Suppose that I imposed The following deformation field on a

Suppose that I imposed The following deformation field on a sample that is initially a unit cube with one corner at The origin and one at (1, 1,1): If X1 > X2, X1 = X1cos(tan-1[X2/X1]) x2 = X1sin(tan-1[X2/X1]) x3 = (4/pi)*X3 if X1 le X2 x1 = X2cos(tan-1[X2/X1]) x2 = X2sin(tan-1[X2/X1]) x3 = (4/pi)*X3 What does The deformed shape look like? Calculate F for The deformation. Note that F will be a function of X. This deformation preserves The volume of The overall shape (i.e., The final shape has a volume of 1, just like The initial cube). Is volume preserved at every point in The domain, or are some points stretched and some compressed?

Solution

FOR SIMPLICITY & TO AVOID CONFUSION USING .. MATRIX X FOR X1,X2,X3 AND MATRIX Y FOR Y1=x1 , Y2=x2 , Y3=x3 ….. WE HAVE .. Y= X = IF …X1 > X2 … IF …X1 < = X2 … Y1 X1 Y1 = X1*COS[ATAN(X2/X1)] Y1 = X2*COS[ATAN(X2/X1)] Y2 X2 Y2 = X1*SIN[ATAN(X2/X1)] Y2 = X2*SIN[ATAN(X2/X1)] Y3 X3 Y3 = [4 / PI ] *X3 Y3 = [4 / PI ] *X3 b).. SIZE OF CUBE = 1 UNIT CORNERS ARE AT [0,0,0 ] & [1,1,1,1] SO IF WE CONSIDER ITS 4 VERTICES AS O,A,B,C CORRESPONDING TO ITS 3 COTERMINOUS EDGES OA , OB , OC THEN THEY ARE GIVEN BY ... ORIGINAL DEFORMED X1 X2 X3 Y1 Y2 Y3 O 0 0 0 #DIV/0! #DIV/0! 0 A 0 1 0 0 1 0 B 0 0 1 #DIV/0! #DIV/0! 1.27324 C 1 0 0 1 0 0 AO= 0 1 0 1.177 1 1.177 BO= 0 0 1 #DIV/0! #DIV/0! 2.45024 CO= 1 0 0 2.177 0 1.177 VOLUME = [OA , OB , OC ] = 1             V \' = [O;A\' , O\'B\' , O\'C\' ] = #DIV/0! THE TRANSFORMATION IS NOT DEFINED FOR THE CORNER AT ORIGIN AND THE VERTEX DIRECTLY ABOVE IT [ B ] WITH X1 = X2 = 0 …. PLEASE DEFINE THE FUNCTION FOR THESE CASES WHEN X 1 = X2 = 0 TO PROCEED .ON HEARING FROM YOU I SHALL EDIT THE RESPONSE ALSO PLEASE ELABORATE ON ALL SYMBOLS USED ….