I have a question about a particular portion of my multivariable calculus book in the chapter on Green's Theorem.
Let me first set the theory up. If you want to jump to the question it is further below:
Here is the version of Green's Theorem that is being used:
Let C be a positively oriented piecewise-smooth simple closed curve that bounds the region R in the plane. Suppose that the functions $P(x,y)$ and $Q(x,y)$ have continuous first-order partial derivatives on R. Then $$\oint_C Pdx + Qdy=\int \int_R (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y})dA $$
This theorem at this point in the exposition of the book is valid for regions that can be divided into subregions that are both vertically and horizontally simple.
Now take a region R with boundaries that consist of two simply closed curves, ie a region R with a hole in it, an annular region R for example:
We know that $$\int \int_R (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y})dA = \oint_C Pdx + Qdy = \oint_{C_1} Pdx + Qdy + \oint_{C_2} Pdx + Qdy$$
We get this because when we calculate the line integrals for the curve around $R_1$ and then do the same for the curve around $R_2$, the portions with the arrows cancel out. We are left with a line integral over the outer portion of $C$, ie $C_1$, plus a line integral over the inner portion of $C$, ie $C_2$.
My question refers to the following problem: calculate $$\oint_C \frac{-y}{x^2+y^2}dx + \frac{x}{x^2+y^2}dy$$
where C is a positively oriented piecewise-smooth simple closed curve that encloses the origin $(0,0)$
First thing we note is that $$P = \frac{-y}{x^2+y^2}$$ $$Q = \frac{x}{x^2+y^2}$$ and we have that $$\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 0$$ everywhere except at the origin. Using Green's Theorem, this means that $$\oint_C Pdx +Qdy = 0$$ for any C not enclosing the origin.
But our $C$ does enclose the origin. So how do we calculate the line integral in this case? We enclose the origin in a small circle of radius a, such that the circle is contained within our region R (as in the picture above). We parametrize the circle as $$x=a \cos t, y = a \sin t, 0 \leq t \leq 2 \pi$$ which is a counterclockwise orientation by the way (this is the reason for the negative sign in the following expression).
$$\oint_C \frac{-y}{x^2+y^2}dx + \frac{x}{x^2+y^2}dy - \oint_{C_a} \frac{-y}{x^2+y^2}dx + \frac{x}{x^2+y^2}dy = \int \int_R 0 dA = 0$$
$$\implies \oint_C \frac{-y}{x^2+y^2}dx + \frac{x}{x^2+y^2}dy = \oint_{C_a} \frac{-y}{x^2+y^2}dx + \frac{x}{x^2+y^2}dy$$
Finally, the question
The book goes ahead and calculates the integral on the right side, ie over $C_a$. This is equal to $2 \pi$. Why are we allowed to calculate this integral without somehow taking some sort of limit because we have a singularity at the origin. Ie, the vector field is not defined at $(0,0)$. The book calculates the integral by changing coordinates to polar coordinates, no limit is taken, and in polar coordinates there is no singularity anymore. Don't we have to take some sort of limit as $a$ goes to zero?
My experience with this so far is with single-variable calculus where we would need to integrate from say $a$ to 1, and take the limit as $a$ goes to $0^{+}$.