I want to find the following area under the folium of Descartes: 
My attempt: A parametrization of the folium of Descartes is
$$\alpha(t)=\left(\frac{3at}{1+t^3},\frac{3at^2}{1+t^3}\right).$$
It is reasonable to rotate the curve $\pi/4$ radians counterclockwise so that the assymptote $x+y+a=0$ lies on the $x$ axis:
$$\mathcal{R}_{\frac{\pi}{4}}\alpha(t)=\begin{pmatrix}\cos \frac{\pi}{4}&-\sin \frac{\pi}{4}\\
\sin \frac{\pi}{4}&\cos \frac{\pi}{4}\end{pmatrix}\alpha(t).$$
Simplifying, we get
$$(u(t),v(t)):=\mathcal{R}_{\frac{\pi}{4}}\alpha(t)=\left(\frac{3a(1-t)t}{\sqrt{2}(1+t^3)},\frac{3at}{\sqrt{2}(1-t+t^2)}\right).$$
Now, the desired area should be
$$\text{Area}=\int_{-1}^0v\,du=\int_{-1}^0v\frac{du}{dt}dt,$$
since $\alpha(t)$ has $x+y+a=0$ as assymptote when $t\to -1$ and $\alpha(t)\to (0,0)$ when $t\to 0$. Simpliflifying again we get
$$\text{Area}=\frac{9a^2}{2}\int_{-1}^0\frac{ t \left(t^4-2 t^3-2 t+1\right)}{(t+1)^2
\left(t^2-t+1\right)^3}.$$
Since this integral does not converge, the area should be $+\infty$. Is this correct?
Best Answer
The integral for the area should be
$$\int_{-1}^0 \left\lvert v(t)+\frac a{\sqrt2} \right\rvert \, dt = \frac{|a|}{\sqrt2} \int_{-1}^0 \frac{(1+t)^3}{1+t^3}\, dt$$
which does converge. The figure below shows the part of the original curve (blue) and rotated curve (red) plotted with $t\in(-1,0)$ and $a>0$, along with their asymptotes (dashed). Upon rotation, the new asymptote in the $uv$ plane is the line $v=-\frac a{\sqrt2}$.