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3DPlot triple integrals volume (x,y,z) under a curve?

vick ratnam
Posted 9 years ago

Hi,

I have a triple integral and i would like to plot it in Mathematica to understand the object and the implications of its volume under the curve.

It's easy to integrate and the answer is 57375 / 64.

The triple Integral is as follows:

? ? ? x^3 * y^5 * z dxdydz z = (1,2); y = (z, 4z); and x = (z/y, 2z/y)

I have done the following, most of the syntax i copied from another demonstration but i still think it is not the real shape of the object in question:

r = ImplicitRegion[{z > 0, y > 0, z >= 0}, {x, y, z}];
 i = HoldForm[
    Integrate[x^3*y^5*z, {z, 1, 2}, {y, z, 4*z}, {x, z/y, 2*z/y}]];
 cp = ContourPlot3D[{z, y, z}, {x, 0, 40}, {y, 0, 40}, {z, 0, 40}, 
    Mesh -> None, ContourStyle -> {Red, Green, Blue, Orange}, 
    PlotLegends -> "Expressions"];
 rp = RegionPlot3D[r, PlotPoints -> 100, Background -> Black];
 TraditionalForm[
  Column[{r, Row[{i, "=", ReleaseHold[i]}], 
    Row[{"Volume[r]= ", Volume[r]}], Row[{cp, rp}]}, 
   Alignment -> Center]]

Can someone help me to plot the actual object in 3D?

Thanks
POSTED BY: vick ratnam
6 Replies
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vick ratnam
vick ratnam
Posted 9 years ago

Thanks a lot Jason!

Cheers
POSTED BY: vick ratnam
vick ratnam
vick ratnam
Posted 9 years ago

What about plotting the curve? You said it was difficult!

What we have plotted is only the range of the object, isn't it?

By the way how do you get the plot bigger in your last reply?

Thanks a lot for your help!
POSTED BY: vick ratnam

You can look into the various ways of plotting multivariable functions, look here and here.

To get a plot larger, use the ImageSize option.

Here is one method, introduced in version 10.2,

DensityPlot3D[x^3*y^5*z, {z, 1, 2.5}, {y, 0, 7}, {x, 0, 2.5}, 
 RegionFunction -> 
  Function[{x, y, z}, 
   1 <= z <= 2 && z <= y <= 4 z && z/y <= x <= 2 z/y], 
 PlotPoints -> 100]

Mathematica graphics
POSTED BY: Jason Biggs

Apparently ImplicitRegion is new to version 10 - in your original code you had ImplicitRegion, you should notice that it is in blue, meaning it isn't defined.

This will also work,

RegionPlot3D[
 1 <= z <= 2 && z <= y <= 4 z && z/y <= x <= 2 z/y, {z, 1, 2.5}, {y, 
  0, 7}, {x, 0, 2.5}, PlotPoints -> 100]  
POSTED BY: Jason Biggs

Plotting your curve is difficult, since it is a function of 3 dimensions, you would need to do a ContourPlot3D. But plotting the region in question is not too difficult. The straightforward way to do this is with RegionPlot3D

RegionPlot3D[
 ImplicitRegion[
  1 <= z <= 2 && z <= y <= 4 z && z/y <= x <= 2 z/y, {x, y, z}], 
 PlotPoints -> 100]

Mathematica graphics

But the trouble is that it looks awful, so you can use a function created by Simon Woods, detailed here,

contourRegionPlot3D[region_, {x_, x0_, x1_}, {y_, y0_, y1_}, {z_, z0_, z1_}, 
  opts : OptionsPattern[]] := Module[{reg, preds},
  reg = LogicalExpand[region && x0 <= x <= x1 && y0 <= y <= y1 && z0 <= z <= z1];
  preds = Union@Cases[reg, _Greater | _GreaterEqual | _Less | _LessEqual, -1];
  Show @ Table[ContourPlot3D[
     Evaluate[Equal @@ p], {x, x0, x1}, {y, y0, y1}, {z, z0, z1}, 
     RegionFunction -> Function @@ {{x, y, z}, Refine[reg, p] && Refine[! reg, ! p]},
     opts], {p, preds}]]
contourRegionPlot3D[
 1 <= z <= 2 && z <= y <= 4 z && z/y <= x <= 2 z/y, {z, 1, 2.5}, {y, 
  0, 7}, {x, 0, 2.5}, AxesLabel -> {"x", "y", "z"}]

Mathematica graphics
POSTED BY: Jason Biggs
vick ratnam
vick ratnam
Posted 9 years ago

Hi,

Thanks for replying!

The second plot works fine. However, i'm not able to duplicate your first plot. I'm using Mathematica 9 Home Edition.

The error I'm getting is as follows:

RegionPlot3D::argr: RegionPlot3D called with 1 argument; 4 arguments are expected. >>

Can you help with this one?

Thanks
POSTED BY: vick ratnam
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