I have the expression of a vector e, with 3 parameters tt1 tt2 and tt3:

e1 = -(Sin[tt1]*Sin[tt2])/2 - (Sin[tt2]*Sin[tt3])/2
e2 = (Cos[tt1]*Sin[tt2])/2 - (Cos[tt3]*Sin[tt2])/2
e3 = (Cos[tt1]*Cos[tt2]*Sin[tt3])/2 - (Cos[tt3]*Sin[tt1])/2 - 
       (Cos[tt1]*Sin[tt3])/2 + (Cos[tt2]*Cos[tt3]*Sin[tt1])/2
e = {e1, e2, e3};

and the range of tt1 to tt3 are all from 0 to 2 Pi. In this case, since we have 3 parameters, the endpoints of e should form a region rather than a surface, and I'd like to plot this region, or its boundary (preferably both).

The only relevant command I found is the RegionPlot3D command, but seems that one handles the Cartesian coordinates with inequalities as constraints, which is not true in my case.

Could someone give me a hand?


When I try this code:

  {e1 == -(Sin[tt1]*Sin[tt2])/2 - (Sin[tt2]*Sin[tt3])/2, 
   e2 == (Cos[tt1]*Sin[tt2])/2 - (Cos[tt3]*Sin[tt2])/2, 
   e3 == (Cos[tt1]*Cos[tt2]*Sin[tt3])/2 - (Cos[tt3]*Sin[tt1])/2 - 
            (Cos[tt1]*Sin[tt3])/2 + (Cos[tt2]*Cos[tt3]*Sin[tt1])/2, 
   Cos[tt1]^2 + Sin[tt1]^2 == 1, 
   Cos[tt3]^2 + Sin[tt3]^2 == 1}, 
  {e1, e2, e3}, 
  {Cos[tt1], Sin[tt1], Cos[tt3], Sin[tt3]}]

it gives me an error:

General::ivar: -(1/2)Cos[tt3]Sin[tt1] + 1/2Cos[tt2]Cos[tt3]Sin[tt1] - 1/2 Cos[tt1]Sin[tt3] + 1/2Cos[tt1]Cos[tt2]Sin[tt3] is not a valid variable.

I thought that Groebner basis is used for solving equations, but in my case, I'm not trying to solve e1 == e2 == e3 == 0, instead, I'm trying to plot the region that can be traced by the endpoints of e. Besides, tt1, tt2 and tt3 are independent variables, so I doubt if we could eliminate some of them, when I'm trying to plot the boundary of the region?

Update II

To Andre: thanks a lot, I tried the code in the link, and it probably gives me the correct result. Now that I have this region, I wonder if I can do some further operations, such as obtaining some intersection of this region about some plane, for example the plane $x+y=0$? Moreover, can I get the volume of this region?

The above answer should have already solved my problem, but I'm kind of curious about the method using the Grobener basis, as well, because I can hardly understand the code in the above link (sorry I'm quite new to Mathematica).

You guys are right, the code works after I clear the variables, sorry for the mistake.. Now the code gives me:

{e1^4 + 2 e1^2 e2^2 + e2^4 - 2 e1^4 Cos[tt2] - 4 e1^2 e2^2 Cos[tt2] - 
  2 e2^4 Cos[tt2] + e1^4 Cos[tt2]^2 + 2 e1^2 e2^2 Cos[tt2]^2 + 
  e2^4 Cos[tt2]^2 - e1^2 Sin[tt2]^2 - e2^2 Sin[tt2]^2 + 
  2 e1^2 Cos[tt2] Sin[tt2]^2 + 2 e2^2 Cos[tt2] Sin[tt2]^2 - 
  e1^2 Cos[tt2]^2 Sin[tt2]^2 - e2^2 Cos[tt2]^2 Sin[tt2]^2 + 
  e3^2 Sin[tt2]^4}

I wonder what I can do with this to plot the 3D region? Sorry if my question is silly.


closed as off-topic by MarcoB, m_goldberg, Yves Klett, Öskå, Jens Jun 4 '16 at 16:56

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question arises due to a simple mistake such as a trivial syntax error, incorrect capitalization, spelling mistake, or other typographical error and is unlikely to help any future visitors, or else it is easily found in the documentation." – MarcoB, m_goldberg, Yves Klett, Öskå, Jens
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Using GroebnerBasis[] on your equation system, and arbitrarily leaving out tt2 in the elimination list yields an expression in terms of e1, e2, e3, and tt2 (which you can use as a parameter). GroebnerBasis[{e1 == ..., e2 == ..., e3 == ..., Cos[tt1]^2 + Sin[tt1]^2 == 1, Cos[tt3]^2 + Sin[tt3]^2 == 1}, {e1, e2, e3}, {Cos[tt1], Sin[tt1], Cos[tt3], Sin[tt3]}] $\endgroup$ – J. M. is away May 10 '16 at 1:19
  • $\begingroup$ First clear your settings for e1, etc. $\endgroup$ – Daniel Lichtblau May 10 '16 at 1:40
  • $\begingroup$ It gives you an error because you previously assigned expressions to your ek. Use Clear[] and try again. In any case, you can then use ContourPlot3D[] on a list of contours with varying ttk. $\endgroup$ – J. M. is away May 10 '16 at 1:40
  • $\begingroup$ strongly related: $\endgroup$ – andre314 May 10 '16 at 1:47