We were asked to find the area under the given function, as a challenging exercise:
\begin{gather*} 6x^{3} +11x^{2} y+6xy^{2} +y^{3} =x \end{gather*} The first thing I tried to do, was probably naive, and I immediately thought of finding the solution to y in terms of x as a cubic equation.
Needless to say, I really made a mess out of it.
Could someone please provide some hints on how to approach this kind of problem? And it would also be helpful if anyone described a general method to deal with the area under an implicit function (if such a method exists).
$$6 x^3+11 x^2 y+6 x y^2+y^3=x$$ convert in polar coordinates $$x=r\cos t;\;y=r\sin t$$ we get $$r^3 \sin ^3 t+6 r^3 \cos ^3 t+11 r^3 \sin t \cos ^2 t+6 r^3 \sin ^2 t \cos t=r \cos t$$ $r$ can be simplified and the equation can be solved wrt $r^2$ $$r^2 \sin ^3 t+6 r^2 \cos ^3 t+11 r^2 \sin t \cos ^2 t+6 r^2 \sin ^2 t \cos t- \cos t=0$$ $$r^2=\frac{\cos t}{\sin ^3 t+6 \cos ^3 t+11 \sin t \cos ^2 t+6 \sin ^2 t \cos t}$$ which can be simplified dividing numerator and denominator by $\cos^3 t$ $$r^2(t)=\frac{1}{\cos^2 t}\cdot \frac{1}{\tan^3 t+6\tan^2 t+11\tan t +6}$$ Any area $S$ we want to calculate can be solved by the integral $$S=\frac12\int_a^b r^2(t) dt$$ Setting $\tan t=u$ we get $$\int \frac{du}{u^3+6 u^2+11 u+6}=\int \left(-\frac{1}{u+2}+\frac{1}{2 (u+3)}+\frac{1}{2 (u+1)}\right)\,du=$$ $$=\frac{1}{2} \log (u+1)-\log (u+2)+\frac{1}{2} \log (u+3)+C=$$ $$=\frac{1}{2}\left(\log \left(u^2+4 u+3\right)-2 \log(u+2)\right)+C=$$ $$=\frac{1}{2} \log \left(\frac{u^2+4 u+3}{(u+2)^2}\right)+C=$$ $$=\frac{1}{2} \log \left(\frac{\tan^2 t+4 \tan t+3}{(\tan t+2)^2}\right)$$ $$S=\left[\frac{1}{4} \log \left(\frac{\tan^2 t+4 \tan t+3}{(\tan t+2)^2}\right)\right]_a^b$$ To get the area in the picture below we compute $$S=\left[\frac{1}{4} \log \left(\frac{\tan^2 t+4 \tan t+3}{(\tan t+2)^2}\right)\right]_0^{\pi/2}=\frac{1}{4} \log \left(\frac{4}{3}\right)\approx 0.072 $$