Four - Leaved Rose 

           By: Ali Mohamed Abu Oam       asam295@hotmail.com 

     The puepose of this worksheet is to show the powerfull of  Algebra compuetr system ( CAS ) in modeling and calculating the area and the volume in case of the Four-Leave Rose.      

 The Traditional Method: 

The area of a loop of the four-leaved rose using double and single integrals ,the following equation in polar coordinates r = cos( 2) generates a graph called the four-leaved rose. We have seen that the area of the loop generated when goes over
     is   . 

   The area of the loop is given by the double integral over the domain D expressed in polar coordinates as 

                  D = { ( r, ) |  , , , }      

               So,       =  =  =  

                Above, we used the trigonometric identity   (x) =1/2 ( ) . Next, we try to estimate the area of the loop using a simple integral 

                     .  

              Since the loop generated by belongs to the closed interval       , is symmetric with respect to the x-axis, it follows that the area
we are interested in is twice the area of the graph given in the picture below ( under the red curve ) . 

     To compute its area we need to nd an expression of  the variable y as a function of the variable x, y = f(x). Then to compute the are we could use   

                   =   .  Given that    ,      ,    . from the equation of the loop in polar coordinates 

       = ( ) ( ) , we have    .  

                 To find a formula for f(x) we would have to solve the above equation for y in terms of x, and this is rather complicated, but there is still a way to use just a single integral for computing the desired area.  

                   The four-leaved rose can be viewed as a particular hypocycloid  with equations: 

              x =  ,   y =       ................................................................................  ( 0 , 1 ) 

         with    ,   ,  for     , ,                                                                                                                                                                                                              

                  we obtain the loop for which we want to compute the area, and to get just the upper half of that loop we let t belong to [,0] Note that in the above equations we have dependence of just one parameter t. The point (x; y) = (0; 0) corresponds to the value of the parameter t = , and the point (x; y) = (1; 0) corresponds to t = 0. The area of half of the loop is still given by the integral , but now in order to estimate it, we can use the change of variables : 

                                                                                                                            x =    ,                                                                                                                                        

   which gives us an expression for y in terms of t, from equation (0.1) and also     = ( ) .                                                                                                                                                                                                                                                                                                                                                                           

                  The area is then :                                                                                                                                                                                                                                                                                                                                                                                                                                                                                               

           = . (   ( ) ) =                                                                                                                                                                                                                                                                                                                                                                                                                                                                                         

           If you don't just want to give up here, the half angle formula helps us with the rst and the third terms in the sum under the integral.                                                                                                                                                                                                                                         

            = = | -..0 = = .                                                                                                                                                                                                                                                                                                                                                                                                                                                                            

            =  = | -..0 = = .                                                                                                                                                                                                                                                                                                                                                                                                                                                                                  

         Next, we use the following trigonometric identity :                                                                                                                                                                                                                                                                                                                                                                                                                                                                                            

            to evaluate the remaining term : 

                              =  = |-..0 = = 2 .                                              

                           Now putting all the terms together, we obtain that the area of half of the loop is :                                                                                                                                                                                                                     

                                                                                                                      ( - 1 + 2 - ( )) =  .                                                                                                                                                                                                                                                                                                                                                                                                              

          Maple Code :                                                                                                                                                                                                                                          

                               Now we will model and solve a similar problem using maple, and to caculate the area and the volume. The procedure followed by some animations pointed to the mentioned area or a volume .                                                                                                                                                                                                                     

(1)

Animation : boundary of the solid of r?volution : 

Evaluation of the volume by disks and by cylindrical shells : 

The end .