Perimeter of Curve Problem?

Question

Show that the perimeter of the curve with equation x^2/3 + y^2/3 = 1 is 6.

**Note: This is to be solved using length of arc length (integration) method.

Answer 1

this curve is called asteroid which is symmetrical in all four quadrants.

see the figure of your asteroid

http://i38.tinypic.com/mmbcbm.jpg

perimeter of asteroid =  4 times length of asteroid in the first quadrant

=4 ∫ √[ + (dy/dx)²]dx   (limit of integration is from x = 0 to x = 1)

dy/dx = - (y/x)^1/3

=4 ∫ √[ + ((y/x)^2/3] dx

=4 ∫ √[ (x^2/3  + y^2/3)/x^2/3] dx

From the equation of the curve ; x^2/3  + y^2/3  =  1 .

=4 ∫ √[ (x^2/3  + y^2/3)/x^2/3] dx  =   4 ∫ √[ (1/x^2/3] dx

=4 ∫ dx/(x^1/3)  (limit of integration is from x = 0 to x = 1)

= 6 x^2/3

= 6 [( 1)^2/3 – (0)^2/3]  = 6