Proving that $f(x,y) = \frac{1}{x^2 + y^2}$ is not integrable both on $B^2 - \vec 0$, and $\mathbb{R}^2 - \bar B^2$ In the book of Analysis on Manifolds by Munkres, at page 151, question 3, it is asked that

Let $U$ be the open set in $\mathbb{R}^2$ consisting of all $x$ with $\|x\|<1$.  Let
  $$ f(x,y) = \frac{1}{x^2 + y^2} $$
  for $(x,y) \ne 0$.  Determine whether $f$ is integrable over $U\setminus 0$ and over $\mathbb{R}^2\setminus \overline{U}$; if so, evaluate.

I have given the following answers, but, as you can see below, in both cases I have found that the integral does not exists, so is everything correct in my answers, because the author says "if so evaluate", which means in general at least one of the integral exists and we are going to evaluate it :).
My answers:
Let $g(r,\theta) = (r \cos\theta, r \sin \theta),$
where $g: (0,1)\times (0,2\pi) \to U - \{(x,y) \in \mathbb{R}^x | y = 0, x \geq 0\}.$
Note that the excluded set $\{(x,y) \in \mathbb{R}^x | y = 0, x \geq 0\}$ is of measure zero, hence it does not contribute the integral that we are going to evaluate.
For $U - \vec 0,$ let $C_n = [1/n, 1-1/n] \times [1/n, 2\pi - 1/n]$ be a rectifiable set in $(0,1)\times (0,2\pi)$ s.t $C_n \subset Int (C_{n+1}) $ and their union is $(0,1)\times (0,2\pi)$. Since for each $n \geq 2$, this set is a rectifiable set s.t $n+1$-th set contains $n$-th one and the union of these sets form a cover for $(0,1)\times (0,2\pi)$, we can evaluate the improper integral of any function over $(0,1)\times (0,2\pi)$ by first integrating it on $C_n$, and then taking the limit as $n \to \infty$.
Now, by Fubini's theorem, we have 
$$\int_{1/n}^{2\pi - 1/n} \int_{1/n}^{1-1/n} \frac{1}{r^2} * |r| = ln(\frac{1 - 1/n}{1/n} ) * (2\pi - 2/n),$$
but this function blows up as $n\to \infty$, hence it is not bounded, hence the improper integral does not exists.
For $\mathbb{R}^n - \bar U$, we form the partition 
$$[1+ 1/n, n] \times [1/n, 2\pi - 1/n],$$
for $n\geq 2$, and apply the same reasoning, and get 
$$ln(\frac{n}{1 + 1/n} ) * (2\pi - 2/n),$$
but this also blows up as $n \to \infty$, hence it is not bounded, hence the integral does not exists.

Question:
Is there anything wrong with the answers given above ? Is there any flaw ? 


Since for each $n\ge 2$, this set is a rectifiable set s.t $n+1$-th set contains $n-th$ one and the union of these sets form a cover for $(0,1)\times (0,2\pi)$, we can evaluate the improper integral of any function over $(0,1)\times (0,2\pi)$ by first integrating it on $C_n$, and then taking the limit as $n \to \infty$

This is wrong...
e.g. take $f(x) = \frac{1}{x}$ and the domain $C = (-1,1)\setminus\{0\}$ then 
$f$ is not integrable over $C$ because $$\int_C \, |f(x)| \, dx = \infty$$ but by your "assumption" it would hold with $$C_n = (-1,1)\setminus\left(\frac{-1}{n},\frac{1}{n}\right)$$ that:
$$\int_D\; f(x) \, dx = \lim_{n\to\infty} \int_{C_n} f(x) \, dx = \lim_{n\to\infty} 0 = 0$$ because $$ \int_{C_n} f(x) \, dx = 0$$ for all $n\ge 1$
