@hitstudent 

A:=plot3d([x,40*cos(t),40*sin(t)], x=-60..60, t=0..2*Pi, style=surface, color=green):
B:=plot3d([20*cos(t),20*sin(t),z], z=sqrt(40^2-20^2*sin(t)^2)..100, t=0..2*Pi, style=surface, color=yellow, transparency=0.2):
plots[display](A, B, scaling=constrained, axes=normal);
F:=sqrt(40^2-y^2):
int(r*eval(sqrt(1+diff(F,x)^2+diff(F,y)^2), y=r*sin(t)), [r=0..20, t=0..2*Pi]);  # Exact value of the area
evalf(%);  # Approximate value of the area

Addition.  We can see that the exact value of this area is expressed in terms of elliptic integrals. This shows that this figure is not an ellipse, as ellipse area is easily expressed through its axis.

@hitstudent

plot3d([r*cos(t), r*sin(t), r^2], r=0..3, t=0..2*Pi, axes=normal);

Of course Maple found the area only of one side of this surface.

@asa12  I am not a Maple developer and do not know their plans for future versions.

@J4James 

1. I do not understand the meaning of this phrase  "The starting point for f(x)=x in 0<x<4 has to be x-x/4*4 ".

2. type(x/4, integer)=false  means that  x  is not divisible by 4 .

3. Your code is correct, but it can be a bit easier:

f:=x->piecewise(type(x/4, integer)=false,x-floor((x)/4)*4, undefined);
A:=plot(f, -12..12, scaling=constrained, discont):
B:=plot([seq([-8+4*k,0],k=0..4), seq([-8+4*k,4],k=0..4) ],style=point, symbol=circle):
plots[display](A, B);

Note that  f(x)=f(x-4)=f(x+4)
 

@J4James  

1. If we shift the graph of a function to the right by 2 units, simply replace  f(x)  by  f(x-2) .

2. The sequences of  x-coordinates of the circles  (below and above  x-axis)  form two arithmetic progressions.

f:=x->piecewise(type((x+2)/4,integer)=false,x-floor((x+2)/4)*4, undefined):
A:=plot(f(x-2), x=-10..10, scaling=constrained, discont):
B:=plot([seq([-8+4*k,-2],k=0..4), seq([-8+4*k,2],k=0..4) ],style=point, symbol=circle):
plots[display](A, B);
 

@Markiyan Hirnyk

6=sqrt(36) 

@kharonsen  You forgot to copy a minus sign in front of  evalindets :

A:=2*Pi^2*(-1+delta)/r^(1+delta);
-evalindets(A, 'And(`+`, satisfies(f->sign(op(1,f))=-1))', t->-t);

@Preben Alsholm  for the useful information

@acer  Thank you very much for your solution. Your procedure also fixes the bug in  NumberTheory:-RepeatingDecimal(p)  if  -1<p<0

@AmirHosein Sadeghimanesh I don't see any bugs in the work of  

T := combinat[cartprod]([[1, 2, 3]$n]);

Example of work:

T := combinat[cartprod]([[1, 2, 3]$3]);

i:=0:

while not T[finished] do

i:=i+1: L[i]:=T[nextvalue]();

od:

convert(L, list);

I have not seen your procedure and therefore do not know why it produces a wrong result.

@Markiyan Hirnyk 

https://en.wikipedia.org/wiki/Repeating_decimal

@Markiyan Hirnyk  Now I write a procedure in which is displayed a repeating decimal with highlighted period. This can be done by using a line above repeating group of digits, but it is possible also by parentheses or by dots as in my question. I know how to implement the first 2 methods, but for completeness I would like to learn, and the latter method. 

@emendes  Use this modification of Carl's procedure:

MyTest:= (F::list(algebraic), V::And(list(name), satisfies(V-> nops(V)=nops(F))))->
not ormap(k-> andmap(f-> f::freeof(V[k]), [F[..k-1][], F[k+1..][]]), {$1..nops(V)}):

Example of use:

MyTest([x*y*theta[5]+x*theta[2], x*y*theta[15], x*theta[22]+y*theta[23]+z*theta[24]], [x,y,z]);

                                                    false

@Markiyan Hirnyk 

The original equation is reduced to the equation  tan(Pi*t) = tan(2*Pi*t^2)  by the 4 steps:
1. The second and third terms of the original equation is transported to the right.
2. On the right the term  tan(Pi*t)  is factored out.
3. Divide both sides by the difference in parentheses.
4. On the left apply the formula for the tangent of the double angle.

A further solution is obvious.

If a fraction  -1< q <0  then an error occurs  (the minus sign disappears):

NumberTheory[RepeatingDecimal](-1/7);