@nm  I don't know the answer to the last question. I've never used parentheses when working with matrices.

@oggsait  Use the  legend  option for this.

@ik74  In your drawing, the center of the circle should be to the right and not to the left of the centerline.
The code for this case:

restart;
a:=2: R:=-8: z:=sqrt(R^2-(r-a+R)^2):
plot3d([[r,phi,z],[r,phi,-z]], phi=0..2*Pi, r=2..3, coords=cylindrical, scaling=constrained);

@daljit97  Sorry, I find it difficult to explain such obvious things (English is not my native language). Probably the best option for you is to take a tutorial and read about the simplest for-loops  do ... od  and conditionals  if ... fi .

This question has already been answered here  https://www.mapleprimes.com/questions/231357-Plotconvex-Or-Concave-Shell-#answer276406

@erik10  When solving nonlinear systems at specific intervals, the  DirectSearch  package can be useful.  It can be downloaded freely from the Maple Application Center. 

I think that the position of the rectangle surrounding these legends should be made one of the formal parameters of the procedure . You will specify it each time you use the procedure, depending on the situation (use  plottools:-rectangle  command). The rest can be automated in the procedure body.

@nm  You seem to have misunderstood the meaning of my answer. My code does exactly the same as  tomleslie's one or almost the same as  vv's code. I edited my answer a bit.

@priya_priti  Now the syntax is correct, but Maple simply cannot symbolically solve your system. By the way, your system has the obvious null solution: u(x)=0, v(x)=0, w(x)=0 .

@manju  In the new file, you did not do what I wrote about above.

@Keshav4  If a> = 0 then obviously your function  f  is increasing and it takes its smallest value at  x = 0 . For negative values of the parameter  a , the example can be solved only numerically.

@Prakash J 

plot(abs@h, 1..2, labels=["t","h(t)"]);

@Scot Gould  The disadvantage is that the syntax is longer, and you yourself have already written about the advantages. I usually don't use this option because it is quite possible to do without it.

@vv  and @Carl Love  Thank you for answers.

You can arrange this calculation without assignments and conversions:

restart; 
Sum(GAMMA(2*b-1+2*n)*GAMMA(2*n-s)*(b-1/2+2*n)/(GAMMA(2*b+2*n+s)*GAMMA(2*n+1)), n = 0 .. infinity); 
subs(Sum = sum, %) = eval(%, Sum = sum);