Integrate area enclosed by $x^4 + y^4 = x^3 + y^3 $

Question

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?!

Answer 1

There is no simple method for such problems. In your example you can at least do the following, due to symmetry: Introduce new coordinates $(u,v)$ whose axes are rotated $45^\circ$ with respect to the $(x,y)$-axes. This gives$$x={1\over\sqrt{2}}(u-v),\quad y={1\over\sqrt{2}}(u+v)\ .$$ In this way your equation becomes $$-\sqrt{2} u^3 + u^4 - 3\sqrt{2} u v^2 + 6 u^2 v^2 + v^4=0\ .$$ Since this equation is biquadratic in $v$ you can solve for $v$ (choose the right branch). The circumference of your shape is then of the form $$v=\pm \psi(u)\qquad\bigl(0\leq u\leq \sqrt{2}\bigr)\ ,$$ where $\pm\psi(u)$ describes the upper and the lower half of the circumference. Unfortunately the resulting expression
$$\psi(u):=\sqrt{{3u\over\sqrt{2}}-3u^2+{\sqrt{18u^2-32\sqrt{2}u^3+32u^4}\over2}}$$ is so complicated that you cannot integrate it elementarily. The numerical value for the enclosed area is $1.6663914997252287$.

Quanto has found the exact area value $A={3\sqrt{2}\pi\over8}=1.6660811\ldots\ $.

Answer 2

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,

$$A= \frac{3\sqrt2}{8}\pi$$

Answer 3

The curve has genus zero, so we know that it has a rational parametrization and therefore the area integral is elementary. In this case a parametrization can be found by taking $y = t x$. Then $$x^4 + y^4 - x^3 - y^3 = x^3 (x + t^4 x - 1 - t^3), \\ (x, y) = \left( \frac {t^3 + 1} {t^4 + 1}, \frac {t (t^3 + 1)} {t^4 + 1} \right), \\ S = \frac 1 2 \int_{\mathbb R} (x \dot y - \dot x y) dt.$$ The integral of $\sqrt {\dot x {}^2 + \dot y {}^2}$ is apparently not elementary.