Find the area bounded by a quartic curve

Find the area bounded by the curve defined by: $$x^4+y^4=x^2+y^2$$ .

The final solution should be $$\sqrt{2}\pi$$. I tried to change it into polar coordinates, but I couldn't calculate the integral.

Here is the solution:

Integrate to get the area. By symmetry, it is eight times the integral $$\theta = \pi/4$$ and $$\pi/2$$. In (simpler) rectilinear coordinates:

$$\int\limits_{x=0}^1 \frac{\sqrt{\sqrt{-4 x^4+4 x^2+1}+1}}{\sqrt{2}} - x\ dx = 0.55536$$

where

$$y(x) = \frac{\sqrt{\sqrt{-4 x^4+4 x^2+1}+1}}{\sqrt{2}}$$

is the solution to the original stated equation in the relevant region.

So the total area is: $$8 \times 0.55536 = 4.44288 = \sqrt{2} \pi$$.

$$x=r\cos{\theta};y=r\sin{\theta}$$

$$A=\int_0^{2\pi}\frac{r^2}{2}\text{d}\theta$$

Solve for $$r^2$$

$$r^4(\cos^4{\theta}+\sin^4{\theta})=r^2$$

$$r$$ is non-zero, so:

$$r^2=\frac{1}{\cos^4{\theta}+\sin^4{\theta}}=\frac{1}{1-2\sin^2{\theta}\cos^2{\theta}}=\frac{2}{2-\sin^2{2\theta}}=\frac{2}{2\cos^2{2\theta}+\sin^2{2\theta}}=\frac{2\sec^2{2\theta}}{2+\tan^2{2\theta}}$$

$$\frac{r^2}{2}=\frac{\sec^2{2\theta}}{2+\tan^2{2\theta}}$$

As you stated the area is eight times a single octant of the region so:

$$A=4\int_0^{\frac{\pi}{4}} \frac{2\sec^2{2\theta}\text{d}\theta}{2+\tan^2{2\theta}}=4\int_0^{\frac{\pi}{4}}\frac{\text{d}(\tan{2\theta})}{2+\tan^2{2\theta}}=4\int_0^\infty \frac{\text{d}t}{2+t^2}=4\frac{\pi}{2\sqrt{2}}=\sqrt{2}\,\pi$$

• Nice... keeping it in radial coordinates (+1). – David G. Stork Mar 26 at 0:46