$\omega = \frac{x(x^2+y^2 - 1)}{(x^2+y^2 + 1)^2-4x^2}dy+ \frac{y(x^2+y^2 + 1)}{(x^2+y^2+1)^2 - 4x^2}dx$. Calculate $3$ integrals over $3$ circles.

I found such problem with solution.

Problem:

We have:

• $C1 = \{ (x, y) : (x-1)^2 + y^2 = \frac{1}{4} \}$
• $C2 = \{ (x, y) : x^2 +y^2 = 4 \}$
• $C3 = \{ (x, y) : (x + 1)^2 + y^2 = \frac{1}{4} \}$

Those are counterclockwise oriented circles. Then, we have:

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

Calculate $3$ integrals: $\int_{C_1}\omega$ , $\int_{C_2}\omega$ , $\int_{C_3} \omega$

Solution:

Decomposing the denominators into simple fractions we get:$$\omega = \frac{1}{2} \left( \frac{x-1}{(x-1)^2 + y^2}dy - \frac{y}{(x-1)^2 + y^2}dx \right) + \frac{1}{2} \left( \frac{x+1}{(x+1)^2 + y^2} - \frac{y}{(x+1)^2 + y^2} \right)$$

Now we can call the first form as $\omega_1$ , and the second form as $\omega_2$. We check easily that both forms are closed, i.e., equality occurs:

$$ \frac{\partial}{\partial x} \left( \frac{x-1}{(x-1)^2 + y^2} \right) = \frac{\partial}{\partial y} \left( \frac{-y}{(x-1)^2 + y^2} \right) \ \text{ and } \ \frac{\partial}{\partial x} \left( \frac{x+1}{(x+1)^2 + y^2} \right) = \frac{\partial}{\partial y} \left( \frac{-y}{(x+1)^2 + y^2} \right)$$

From that we get:

$$\int_{C_1} \omega_2 = \int_{C_3} \omega_1 = 0$$

Therefore, we can calculate that:

$$\int_{C_1} \omega = \pi = \int_{C_3} \omega$$

We expect the integral after $C_2$ to be $2 \pi$ in view of this, which we check by direct calculation after simplification.

$$\int_0^{2 \pi} \frac{12 + 8sin^2t}{25 - 16cos^2 t} dt = 2 \pi$$

There are three things that I don't quite understand here.

1. How is it deducted that $\int_{C_1} \omega_2 = \int_{C_3} \omega_1 = 0$ ? As I understand it, we know that those forms are closed and the curves over which we evaluate the integrals do not enclose any points from outside of our domain. Therefore the forms are exact and integrals of exact forms over closed curves are equal to $0$. Is that correct?

2. Then it is said that we can calculate that $\int_{C_1} \omega = \pi$. As I understand it, $\int_{C_1} \omega = \int_{C_1} \omega_1 + \omega_2 = \int_{C_1} \omega_1 + \int_{C_1} \omega_2 = \int_{C_1} \omega_1$. To calculate that I need to parametrize $C1$ with $x = \frac{1}{2}cos\alpha + 1$ and $y = \frac{1}{2}sin\alpha$. Jacobian of such parametrization is equal to $\frac{1}{2}$ so I get:

$$\int_C \frac{1}{2} \left( \frac{x-1}{(x-1)^2 + y^2}dy - \frac{y}{(x-1)^2 + y^2}dx \right) =$$$$= \frac{1}{2} \int_0^{2 \pi} \left( \frac{\frac{1}{2}cos\alpha}{ \left(\frac{1}{2}cos\alpha \right)^2 + \left( \frac{1}{2}sin\alpha \right)^2} \left[\frac{1}{2}sin\alpha \right]'_{\alpha} - \frac{\frac{1}{2}sin\alpha}{ \left( \frac{1}{2}cos\alpha \right)^2 + \left( \frac{1}{2}sin\alpha \right)^2} \left[\frac{1}{2}cos\alpha + 1 \right]'_{\alpha} \right) \frac{1}{2} \ d \alpha = $$$$= \frac{1}{16} \int_0^{2 \pi} \left( \frac{cos\alpha}{ \left(\frac{1}{2}cos\alpha \right)^2 + \left( \frac{1}{2}sin\alpha \right)^2} cos \alpha + \frac{sin\alpha}{ \left( \frac{1}{2}cos\alpha \right)^2 + \left( \frac{1}{2}sin\alpha \right)^2} sin \alpha \right) \ d \alpha = $$

$$= \frac{1}{16} \int_0^{2 \pi} \left( \frac{cos^2 \alpha}{ \frac{1}{4}cos^2 \alpha + \frac{1}{4}sin^2 \alpha} + \frac{sin^2\alpha}{ \frac{1}{4}cos^2 \alpha + \frac{1}{4}sin^2 \alpha} \right) \ d \alpha = $$

$$= \frac{1}{4} \int_0^{2 \pi} \left( \frac{cos^2 \alpha}{ cos^2 \alpha + sin^2 \alpha} + \frac{sin^2\alpha}{cos^2 \alpha + sin^2 \alpha} \right) \ d \alpha = $$

$$= \frac{1}{4} \int_0^{2 \pi} 1 \ d \alpha = 2 \pi$$

If I didn't include the Jacobian ($\frac{1}{2}$) in my calculations, I would have the right value. Therefore here comes my question - should I write my Jacobian in such situations?

3. In the solution, it is written that "We expect the integral after $C_2$ to be $2 \pi$ in view of this". But where does that expectation come from before calculations?