Integrate area enclosed by $x^4 + y^4 = x^3 + y^3 $ I want to calculate the area bounded by the region 
$$x^4 + y^4 = x^3 + y^3 $$
with integral. And also it's perimeter. 
Can somebody please help me with it?! 


Express the curve $x^4 + y^4 = x^3 + y^3$ in its polar coordinates,
$$r(\theta) = \frac{\cos^3\theta+\sin^3\theta}{\cos^4\theta+\sin^4\theta}\tag{1}$$
Recognizing that a complete loop is formed starting at origin and varying $\theta$ from $-\frac{\pi}{4}$ to $\frac{3\pi}{4}$. the area integral is then,
$$A= \int_{-\frac\pi4}^{\frac{3\pi}{4}}\int_0^{r(\theta)}rdrd\theta=\frac12\int_{-\frac\pi4}^{\frac{3\pi}{4}}r^2(\theta)d\theta$$
Simplify the integrand $r(\theta)$ given by (1),
$$r(\theta)=\frac{(\cos\theta+\sin\theta)(\cos^2 -\cos\theta\sin\theta +\sin^2\theta)}{(\cos^2\theta+\sin^2\theta)^2 - 2\cos^2\theta\sin^2\theta}$$
$$=\frac{\sqrt2\sin(\theta+\frac\pi4)(2-\sin2\theta)}{2- \sin^22\theta}$$
and apply the variable change $t=\theta + \frac{\pi}{4}$ to reduce the integral to, 
$$A=\int_0^\pi \sin^2t \left(\frac{2+\cos 2t}{2-\cos^2 2t}\right)^2dt$$The area can then be integrated analytically to yield,
$$A= \frac{3\sqrt2}{8}\pi$$
