Geometric expression can perhaps figure out the solution for the overlapping circles , but i did not yet figure out how to to this. 
 

@Ronan 

Hello,

It is a Curvilinear Polygon

You draw the two circles and add the two intersections points ( via contstruct panel: intersection) and these two points are used for the definition of the overlapping area of the two circles.

Choose the arc- drawing  in the drawing panel , then click on a intersection point and drag a curvelinear line to the other intersection point.

Now this a half part and back again for the other part.

Then select the area and then calculate symbolic area ( see panels ).

@Rouben Rostamian  

Thanks

Clever done this by you and experienced user i think.

It started with a (big) circle c1 with his midpoint in the center of a xoy  coordinate system.
c2 is a circle with his midpoint on x-axis ( R, 0 )

We can calculate half of the blue area with two integrals for the upperparts of the two circles
For circle c1 there can be a area integrated from X (intersection ) to X= L along the x-axis. 

For circle c2 there can be a area intergrated from XOZ (x=0) to X (intersection )  along the x-axis
 

This two values of the integrated areas, doubled gives the blue area  sketched in Geometric Expression. = A
We get a equation.

Note: never have seen this %[1] : if % can have multiple output ? ( depends on input : here on one line with mutiple input )

Only the integration with ( integration interval , ? , ? ).

That's it.

@Mariusz Iwaniuk 

Hi, this is the formulae and useless 
 

Download de_koe_in_het_weiland1.mw

@Rouben Rostamian  

Thanks

Found this in old studymaterial and there it can be solved on a standard way

http://www.davdata.nl/math/grazinggoat.html

 Or advanced with a double integral and polar coordinates , what you did i think.
Unfortanely i must overhaul my old studymaterial ( made in Maple V )  about the double integrals

@Mariusz Iwaniuk 

Thanks

I think it has to do that the formulae is not correct, that i did not defined it correct in Geometry Expression.
Must look again to it. 

Looking a the midpoints of triangle and inscribed ellips can be also studied with Geometry Expressions

Geometry Expressions is the world's first application that lets you visualize geometry and formulate symbolic expressions by constructing a sketch. 
Simply sketch your geometry, add symbolic constraints, and then automatically output symbolic measurements and expressions. 

Its only 10 USD for program now..a bargain.

Results found in GE can be copied and pasted into Maple for further simplification 
A solid knowledge of the old Greek geometry is needed as base.. 

Note: a good example for me to see , because i do see some programing constructs in it what i encountered  here on the forum asking about programming.

You can go further with advanced mathematics for exploring this ellips 
Is it possible to get the code behind this worksheet ?

314301_mardentheorem.mw

@tomleslie 

Thanks!

For the extensive answer.

Was it in the Riemann sum example that a function what is passed to the procedure was already defined as a function with a arrow operator.

X:= Array(0..N, [seq(j, j=a..b, (b-a)/N)]): # x values stored in array 
 Y:= Array(0..N, f ~ (X)):   # y -values stored in array
 

Now in this example of Polygonal approximation the function is defined as a expression , so this must first defined as a function in order to use the ~ elementwise operator.

xv := Array ()  # x- values stored in array 
yv:=unapply( f, indets(f, 'name')[]) ~ (xv);   y-values stored in a list ?  ..no as a array as showed the ouput 

I noticed the difference between '[ ]' and [ ] 

Think i understood it now for the f part 
In the book the functions are given as expressions. 

return display ( ) .. to make sure there will be a plot ?

I try the attached functions in the book example...but noticed the procedure definition by p,L := approxL ( ):

Why is that ?

@tomleslie 
Thanks

First there is a series x-values stored in Array xv 

Its a difficult one to decipher. yv:=unapply( f, indets(f, 'name')[])~(xv);  What is the idea behind this statement    

yv := xv^3 - xv*sin(xv)  outside the procedure call 
unapply( ) should make a function from the expression in  ( ) 

Problem to check the outcome from statements in procedure , beause it are local variables ?

Complicated this.

@vv 

Thanks

Yes, you do have absolutely right, its my lack of knowledge of how to handling series precisely. 
Thought that with some basic calculus, partial deratives , etc i could follow it.
But no.

Don't know if the calculus textbook has this sort of example i posted here ? 
I do have seen some examples ..but complicated

Could not find a solution strategy .

Download forum_excset3_task_3b.mw

@vv 
Thanks

Can you elaborate what the meant here  : You can also compute int(int(1/(1-x),x),x) + c1*x+c2; and adjust the constants such that f(0)=f'(0)=0.

@vv 

Thanks

Using the right commands and if i  look at the integral : i should never came on this one
Seems to be a double integral trick  ? 

                      -int(1/(x - 1), [x = 0 .. y, y = 0 .. z])

But don't get answer  :  ?  ..yes with value( )   

@Carl Love 

Thanks

Using this limit gives the radius of convergence for any algebraic function (not complex).
This limit outcome can be interval , real number axis  or one real number ?

Now to find a example of this.

Integrating and differentiating a powerseries makes now sense after have seen a example 
Not adding a constant of integration to a powerseries makes that it not good fit with a function as it seems in a seen example. 
 

@janhardo 

i bought years back some used calculus books : thomas calculus, early trancedentals and another calculus book

There is lot to find there in those books about powerseries, so i  want to dive further with Maple in the powerseries...