For easy use rewrote the program as a procedure. Formal parameters: a  and  b  - sizes of the table,  T - the  duration of the animation,  P - the coordinates of the initial point ,  V - the velocity of the ball, N - (optional) the number of the frames (by default  N=100).

The code of the procedure:

restart;

Billiard:=proc(a::positive, b::positive, T::positive, P::list, V::list, N::posint:=100)

local A, B;

if (P[1]<=0 or P[1]>=a) or (P[2]<=0 or P[2]>=b) then error "Should be 0<=P[1]<=a and 0<=P[2]<=b" fi;

A:=plots[animate](plot,[[2*a/Pi*abs(arcsin(sin(Pi/2/a*(P[1]+V[1]*t)))), 2*b/Pi*abs(arcsin(sin(Pi/2/b*(P[2]+V[2]*t)))), t=0..tau], thickness=5, color=red, numpoints=1000], tau=0..T, frames=N):

B:=plottools[rectangle]([0,b], [a,0], thickness=2, color=green):

plots[display](A, B, scaling=constrained, axes=none)

end proc: 

Examples 1.

Billiard(2, 1, 10, [0.5,0.5], [0.5,-sqrt(3)/2]);

Example 2.  I wonder in what proportions of the parameters the trajectory is closed?

Billiard(4, 2, 40, [1,1], [3/4,2]); 

Here is a simple example of an animation path of a billiard ball on a rectangular table:

restart;

A:=plots[animate](plot,[[3/Pi*abs(arcsin(sin(Pi/2*(0.5+0.2*t)))), 2/Pi*abs(arcsin(sin(Pi/2*(0.5+sqrt(3)/5*t)))), t=0..a], thickness=5, color=red, numpoints=1000], a=0..25, frames=100):

B:=plottools[rectangle]([0,1], [3/2,0], thickness=2, color=green):

plots[display](A, B, scaling=constrained, axes=none); 

restart;

p:=prevprime(41): n:=0:

while p<107 do

p:=nextprime(p);  n:=n+1;  sol:=msolve(y^2 + y - 11=0, p);

if sol<>NULL then L[n]:=[p,sol] fi;

od:

L:=convert(L,list):

op(L);

[41, {y = 21}, {y = 19}], [59, {y = 17}, {y = 41}], [61, {y = 8}, {y = 52}], [71, {y = 25}, {y = 45}], [79, {y = 9}, {y = 69}], [89, {y = 28}, {y = 60}], [101, {y = 33}, {y = 67}]

The conjecture is obvious.

plots[implicitplot]  command makes it possible to build any straight lines. If the coefficient of the y is equal to 0, and the coefficient of the x is not equal to 0, we obtain a vertical line.

Example:

plots[implicitplot]([seq(seq(a*x+b*y = 1, a = -1 .. 1), b = -1 .. 1)], x = -2 .. 2, y = -2 .. 2);

You do not have to convert these numbers into polar form, can immediately build these points:

z0 := sqrt(3)+I;
z1 := -1+I*sqrt(3);
z2 := -sqrt(3)-I;
z3 := 1-I*sqrt(3);

Points := seq(plots[pointplot]([Re(z||i), Im(z||i)], color = red, symbol = solidcircle, symbolsize = 15), i = 0 .. 3):
Names := seq(plots[textplot]([Re(z||i)+.2, Im(z||i), cat('z', i)], color = blue, font = [TIMES, ROMAN, 16]), i = 0 .. 3):

plots[display](Points, Names);

Of course, the same can be build by  plots[polarplot]  command:

restart; 

z0 := sqrt(3)+I: 
z1 := -1+I*sqrt(3): 
z2 := -sqrt(3)-I: 
z3 := 1-I*sqrt(3): 

Points := plots[polarplot]([seq([abs(z||i), argument(z||i)], i = 0 .. 3)], style = point, color = red, symbol = solidcircle, symbolsize = 15):

Names := seq(plots[textplot]([Re(z||i)+.2, Im(z||i), cat('z', i)], color = blue, font = [TIMES, BOLD, 16]), i = 0 .. 3):

plots[display](Points, Names);

Your body is limited to 5 surfaces:

1. y=x  and  y=-x  - two planes (green).

2. x^2+y^2+z^2 = 4  and  x^2+y^2+z^2 = 9  - two spherical surface (yellow).

3. z = sqrt(x^2+y^2)  -  conical surface (cyan).

All surfaces are defined by the parametric equations:

A:=plot3d([r*cos(t)/sqrt(2),r*cos(t)/sqrt(2),r*sin(t)], r=2..3, t=Pi/4..Pi/2, color=green):

B:=plot3d([-r*cos(t)/sqrt(2),r*cos(t)/sqrt(2),r*sin(t)], r=2..3, t=Pi/4..Pi/2, color=green):

C:=plot3d([r*cos(t),r*sin(t),r], r=2/sqrt(2)..3/sqrt(2), t=Pi/4..3*Pi/4, color=cyan):

E:=plot3d([2*cos(phi)*sin(theta),2*sin(phi)*sin(theta),2*cos(theta)], phi=Pi/4..3*Pi/4, theta=0..Pi/4, color=yellow):

F:=plot3d([3*cos(phi)*sin(theta),3*sin(phi)*sin(theta),3*cos(theta)], phi=Pi/4..3*Pi/4, theta=0..Pi/4, color=yellow):

plots[display](A,B,C,E, F, view=[-2.5..2.5,0..2.5,0..3.5],scaling=constrained, axes=normal, orientation=[25,75], labels=[x,y,z]);

You must use the geometric meaning of the definite integral. Your integral is precisely the area of a quarter of a circle of radius 4, because the plot of the function  f(x)=sqrt(16-x^2), x=-4..4, is the upper semicircle of the radius 4.

The name of a function or expression should be  name  type. But

type(omega^2, name); 

                     false

Instead of  omega^2  you can write  `omega^2`

type(`omega^2`, name);

                      true 

restart;

plots[implicitplot](x^3+y^3-3*x*y=0, x=-3..3, y=-3..3, thickness=2, gridrefine=4);

I believe this is the simple way:

[7,9,1]:

add(%[i]*10^(i-1), i=1..nops(L));

                       197

Should be  add  instead of  sum . Write as follows:

Omega := n-> R^n*P(n, z/R)*ln(r)+R^n*add((2*n-4*i+1)*P(n-2*i, z/R)/(i*(2*n-2*i+1)), i = 1 ..

(1/2)*(n-(n mod 2))) ;

restart;

ee := -ln(-a/(b-c)):

ff:=combine(ee) assuming a/(c-b)>0;

gg:=op(1, ff);

subs(gg=numer(gg)/denom(gg), ff);

In your expression you forgot to put the separators (colons or semicolons).

restart;

m := 2:

k := 1.0931:

a := k-m:

b := k+m-1:

z := k*m/10^(.1*10):

simplify((10^(.1*yo))^((b-a+2*p-1)*(1/2))*z^((b-a+2*p+1)*(1/2))*GAMMA((1/2)*(1-b+a-2*p))/(p!*GAMMA(p-a+1)*GAMMA(1+(1/2)*(1-b+a-2*p))));

[seq(limit(%, p = k), k = 0 .. 10)];

Example:

cat(op(0, A[1]), op(A[1]));

                  A1

The inequality  a<b<c  is equivalent to  a<b and b<c . The last entry may be used to manipulate in Maple with this inequality and for its solution.

Example:

x/3-1>1 and x/3-1<5;  #  It's equivalent to 1 < x/3-1 < 5

map(x->3*x, %);  #  Both inequalities are multiplied by 3

solve(%);