Skip to main content

plotting - Calculating volume between two surfaces of revolution



Once again, I'm very new to Mathematica and winging it as I go. I probably don't have the math education at this point to even justify buying it in the first place, but that's off topic.


I wanted to model a simple curve by rotating it around the x axis, and then actually use it as a function and take its integral, etc.


To be more specific, all I want to do is rotate the area enclosed by y == 10 - 6 x + x^2 and y == x around the x axis to get a 3D model and calculate its volume. However I either can't find the related Help topic, or did find it but found that it so quickly jumped from the basic explanation to complicated examples that I couldn't understand it, something I'm finding more common. Does anyone have any advice?


Sorry if this question is too obvious or dumb.


EDIT: Oops, my mistake, the function I was trying to plot was 10 - 6 x + x^2 with a negative 6x. This way y=x intersects it and what's being plotted, assuming the bounds are 2 to 5, is just the area in between the curves. I haven't yet been able to figure out how to do this. Everything just seems to plot both functions on top of each other.



Answer



Mathematica is a good for learning support. Sometimes it is very useful to visualize the problem.


Response after OP's edit: This is analogous to the first problem You have stated.


Plot of the boundary for 2 < x < 5:


 RevolutionPlot3D[{{x},{10 - 6 x + x^2}}, {x, 0, 7}, {th, Pi, 2 Pi},

        RevolutionAxis -> {1, 0, 0}, BoxRatios -> {1, .5, 1}, Lighting -> "Neutral",
        PlotStyle -> {LightRed, Green}, Axes -> True, AxesOrigin -> {0, 0, 0}, 
        AxesStyle -> Thick, Boxed -> False, ImageSize -> 800, 
        TicksStyle -> Directive@{20, Bold}]

enter image description here


   Show[RevolutionPlot3D[{{x}, {10 - 6 x + x^2}}, {x, 0, 7}, {th, Pi, 2 Pi}, 
           RevolutionAxis -> {1, 0, 0}, PlotStyle -> {LightBlue, Orange}, 
           Axes -> True, AxesOrigin -> {0, 0, 0}, AxesStyle -> Thick, Boxed -> False,
           ImageSize -> 800, TicksStyle -> Directive@{20, Bold}, Lighting -> "Neutral"]

        ,
        RegionPlot3D[2(10-6x+x^2)^2 && z^2+y^2           {x,1,6}, {y,-6,6}, {z,-6,6}, BoxRatios -> 1, Mesh -> None, PlotRange->All
           PlotPoints -> 50, PlotStyle -> Directive[{Red, Opacity@.8}]
                    ],
        BoxRatios -> {5, 2, 3}]

(plots not in scale)


enter image description here


If You do not have math skills, use Mathematica to sum up parts of the space which fullfil set of ineaqualities.



 NIntegrate[Boole[
     2 < x < 5 && z^2 + y^2 > (10 - 6 x + x^2)^2 && z^2 + y^2 < x^2
           ], {x, 2, 5}, {y, -30, 30}, {z, -30, 30}] // Chop


73.5133



Later, You can check this with analitical solution (look to the link Jens has added).


Pi (Integrate[x^2 - (10 - 6 x + x^2)^2, {x, 2, 5}]) // N



73.5133



Edit2, I think following line shoud be here since we are trying to learn something. Why integrated function has this form? You can look at it in this way: First we are calculating volume limited by revolution surface of x and then we are subtracting volume limited by 10-6x-x^2. And since integral is linear:


enter image description here


Comments

Post a Comment

Popular posts from this blog

plotting - Filling between two spheres in SphericalPlot3D

Manipulate[ SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, Mesh -> None, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], {n, 0, 1}] I cant' seem to be able to make a filling between two spheres. I've already tried the obvious Filling -> {1 -> {2}} but Mathematica doesn't seem to like that option. Is there any easy way around this or ... Answer There is no built-in filling in SphericalPlot3D . One option is to use ParametricPlot3D to draw the surfaces between the two shells: Manipulate[ Show[SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], ParametricPlot3D[{ r {Sin[t] Cos[1.5 Pi], Sin[t] Sin[1.5 Pi], Cos[t]}, r {Sin[t] Cos[0 Pi], Sin[t] Sin[0 Pi], Cos[t]}}, {r, 1, 2 - n}, {t, 0, Pi}, PlotStyle -> Yellow, Mesh -> {2, 15}]], {n, 0, 1}]
Post a Comment
Read more

plotting - Plot 4D data with color as 4th dimension

I have a list of 4D data (x position, y position, amplitude, wavelength). I want to plot x, y, and amplitude on a 3D plot and have the color of the points correspond to the wavelength. I have seen many examples using functions to define color but my wavelength cannot be expressed by an analytic function. Is there a simple way to do this? Answer Here a another possible way to visualize 4D data: data = Flatten[Table[{x, y, x^2 + y^2, Sin[x - y]}, {x, -Pi, Pi,Pi/10}, {y,-Pi,Pi, Pi/10}], 1]; You can use the function Point along with VertexColors . Now the points are places using the first three elements and the color is determined by the fourth. In this case I used Hue, but you can use whatever you prefer. Graphics3D[ Point[data[[All, 1 ;; 3]], VertexColors -> Hue /@ data[[All, 4]]], Axes -> True, BoxRatios -> {1, 1, 1/GoldenRatio}]
Post a Comment
Read more

plotting - Mathematica: 3D plot based on combined 2D graphs

I have several sigmoidal fits to 3 different datasets, with mean fit predictions plus the 95% confidence limits (not symmetrical around the mean) and the actual data. I would now like to show these different 2D plots projected in 3D as in but then using proper perspective. In the link here they give some solutions to combine the plots using isometric perspective, but I would like to use proper 3 point perspective. Any thoughts? Also any way to show the mean points per time point for each series plus or minus the standard error on the mean would be cool too, either using points+vertical bars, or using spheres plus tubes. Below are some test data and the fit function I am using. Note that I am working on a logit(proportion) scale and that the final vertical scale is Log10(percentage). (* some test data *) data = Table[Null, {i, 4}]; data[[1]] = {{1, -5.8}, {2, -5.4}, {3, -0.8}, {4, -0.2}, {5, 4.6}, {1, -6.4}, {2, -5.6}, {3, -0.7}, {4, 0.04}, {5, 1.0}, {1, -6.8}, {2, -4.7}, {3, -1.
Post a Comment
Read more