Radius of limit cycle for van der Pol oscillator in the limit of $\varepsilon\ll1$ using Green's Theorem 
We want to determine the radius of the nearly circular limit cycle of the van der Pol oscillator $$\ddot x + \varepsilon \dot x \left( x^2 - 1 \right) + x = 0$$ in the limit $\varepsilon\ll1$. Assume that the limit cycle is a circle of unknown radius $a$ about the origin, and invoke the normal form of Green’s theorem (i.e., the $2$-dimensional divergence theorem) $$\oint_C\mathbf{v}\cdot\mathbf{n} \, dl=\iint_A \nabla\mathbf{v} \, dA$$ where $C$ is the cycle and $A$ is the region enclosed. By substituting $\mathbf{v}=\dot x =(\dot x,\dot y)$and evaluating the integrals, show that $a\approx2$.


I first rewrote the van der Pol equation in terms of $x$ and $y$ to get
$$\dot x=y \\ \dot y=-x-\varepsilon y(x^2-1)$$
The double integral then becomes
$$\iint_A \nabla\mathbf{v} \, dA=\iint_A -\varepsilon (x^2-1) \, dA$$
rewriting this in polar coordinates, we get that
$$\iint_A -\varepsilon (x^2-1) \, dA=\int_0^{2\pi}\int_0^a -\varepsilon r(r^2\cos^2(\theta)-1) \, dr \, d\theta \\=\int_o^{2\pi}-\frac{\varepsilon}{4}r^4\cos^2(\theta)+\frac{\varepsilon}{2}r^2\biggr|_0^ad\theta \\=\int_o^{2\pi}-\frac{\varepsilon}{4}a^4\cos^2(\theta)+\frac{\varepsilon}{2}a^2d\theta \\=-\frac{\varepsilon}{4}a^4\left(\frac{\sin(\theta)\cos(\theta)+\theta}{2} \right) + \frac{\varepsilon}{2}a^2\theta\biggr|_0^{2\pi} \\=\varepsilon a^2\pi-\frac{\varepsilon a^4\pi}{4} \\ =\varepsilon a^2\pi(1-\frac{1}{4}a^2)$$
I'm not exactly sure how to get this answer when doing the line integral. It seems the $\varepsilon$ term disappears. My work is as follows:
Using $r(t)=\langle a\cos(t), a\sin(t) \rangle$ such that $r'(t)=\langle -a\sin(t), a\cos(t) \rangle$, we have
$$\oint_C\mathbf{v}\cdot\mathbf{n} \, dl=\int_0^{2\pi}\langle a\sin(t),-a\cos(t)-\varepsilon a\sin(t)(a^2\cos^2(t)-1)\rangle \cdot \langle -a\sin(t), a\cos(t) \rangle \, dt \\\int_0^{2\pi}=-a^2-\varepsilon a^4 \sin(t)\cos^3(t)+\varepsilon a^3\sin(t)\cos(t) \,dt \\=-a^2t+\frac{a^4}{4} \varepsilon \cos^4(t)-\frac{a^2}{2}\varepsilon \cos^2(t)\biggr|_0^{2\pi} \\=-a^22\pi$$
If anyone could point out where I went out in either integral I'd be super appreciative!
 A: There is a way to solve it without calculating the line integral explicitly.
The line integral satisfies
$$\oint_C\mathbf{v}\cdot\mathbf{n} = 0,$$
because the vector field must be tangential to the line $C$ everywhere on the stable limit cycle (or else the limit cycle would not be be stable, a contradiction). As such, the vector field $\mathbf{v}$ is perpendicular to $\mathbf{n}$ everywhere on the line $C$.
Hence by Green's theorem, $\iint_A \nabla\mathbf{v} \, dA = 0$. As your calculations point out, this gives
$$\varepsilon a^2\pi(1-\frac{1}{4}a^2) = 0.$$
Solving for $a$ gives
$$a = 2,$$
assuming $a$ greater than zero.
A: You should use instead $\mathbf{n}=\langle \cos(t), \sin(t) \rangle$ such that $||\mathbf{n}||=1$, and we have
$$\oint_C\mathbf{v}\cdot\mathbf{n} \, dl=\int_0^{2\pi}\langle a\sin(t),-a\cos(t)-\varepsilon a\sin(t)(a^2\cos^2(t)-1)\rangle \cdot \langle \cos(t), \sin(t) \rangle \, dt \\\int_0^{2\pi}=-\varepsilon a^4 \sin(t)\cos^3(t)+\varepsilon a^3\sin(t)\cos(t) \,dt \\=\frac{a^4}{4} \varepsilon \cos^4(t)-\frac{a^2}{2}\varepsilon \cos^2(t)\biggr|_0^{2\pi} \\=0$$
Now you can see that these two are only equivalent if $a=2$
