calculate its volume.

Currently I'm trying the following to get it to plot:

RevolutionPlot3D[{(16 - x^4)}, {x, 2, 3}, RevolutionAxis -> "Y"]

But it isn't giving me the shape I'm looking for. Am I formulating the problem wrong?

I know its going to be a negative solution with pi in it and it would be the integration bound by 2 and 3.


You want to rotate the region shaded in this plot,

Plot[16 - x^4, {x, 2, 3}, Filling -> Axis, PlotRange -> {{1, 4}, All}]

enter image description here

The main problem with your code is that you should set the rotation axis to the z axis, which is interpreted as the vertical axis in a 3D plot

RevolutionPlot3D[{(16 - x^4)}, {x, 2, 3}, RevolutionAxis -> "Z", 
 BoxRatios -> {1, 1, .5}]

enter image description here

(try it without the BoxRatios option to see why I include that)

But that only gives the curve, not the other sides of the solid. For the top, I'll just plot a constant value of z=0 and for the x=3 side, I will use the parametric input form

RevolutionPlot3D[{{(16 - x^4)},
  {3, 16 - x^4}},
 {x, 2, 3},
 BoxRatios -> {1, 1, .5}]

enter image description here

Finally, if you want a nice looking solid with one color, and no mesh lines, you can use

RevolutionPlot3D[{{(16 - x^4)}, {0}, {3, 16 - x^4}}, {x, 2, 3},
 BoxRatios -> {1, 1, .5},
 Mesh -> None,
 PlotPoints -> 100,
 Axes -> False,
 Boxed -> False,
 PlotStyle -> Directive[Orange, Specularity[White, 50]]]

enter image description here

You could get this via RegionPlot3D, but the results are not great

  2 <= Sqrt[x^2 + y^2] <= 3 && (16 - (x^2 + y^2)^2) <= z <= 0, {x, y, 
   z}], PlotRange -> {{-3.5, 3.5}, {-3.5, 3.5}, {-70, 10}}, 
 BoxRatios -> {1, 1, .5}, PlotPoints -> 200, Axes -> True]

enter image description here

| improve this answer | |
  • $\begingroup$ Alright, thank you the formatting on the Rotate 3d really threw me off. $\endgroup$ – Nick Mar 31 '16 at 15:40
  • $\begingroup$ Can you help me set up the plotting another one? I'm not sure how to go about plotting it to infinity. Find the volume of the solid obtained by rotating the region below the graph of y = e^(−x) about the x-axis for 0 ≤ x < ∞ Currently I have this $\endgroup$ – Nick Mar 31 '16 at 15:43
  • $\begingroup$ RevolutionPlot3D[{y = e^(-x)}, {0}, {x, 0, infinity}] RevolutionPlot3D::pllim: "Range specification {0} is not of the form {x, xmin, xmax}." Sorry for multi comment, accidentally hit enter instead of shift enter $\endgroup$ – Nick Mar 31 '16 at 15:44
  • $\begingroup$ @Nick - From your wording it sounds like you want to find the volume of these solids of revolution, if that is the case then you want to use Integrate, not RevolutionPlot3D. Anyway, you can't do a plot with the range extending to infinity; what would that look like? $\endgroup$ – Jason B. Apr 1 '16 at 8:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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