@Markiyan Hirnyk 

If alpha is an algebraic number (over a field)  expressed with RootOfs then

evala( Norm(x - alpha) )

is the minimal polynomial for alpha (by definition).
See ?evala,Norm

@Markiyan Hirnyk 

z>1 was considered implicitly because the ode was considered in an interval containing z=2 and z=3, corresponding to x=1 and x=2.
So, all we need to know is that the hypergeometric function
hypergeom([1/3, 2/3], [3/2], w)
exists (and is C^2) for w<=0.

I prefer to work in cartesian coordonates

restart;
with(plots):with(plottools):
f:=r^2 *cos(theta)+r*sin(theta):
r1:= 0.3+0.1*cos(theta):
r2:= 0.5+0.1*cos(theta):
p:=densityplot(f, theta=0..2*Pi, r=r1..r2,  
colorstyle = HUE, style = patchnogrid):
T:=transform( (u,v) -> [v*cos(u),v*sin(u)]):
display(T(p));

@arman 

When coords=polar, the axes are cartesian.
Why don't you plot directly in cartesian coordinates? Simply use x=r*cos(t), y=r*sin(t).
P.S. Your image is missing.

 

@one man 

Try this simple one which is obvious by hand:

x1^4 + x2^4 + x3^4 - 1 = 0, x3 = 0.

You must be careful with the numbering (0..n) and the fact that in matices indices start at 1.
You should also check the formulae in wiki. Note that the wiki formulae for TSP are completely wrong (they talk there about index 0 which does not exist) and I had to correct them before coding.

Happy New Year!

Thank you for the answer.
So, the reason for opposite orders will remain a mistery.

Happy New Year! (actually also for all the mapleprimes members).

V.A.

@Markiyan Hirnyk 

I don't think that sort could be useful for an answer.
I'd like to know whether there is a consistent/logical rendering order for the objects in display.
The CURVES seem to be on top of POLYGONS and the orders differ.
When using transparency e.g., the order is important.
 

@torabi 

I suspect that the system has no solution (even after trying some other initial conditions).
Unfortunately it is hard to prove that this is the case and even harder to find an existence theorem for such a system. Sorry.
(If the system came from a concrete problem, you should try anoher approach).

@Rouben Rostamian  

Thank you Rouben, I'll keep these worksheets in my "special" collection.
In order to solve the memory problem for animation, one may use Explore. This way a single frame is present in memory but now the animation speed depends on  the computer speed.

anim := proc(tau, lambda, T)
  local R := 6;  # orbit's radius
      display([
      sphere([0,0,0], 1, style=surface, color="Orange"),
      tubeplot([R*cos(t), R*sin(t), 0], t=0..2*Pi, radius=0.04, color=red, style=surface),
      translate(globe(tau, lambda, T), R*cos(T), R*sin(T), 0)],
      view=[-R-1..R+1,-R-1..R+1,-1..1],
      scaling=constrained, lightmodel=light4, axes=none, orientation=[-144,65,0]
    )
end proc:
Explore(anim(tau,lambda,T),
parameters=[[tau=0..2*Pi,animate=false,shown=true], [lambda=0..2*Pi,animate=false,shown=true], [T=0..2*Pi,animate=true]],
initialvalues=[tau=Pi/6,lambda=Pi/4], loop, size=[1000,1000], numframes=240, autorun);

@torabi 

So, what is your conclusion?

I seem to remember that you have posted this system several times, without success.
Are you sure that it is not similar to a simpler one:

dsolve({diff(f(x),x)=1/g(x), diff(g(x),x)=0,  g(0)=0, f(0)=0});

?

@Rouben Rostamian  

Very nice solution, vote up!

PS. An animated 3d plot would be also nice; maybe someone will have the patience to do it.

@one man 
I thought that you want to know how your surfaces really look (globally).
But if you want to test the procedure then please choose examples which satisfy the mentioned limitations.

@one man 
For these implicit surfaces the above procedure cannot be used (for a global representation) because the conditions are not satisfied (about interior point and star-shapeness).
But it is not very difficult to parametrize them.
1.   f:=(x^2+y^2-2/5)^2+(z+sin(x*y+z))^4-1/10;

plot3d([(1/10)*(40+10*10^(1/2)*cos(u))^(1/2)*cos(v),
(1/10)*(40+10*10^(1/2)*cos(u))^(1/2)*sin(v), 
-(1/100)*(40+10*10^(1/2)*cos(u))*cos(v)*sin(v)+
RootOf(-cos(u)*10^(1/2)*cos(v)*sin(v)-10^(1/2)*(10^(1/2)*abs(sin(u)))^(1/2)-4*cos(v)*sin(v)+10*sin(_Z)+10*_Z)],
u=0..2*Pi, v=0..2*Pi, orientation=[60,40]);

2. f := x+y+z(+const);
Probably it's a joke.

3. f:= z-1/2 * exp (sin (x + 5/2 * y + z));

Y:=(k,x,z) -> -(2/5)*x-(2/5)*z+(2/5)*( (-1)^k*arcsin(ln(2*z)) + k*Pi);
plot3d( [seq(Y(k,x,z),k=0..5)], x=0..4*Pi, z=exp(-1)/2 .. exp(1)/2, labels=["x","z","y"],
         scaling=constrained, orientation=[-75,4,-114]);

# The graph has an infinity of connected components, each one being unbounded