Calculate the area of the region $(x + y)^4 = ax^2y$

Question

Calculate the area of the region: $$(x + y)^4 = ax^2y$$

Could you advise how to solving problems of this type?

Answer 1

Since the graph of $(x + y)^4 = ax^2y$ contains a closed loop I will assume that OP wants to find the area of the region bounded by the loop.

First, convert the equation to polar form using $x=r\cos\theta,\,y=r\sin\theta$ yielding

\begin{equation} (r\cos\theta+r\sin\theta)^4=ar^2\cos^2\theta\cdot r\sin\theta \end{equation}

This can be simplified and solved for $r$.

\begin{equation} r=a\cdot\dfrac{\cos^2\theta\sin\theta}{(\cos\theta+\sin\theta)^4} \end{equation}

In polar form the loop in the first quadrant falls between $\theta=0$ and $\theta=\frac{\pi}{2}$. Thus the area can be found by evaluating the integral

\begin{equation} \int_{0}^{\frac{\pi}{2}}\frac{1}{2}r^2\,d\theta=\frac{1}{2}a^2\int_{0}^{\frac{\pi}{2}}\dfrac{\cos^4\theta\sin^2\theta}{(\cos\theta+\sin\theta)^8}d\theta \end{equation}

If we let $F(\sin\theta,\cos\theta)=\dfrac{\cos^4\theta\sin^2\theta}{(\cos\theta+\sin\theta)^8}$ we see that $F(\sin\theta,\cos\theta)= F(-\sin\theta,-\cos\theta)$. Thus we may make the substitution $t=\tan\theta,\,d\theta=\frac{dt}{1+t^2},\,\sin\theta=\frac{t}{\sqrt{1+t^2}},\,\cos\theta=\frac{1}{\sqrt{1+t^2}}$ to obtain, after simplification

\begin{equation} \frac{1}{2}a^2\int_{0}^{\frac{\pi}{2}}\dfrac{\cos^4\theta\sin^2\theta}{(\cos\theta+\sin\theta)^4}d\theta=\frac{a^2}{2}\int_0^\infty \dfrac{t^2\,dt}{(1+t)^8} \end{equation}

Partial fraction decomposition (using Wolfram) gives

\begin{equation} \frac{a^2}{2}\int_0^\infty \left(\frac{1}{(t+1)^6}-\frac{2}{(t+1)^7}+\frac{1}{(t+1)^8}\right)dt\\ \end{equation}

Therefore the area of the region bounded by the loop is

\begin{equation} \frac{a^2}{2}\int_0^\infty \left(\frac{1}{(t+1)^6}-\frac{2}{(t+1)^7}+\frac{1}{(t+1)^8}\right)dt=\frac{a^2}{210}\\ \end{equation}

The graph below is for $a=16$, the value for which the loop contains the point $(1,1)$.