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I am trying to calculate the area of the loop in the folium of Descartes using Green’s theorem. The loop can be parameterized $x = \frac{3at}{1+t^3}$, $y=\frac{3at^2}{1+t^3}$, $0\leq t < \infty$.

$$\iint_F 1=\int_{\partial F}x\,dy=\int_{0}^\infty \left(\frac{3at}{1+t^3}\right)\left(\frac{6at(1+t^3)-3t^2(3at^2)}{(1+t^3)^2}dt\right)$$

$$=\int_{0}^\infty\frac{18a^2t^2 (1+t^3)-27a^2t^5}{(1+t^3)^3}=\int_{0}^\infty\frac{9a^2t^2(2(1+t^3)-3t^3)}{(1+t^3)^3},$$and the book tells me the answer is $$\int_0^\infty \frac{t^2}{(1+t^3)^2}.$$ Where is my error?

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Green’s theorem is:

$$

\int_F\left(\frac{\partial M}{\partial x} – \frac{\partial F}{\partial y}\right)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} (\color{blue}{-ydx} + x dy),\tag{1}

$$

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(-\frac{3at^2}{1+t^3}\Big(\frac{3at}{1+t^3}\Big)’ + \frac{3at}{1+t^3}\Big(\frac{3at^2}{1+t^3}\Big)’ \right)dt.

$$

**Some updates:**

Formula (1) simplifies to

$$

\frac12 \int^{\infty}_0 \frac{9a^2t^2}{(1+t^3)^2}dt, \tag{2}

$$

which is what you have in your book. If we use yours

$$

\int_F 1\,dxdy = \oint_{\partial F} xdy = \int_{0}^\infty\frac{9a^2t^2(2 -t^3)}{(1+t^3)^3}dt,

$$

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}^\infty\frac{9a^2t^2(2t^3 -1)}{(1+t^3)^3}dt.

$$

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.

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