Evaluate area of the field defined by $\left (\frac{x^2}{4}+y^2 \right )^2=x^2+y^2$ 
Evaluate the area of the field defined by $\left (\frac {x^2}{4}+y^2 \right )^2=x^2+y^2$

I tried to turn it into a function $y(x)$, bus I was unable to do it. Is it actually possible to solve it explicitly?
 A: Polar coordinates are a good choice in order to find the area enclosed by the peanut-shaped region given by your equation:

By setting $x=\rho\cos\theta$ and $y=\rho\sin\theta $ we have:
$$ A = 32\int_{-\pi}^{+\pi}\frac{d\theta}{(5-3\cos(2\theta))^2} = 5\,\pi.$$
Also notice that your peanut-curve is an epitrochoid.
A: It is a quadratic in $y^2$ because $(y^2)^2+(\frac{x^2}{2}-1)y^2+(\frac{x^4}{16}-x^2)=0$.
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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We'll calculate the area $\ds{{\cal A}}$ of the region defined by
$$
{\cal A}\equiv
\braces{\pars{x,y} \in {\mathbb R}^{2}\ \mid\ x^{2} + y^{2} > \pars{{x^{2} \over 4} + y^{2}}^2}
$$

It's equivalent to:
  \begin{align}{\cal A}&\equiv\color{#66f}{\large\iint_{{\mathbb R}^{2}}
\Theta\pars{x^{2} + y^{2} - \bracks{{x^{2} \over 4} + y^{2}}^2}\,\dd x\,\dd y}
\\[3mm]&=\int_{0}^{2\pi}\int_{0}^{\infty}\Theta\pars{r^{2}
-\bracks{{1 \over 4}\,r^{2}\cos^{2}\pars{\phi} + r^{2}\sin^{2}\pars{\phi}}^{2}}
r\,\dd r\,\dd\phi
\\[3mm]&=\int_{0}^{2\pi}\int_{0}^{\infty}
\Theta\pars{16 -r^{2}\bracks{-3\cos^{2}\pars{\phi} + 4}^{2}}r\,\dd r\,\dd\phi
\\[3mm]&=\int_{0}^{2\pi}\int_{0}^{\infty}
\Theta\pars{{4 \over 4 - 3\cos^{2}\pars{\phi}} - r}r\,\dd r\,\dd\phi
\\[3mm]&=\int_{0}^{2\pi}\int_{0}^{4\bracks{4 - 3\cos^{2}\pars{\phi}}^{-1}}
r\,\dd r\,\dd\phi
=8\ \underbrace{\int_{0}^{2\pi}{\dd\phi \over \bracks{4 - 3\cos^{2}\pars{\phi}}^{2}}}_{\ds{=\ \color{#c00000}{{5 \over 8}\,\pi}}}
=\color{#66f}{\Large 5\pi} \approx {\tt 15.7080}
\end{align}

