Vtyjkyuk

I am trying to calculate the area of the loop in the folium of Descartes using Green's theorem. The loop can be parameterized $x = frac3at1+t^3$, $y=frac3at^21+t^3$, $0leq t < infty$.

$$iint_F 1=int_partial Fx,dy=int_0^infty left(frac3at1+t^3right)left(frac6at(1+t^3)-3t^2(3at^2)(1+t^3)^2dtright)$$

$$=int_0^inftyfrac18a^2t^2 (1+t^3)-27a^2t^5(1+t^3)^3=int_0^inftyfrac9a^2t^2(2(1+t^3)-3t^3)(1+t^3)^3,$$and the book tells me the answer is $$int_0^infty fract^2(1+t^3)^2.$$ Where is my error?

Green's theorem is:
$$
int_Fleft(fracpartial Mpartial x - fracpartial Fpartial yright)dxdy = oint_partial F (L dx + M dy).
$$
The textbook uses one possible choice for the differential form $omega = (L dx + M dy)$:
$$
int_F 1,dxdy = frac12 oint_partial F (colorblue-ydx + x dy),tag1
$$
not just $x,dy$, you need that extra blue term to get to what is similar to your book's answer. Now plugging the parametrization gives
$$
frac12 int^infty_0 left(-frac3at^21+t^3Big(frac3at1+t^3Big)' + frac3at1+t^3Big(frac3at^21+t^3Big)' right)dt.
$$

Some updates:
Formula (1) simplifies to
$$
frac12 int^infty_0 frac9a^2t^2(1+t^3)^2dt, tag2
$$
which is what you have in your book. If we use yours
$$
int_F 1,dxdy = oint_partial F xdy = int_0^inftyfrac9a^2t^2(2 -t^3)(1+t^3)^3dt,
$$
above formula will yield the same value with (2). Similarly, choosing $-ydx$ will give you same value as well:
$$
int_F 1,dxdy = oint_partial F -ydx = int_0^inftyfrac9a^2t^2(2t^3 -1)(1+t^3)^3dt.
$$
The choice is sure not unique. Furthermore, w/o computing the value of above three integrals, you could always prove:
$$
oint_partial F -ydx = oint_partial F xdy
$$
through integration by parts. Reason why choosing $omega = -ydx + x dy$ is probably like you said, leading to a more manageable formula.