I was looking through past exams for a multi course I'm taking and came across this problem which I'm not sure how to approach.
Find the area of the two regions formed by the part of the folium of descartes which passes through the triangle formed by the vertical tangent to the leaf, the x-axis and the line from the intersection of the tangent to the origin.
The equation of the folium of descartes is:
$$x^3+y^3-3xy=0$$
I've found the tangent line which intersects at $(\sqrt[3]{4}, \sqrt[3]{2})$, so there are two regions within the triangle with vertices $(\sqrt[3]{4}, \sqrt[3]{2}), (0,0), (\sqrt[3]{4}, 0)$ as it is divided by the part of the folium that passes through the triangle. But how do I find the area of either of these parts?
Here's a picture to make things clearer. I'm interested in the areas labelled A1 and A2:
Best Answer
In polar coordinates the equation of the folium is $$ r=\frac{3\sec \theta \tan \theta}{1+\tan^3 \theta} $$ For the area of the region $A_1$ the limits of integration for $\theta$ are: $$ 0\le \theta \le \arctan \left( \frac {\sqrt[3]{2}}{\sqrt[3]{4}}\right)=\arctan \sqrt[3]{\frac{1}{2}}=\alpha $$ and the area is given by: $$ A_1=\frac{1}{2}\int_0^ \alpha r^2 d\theta=\frac{1}{2}\int_0^ \alpha \left[\frac{3\sec \theta \tan \theta}{1+\tan^3 \theta}\right]^2 d\theta $$ with the substitution $u=1+\tan^3 \theta$ we have: $$ du=3\tan^2 \theta \sec^2 \theta d \theta $$ and the limits of integration becomes: $$ 1\le u \le 1+\frac{1}{2} $$ So the area is $$ A_1=\frac{3}{2}\int_1^{\frac{3}{2}} \frac{du}{u^2}=\frac{1}{2} $$