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.
Another numerical method is just to evaluate x^3+y^3-3xy in the first quadrant over a grid extending to about (2,2) or so and add up how many points had this evaluate as negative and adjust for the size of the grid. A program to do that follows; it's coordinate conversions are the result of originally using screen pixel points to get a graphical view of what the curve looked like.
The results of the program were:
1.587 1.499 2.11990612999727 1.499983
indicating the maximum x value on the loop was about 1.587; the x coordinate (and y coordinate) at the maximum distance of the loop from the origin was 1.499 (looks like 1.5); the distance from the origin of this point as this value times sqrt(2), or 2.1199...; and the area, about 1.499983 (looks like 1.5).
DEF fnx (xvl) = 20 + xvl * 250
DEF fny (yvl) = 440 - yvl * 250
DEF fnsx (c) = (c - 20) / 250
DEF fnsy (r) = (440 - r) / 250
FOR row = 0 TO 479 STEP .25
posCt = 0: hitLoop = 0
FOR col = 0 TO 639 STEP .25
xv = fnsx(col): yv = fnsy(row)
good = 1
ans = xv * xv * xv + yv * yv * yv - 3 * xv * yv
IF xv >= 0 AND yv >= 0 THEN
IF ans < 0 THEN
pxCt = pxCt + 1
IF xv > max THEN max = xv
IF xv = yv AND xv > maxEq THEN xr = xv: reach = xv * SQR(2): maxEq = xv
PRINT max, xr, reach, pxCt / (250! * 250! * 16)
Posted by Charlie
on 2005-12-17 11:33:49