Kitonum

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These are replies submitted by Kitonum

@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 "

@kharonsen  Certainly possible, but you must do it yourself. You now have 2 versions: mine, in which everything is very simple and all the shifts made by one unit above or right or left, and Tom' s one that is closer to your source code. If you study these versions you can easily solve your problem. 

@ANANDMUNAGALA  If you mean a broken line in 2D or 3D  use  plottools[curve]  command.

Example (a hexagon in 3D):

plots[display](plottools[curve]([ [2,1,0], [1,2,0], [0,2,1], [0,1,2], [1,0,2], [2,0,1], [2,1,0] ]), axes=normal, color=red, thickness=3);

 

Addition. See also the procedure  PiecewisePlot  above which automates the plotting of a piecewise curve in 3D.

Edited.

@Scot Gould  This is probably due to some specific evaluative Maple rules.

Workaround:

restart;
z:=A*<x,y>;
plots:-fieldplot(eval(z, A=1.0), x=0..1, y=0..1);

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