Area Between Bernoulli's Lemniscate And a Lissajous Curve

We calculate the area using Stokes' Formula:

Let $c:D\subset\mathbb{R}\rightarrow\mathbb{R^3},\quad \vec{c}(t)=(c_x(t),\:c_y(t),\:0)^T$ be a piece-wise smooth closed curve that surrounds an area exactly once. Then $A=\frac{1}{2}\left|\int_D \left<\vec{r}(t)\times \vec{r}^{(1)}(t);\:\vec{e}_z\right>\:dt\right|$

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Let's call $A$ the area between the given curves $\vec{r}_{1,2}(t)$ . Notice they are symmetric to both $x,\:y-$ axes, so $A=4A_1$ , where $A_1$ is the area between both curves in the first quadrant. If we first walk along $\vec{r}_1(t)$ and then along $\vec{r}_2(t)$ for $t\in [0,\:\frac{\pi}{2}]$ each, we surround $A_1$ exactly once clock-wise and may use Stokes' Formula: $A=4A_1=2\left|\sum_{k=1}^2 \int_0^{\frac{\pi}{2}}\left< \vec{r_k}(t)\times\vec{r}_k^{(1)}(t);\:\vec{e}_z\right>\:dt \right|$ Let's calculate the derivatives for both curves using the short-hands $s_k:=\sin(kt),\:c_k:=\cos(kt)$ and $s_2=2s_1c_1$ : $\begin{aligned} \vec{r}_1(t) &= \frac{1}{1+s_1^2}\begin{pmatrix} c_1\\s_1c_1\\0 \end{pmatrix}, & \vec{r}^{(1)}_1(t)&=-\frac{2s_1c_1}{1+s_1^2}\vec{r}_1(t)+\frac{1}{1+s_1^2}\begin{pmatrix} -s_1\\c_1^2-s_1^2\\0 \end{pmatrix} &&&&& \vec{r}_2(t)&= \begin{pmatrix} s_1\\s_1c_1\\0 \end{pmatrix},& \vec{r}_2^{(1)}(t)&=\begin{pmatrix} c_1\\c_1^2-s_1^2\\0 \end{pmatrix} \end{aligned}$ Now we can tackle the vector products. Recall $\vec{a}\times \vec{a}=0$ : $\begin{aligned} \vec{r}_1(t)\times\vec{r}_1^{(1)}(t) &= \vec{0}+\frac{1}{(1+s_1^2)^2}\cdot (c_1(c_1^2-s_1^2)+s_1c_1^2)\vec{e}_z=\frac{c_1^3}{(1+s_1^2)^2}\vec{e}_z &&&&& \vec{r}_2(t)\times\vec{r}_2^{(1)}(t) &=(s_1(c_1^2-s_1^2) -s_1c_1^2)\vec{e}_z=-s_1^3\vec{e}_z \end{aligned}$ Now we can finally calculate the area. In $(*)$ use the substitutions $u:=s_1$ for the first part and $u:=c_1$ for the second part: $\begin{aligned} A&=2\left| \int_0^{\frac{\pi}{2}}\frac{c_1^3}{(1+s_1^2)^2}-s_1^3\:dt \right|\underset{(*)}{=}2\left| \int_0^1 \frac{1-u^2}{(1+u^2)^2}-(1-u^2)\:du \right|=2\left| \frac{u}{1+u^2}-u+\frac{u^3}{3} \right|_0^1=2\cdot\left|-\frac{1}{6}\right|=\boxed{\frac{1}{3}}\end{aligned}$