Line integral along the "infinity symbol curve"

Question

I am preparing for an exam and working on the following exercise. Note that this is not homework but voluntary.

Let $B$ be the curve in $\mathbb{R}^2$ with the equation $(x^2+y^2)^2 = 2a^2(x^2-y^2)$.

1. What is $\int_{x\in{B}}|x|^2\mathrm{d}x$?
2. What is the supremum of $x^3+y^3$ on $B$ (as a function of $a$)? Is this supremum reached? If yes, where?

I have found out that the curve looks like the infinity symbol and that I need to calculate the line integral along the curve. I am not sure, however, how to parameterize the curve. Any help would be appreciated!

Edit:

Thank you very much, A.G. and Rory Daulton! Knowing that the curve is called a lemniscate, I have have found the following question: Line Integral with Lemniscate. Putting the pieces together, I get

$$ \int_{x\in{B}}|x|^2\mathrm{d}x = \int_{x\in{B}}|x|^2\mathrm{d}S, $$

where

$$ \mathrm{d}S = \sqrt{\Big(\!\frac{\mathrm{d}x}{\mathrm{d}\theta}\!\Big)^2+\Big(\!\frac{\mathrm{d}y}{\mathrm{d}\theta}\!\Big)^2}\mathrm{d}\theta = \frac{2a^2\mathrm{d}\theta}{r}. $$

Note that the definition of the lemniscate used here is slightly different from the linked question and hence the equation above also differs. Here is what I have done next:

$$\begin{align} \int_{x\in{B}}|x|^2\mathrm{d}S &= 2\int_{\theta=-\pi/4}^{\theta=\pi/4}\Big(\!\sqrt{x^2+y^2}\Big)^2\ \mathrm{d}S \\ &= 2\int_{\theta=-\pi/4}^{\theta=\pi/4}r^2\ \mathrm{d}S\\ &= 2\int_{\theta=-\pi/4}^{\theta=\pi/4}r^2\frac{2a^2\mathrm{d}\theta}{r} \\ &= 2\int_{\theta=-\pi/4}^{\theta=\pi/4}2a^2r\ \mathrm{d}\theta \\ &= 2\int_{\theta=-\pi/4}^{\theta=\pi/4}2a^2a\sqrt{2\cos2\theta\ }\ \mathrm{d}\theta \\ &= 4a^3\sqrt{2}\int_{\theta=-\pi/4}^{\theta=\pi/4}\sqrt{\cos{2\theta}\ }\ \mathrm{d}\theta \\ &= \dots \end{align}$$

Whoops, something seems wrong. The integral does not look very nice. Any ideas?

Answer 1

If you use polar coordinates, the left-hand side becomes

$$\begin{align} (x^2+y^2)^2 &= (r^2\cos^2\theta+r^2\sin^2\theta)^2 \\ &= r^4(\cos^2\theta+\sin^2\theta)^2 \\ &= r^41^2 \\ &= r^4 \end{align}$$

and the right-hand side becomes

$$\begin{align} 2a^2(x^2-y^2) &= 2a^2(r^2\cos^2\theta-r^2\sin^2\theta) \\ &= 2a^2r^2\cos 2\theta \end{align}$$

Equating,

$$\begin{align} r^4 &= 2a^2r^2\cos 2\theta \\ r^2 &= 2a^2\cos 2\theta \\ r &= |a|\sqrt{2\cos 2\theta} \end{align}$$

Note that $\cos 2\theta$ must be non-negative. Due to symmetry of both the region and the integrand, you can get your integral by doubling the integral for $-\frac{\pi}4\le\theta\le\frac{\pi}4$.