@rlopez  Thank you for your comment. I am familiar with this command. But I always prefer to use a command from  Maple kernel, if it solves the same problem. Also the code

plot([Re,Im]~([%]), style=point);

is a bit shorter than

plots:-complexplot(convert(V, list), style=point);

@Carl Love  Thank you for your helpful comments.

@umar khan 

1. Maple does not write a constant at indefinite integration, you can easily do it yourself.

2. For the numerical solution of a system of two equations  2 initial conditions are necessary.

3. I wrote  u(R)=0  arbitrarily, you can write your own value as you need.

4. Since you wrote "to integrate of  z(x)  from  0  to  x" , to plot I have picked up the constant  so that  u(0)=0 .

@hitstudent  I don't understand, what is the question? The integral  a can be easily calculated both the polar and the cartesian coordinates:

@hitstudent  Since we have the dependence of  r  of  t , then we have to find the bounds for the variable  t  . In the example above, where the pole is located inside of the curve, it will be  t=0..2*Pi

Finding the area:

solve((r*cos(t)-a)^2+(r*sin(t)-b)^2=R^2, r);
r(t):=eval([%][1], {a=-1,b=0, R=2});
int(1/2*r(t)^2, t=0..2*Pi);

The task will be much more difficult if the pole is outside the curve. Here is an example, and it's solution in this case. Note that the integral is calculated numerically, because Maple has failed with the symbolic computation:

solve((r*cos(t)-a)^2+(r*sin(t)-b)^2=R^2, r):
r(t):=eval([%], {a=5,b=3, R=2});
solve(op([3,1],r(t)[1])>=0):
select(s->is(op(1,s)>0 and op(2,s)<Pi/2),[%])[ ];
Int(1/2*(r(t)[1]^2-r(t)[2]^2), t=op(1,%)..op(2,%));
evalf(%);
identify(%);

Normally closed curves better define not in polar coordinates, but parametrically. Here is a simple solution to the previous example in this way:

x := R*cos(t)+a:  y := R*sin(t)+b:
int(x*diff(y, t), t=0..2*Pi);

                                  R^2*Pi

@student_md   The procedure  ModalMatrix1  returns the modal matrix of  A  and  the matrix  D  which you want :

ModalMatrix1:=proc(A)
uses LinearAlgebra;
op(simplify(fnormal~(evalf~([Eigenvectors(A)[2], DiagonalMatrix(convert(Eigenvectors(A)[1],list))])), zero));
end proc:

Examples of use:

ModalMatrix1(<-4,-2; 3,1>);
ModalMatrix1(<-4,-2,5; -2,1,7; 5,7,9>);

@toandhsp  Very strange!  Try the following variant. I think it should be working:

restart; 
L := simplify([log[2](3) = a, log[3](5) = b, log[7](2) = c, log[140](63)]); 
S := {op(op(solve(L[1 .. 3], [ln(3), ln(5), ln(7)])))}; 
simplify(eval(L[4], S));

@toandhsp 

L:=simplify([log[2](3)=a, log[3](5)=b, log[7](2)=c, log[140](63)]);
simplify(eval(L[4], {solve(L[1..3], [ln(3), ln(5), ln(7)])[ ] [ ]}));

@taro  This function  t->t="."  works properly only in the context of  select  or  remove   commands. If you use it separately, then add at the right  is  command:

f:=t->is(t="."):
f(9);

                                 false
 

@jacksonmeg   It is obvious that if your system has a solution  (c1, c2,c3, c4) , the system will also have the solution  (-c1, -c2, -c3, -c4). So everything is OK.

@jacksonmeg  You wrote  g:=subs(x=1, b) . Therefore you must solve the system  {g[1], g[2], ... } 

@jacksonmeg  You choose to every time the same system. Replace  d  by  g  and so on.

@jacksonmeg  For solving in an explicit form for  f  you can use  allvalues  command:

f := {c1 = 0, c2 = -RootOf(2*_Z^2-1), c3 = 0, c4 = RootOf(2*_Z^2-1)};

allvalues(f);

@Preben Alsholm  Here is an example of the same error in the new worksheet (I use Maple 2016.1):

@one man  In your solution not satisfied OP's condition:   "...to plot this surface with a direct mapping, that is without use of implicit plotting over a volume or rejecting grid points "