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A Self Intersecting Curve (Posted on 2005-12-16) Difficulty: 5 of 5
The curve defined by the relation x^3+y^3=3xy intersects itself at the origin and forms a loop. Find the area enclosed by the loop.

  Submitted by Brian Smith    
Rating: 4.5000 (2 votes)
Solution: (Hide)
The area enclosed by the loop is 3/2.

Use the line x=y as the axis of integration. The intersection of the curve and x=y are (0,0) and (3/2,3/2). The length of the axis is 3/sqrt(2). Define d to be the distance along x=y starting from (0,0). At d=3/sqrt(2) the point (3/2, 3/2) is reached.

The set of lines perpendicular to x=y form the cross-sections of integration. At distance d, the cross-section is the segment on the line x+y=d*sqrt(2). The endpoints of the segment can be found by solving the system {x^3+y^3=3xy, x+y=d*sqrt(2)}

x^3+y^3=3xy can be rewritten as (x+y)^3=3xy(x+y+1)
After substituting y=d*sqrt(2)-x:
2d^3*sqrt(2) = 3x(d*sqrt(2)-x)(1+d*sqrt(2))
Expressing as a quadratic in x:
(3+3d*sqrt(2))x^2 + (-3d*sqrt(2)-6d^2)x + (2d^3*sqrt(2)) = 0
Solving for x gives:
x = ( (3d*sqrt(2)+6d^2) +\\- sqrt( (3d*sqrt(2)+6d^2)^-4*(2d^3*sqrt(2))(3+3d*sqrt(2)) ) )/( 6+6d*sqrt(2) )
Simplifying:
x = ( (3d*sqrt(2)+6d^2) +\\- sqrt( 18d^2+12d^3*sqrt(2)-12d^4 ) )/( 6+6d*sqrt(2) )

Denoting the two roots as x1 and x2, the endpoints are (x1,x2) and (x2,x1)
The length of the segment is then |x2-x1|*sqrt(2), which equals:
2d*sqrt(9+6d*sqrt(2)-6d^2)/(3+3d*sqrt(2))

The integral which will calculate the area of the loop is then:
Integ{0,3/sqrt(2)}[2d*sqrt(9+6d*sqrt(2)-6d^2)/(3+3d*sqrt(2))]dd
First make the substitution u=d*sqrt(2) to simplify the expression:
u=d*sqrt(2), du=sqrt(2)dd, d=0 -> u=0, d=3/sqrt(2) -> u=3
Integ{0,3}[u*sqrt(9+6u-3u^2)/(3u+3)]du
Before the next substitution, rewrite the integral as:
Integ{0,3}[u*sqrt(4-(u-1)^2)/((u+1)*sqrt(3))]du
Now, make the substitution u-1=2*sin(T)
u-1=2*sin(T), du=2*cos(T)dT, u=0 -> T=-pi/6, u=3 -> T=pi/2
Integ{-pi/6,pi/2}[(2*sin(T)+1)*sqrt( 4-4*(sin(T))^2 )/( (2+2*sin(T) )*sqrt(3))]dT
The integral simplifies to:
(2/sqrt(3))*Integ{-pi/6,pi/2}[sin(T)+cos(2T)]dT

The integrand evaluates to (1/2)sin(2T)-cos(T) and the whole integral evaluates to (2/sqrt(3))*(0 + 0 + sqrt(3)/2 + (1/2)*(sqrt(3)/2)) = 3/2.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
AnswerK Sengupta2008-03-19 11:06:39
re: Numerical Integration SolutionRichard2005-12-17 13:31:23
Another numerical method as solutionCharlie2005-12-17 11:33:49
Numerical Integration SolutionRichard2005-12-17 02:44:28
Questionre(3): A solution; Not quiteRichard2005-12-16 14:09:55
re(3): A solution; oopsLarry2005-12-16 12:20:10
re(2): A solution; Not quitegoFish2005-12-16 11:59:29
re: A solution; Not quiteLarry2005-12-16 11:41:27
SolutionA solutiongoFish2005-12-16 11:12:50
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