Area of Ellipse and Circle

 

Motivation:

 

                          The area of a circle with radius r is  r² .  In modern days, almost everybody know this fact since childhood.  However, is there any chance that we can prove the result by evaluating the integral of a function ?  Actually,  when we deal with the derivative of trigonometric function sin(x),  the result of  the area of a circle is  r² has been applied to prove

 

                                with the inequality    sin(x)  x  tan(x)

 

              The knowledge of  the fact  sin(x)  x  tan(x)  is based on the area comparison and an triangle and fan which is part of a circle.   Since that is the foundation that we build upon to have all the other theorems,  it is not logically correct to use those theorems to prove that the area of a circle with radius r is   r²  unless we can find out other ways to prove the relationship sin(x)  x  tan(x) while working on the derivative function of sin(x).

 

             However, it is a good exercise to “confirm” the result  r² by using the established theorems.  Furthermore,  there is nothing wrong to find the area of an ellipse by evaluating an integral of a function.   We will just start with the equation of an ellipse and find the area enclosed by the curve.

 

 

 

 

 

Area Enclosed by Ellipse

             E:   ,  where a>0, b>0  .   E is an ellipse.  We can consider E as the combination of two functions as below:

 

                                      y = b   and  y = -b  

                                       

 

                The are symmetric to x-axis. So,  we only need to find the integral for one of the functions and double the result.

 

                                          ½ area =

 

                              Let    x=asin(t)  ,  t  [ ,  ]   .

                             Then

                                           dx= a cos(t) dt

                                            =  b cos(t)

 

                           ½ area = 

                                         

                          And recall we have the following trigonometric equalities:

                                   cos(2t) = cos(t)cos(t) – sin(t)sin(t)

                                               = cos²(t) – sin²(t)

                                               = cos²(t) – ( 1 – cos²(t)) = 2cos²(t) – 1

 

                           So,

                                      cos²(t) =

 

                             ½ area =

 

                         The cycle of  cos(2t)  is  ;  from  to   , it completes one cycle so that the integral contributed by this term is vanished.  

                            

                                 ½ area =   =  

 

                            So,  

                                        Area of Ellipse = ab

 

 

                     When a=b=r  ,  the curve turns out to be a circle.  The area is   r² .

 

                      However, what is the actual value of  ?  If we would like to find the value of    down to  100 digits after the decimal point,  is there any systematic way to do that?  We might start using what we know so far to tackle the problem.

 

Approaches to

                          tan(  ) = 1 

           So,  using its inverse function :     arctan(1) =       .

           The Taylor series expansion of arctan(x) around 0  is:

 

                            f(x)= arctan(x)  ,  f(0) = 0

                            f’(x) =   ,  f(0) = 1

                            f”(x) =  (-1)  ,  f’(0) = 0

                            f⁽³⁾(x) = (-1)( 2 (1+x²)^-2 + (2x)(-2)(2x)(1+x²)^-3)

                                      = (-1)( 2 (1+x²)^-2 + (-8x²)(1+x²)^-3 )

 

                                        f⁽³⁾(0) = -2

                         …

 

                   arctan(x) =  x -  +  -  + …. + (-1)^k-1  + …

 

 

                  Thus, theoretically,  if we put x=1 in the equality above,  we can get the approximate value of   .   The Taylor theorem also provides the information about the error while using the first k terms to do the approximate.   If we use the first k terms, the error is less than    under the circumstance that we can evaluate all of those k terms without any errors.

 

 

 

                  This is just one of the approaches.  Probably you might feel that it does not converge to the right value very quickly: for example, if you use the first 10 terms,  the error of this estimate for  is  ;  so the estimate error for   is around  ; for k=50, the error is around  given that we can evaluate the first 50 terms without introducing any errors.  So, it does not make sense to use this approach to find the value of   .  There exists other way that can quickly converges with fewer terms with less error.

 

 

 

                 Using the following equality might converge more quickly:

 

                                      = 4 arctan(  ) – arctan(  )