Why is $(\sqrt{x^2 + y^2})^{-1} \in L^1(D_1)$ but $(\sqrt{x^2})^{-1} \notin L^1(-1,1)$? Why is $(\sqrt{x^2 + y^2})^{-1} \in L^1(D_1)$, where $D_1 \subset \mathbb{R}^2$ is the unit ball centered at 0, but $(\sqrt{x^2})^{-1} \notin L^1(-1,1)$?
I recognize that the integral
\begin{align}
\int_{-1}^1 \frac{1}{\sqrt{x^2}} dx
\end{align}
diverges, while we can compute the double integral
\begin{align}
\int_{-1}^1\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \frac{1}{\sqrt{x^2 + y^2}} dydx = \int_0^{2\pi}\int_0^1 \frac{1}{r} \cdot rdrd\theta = 2\pi.
\end{align}
My question is: Why should that make sense?
Why should this converge:

and this diverge:

The graph of $\frac{1}{\sqrt{x^2}}$ is a subset of the graph of $\frac{1}{\sqrt{x^2 + y^2}}$. The area under $\frac{1}{\sqrt{x^2}}$ is a subset of the volume under $\frac{1}{\sqrt{x^2+y^2}}$. Or, put one more way, I would image that the volume under $\frac{1}{\sqrt{x^2 + y^2}}$ could be thought of as a solid of revolution obtained by revolving $\frac{1}{\sqrt{x^2}}$. So why should the volume of the solid of revolution exist, when the area under $\frac{1}{\sqrt{x^2}}$ does not?
Every bit of intuition I can muster is telling me that it should be harder for the 2D integral to converge; yet somehow going to 2D gives "extra convergence power". Why does that make sense?
 A: This is the way I look at this type of Paradox or Counter-Intuitive Case.
(A) Consider the Curve $e^{-x}$ & check the Area between the x-axis till "infinity" :
The Curve has infinite length , the Area is finite.

$O-A-B1-X1$ has finite Perimeter & finite Area.
$O-A-B2-X2$ has finite Perimeter & finite Area.
$O-A-B3-X3$ has finite Perimeter & finite Area.
In the limiting Case , we have infinite Perimeter with finite Area !
Length of $O-A-B1-B2-B3------X3-X2-X1-O$ will not converge , but the Area enclosed will converge !
(B) Consider the line of length 1 unit :
It has infinite "Points" but the length is finite.

We can not get length of two or 3 Points (between 0&1)
We can not get length of hundreds or thousands or millions of Points (between 1&2)
When we consider all (infinite) Points (between 2&3) then we can make that 1 unit (our Choice) of length.
We can then "measure" the length of L1-L2 in terms of this new higher level object.
(C) Consider the Square of length 1 unit :
It has finite Area , but infinite length of line.

This is a "space-filling curve" in 2D , which , in the limit , has infinite length.
We can then take that limiting Case to be the unit of Area.
We then have the object with infinite Points & infinite length , but finite Area.
We can then measure other Areas with this new object.
(D) Consider the Cube of length 1 unit :
It has finite Volume , but infinite Points , infinite length of line & infinite Area.
That is Similar to "space-filling curve" in 3D.
With this Volume , we can measure other volumes.
Basically what we have is some object of infinite measure , having infinite sub-objects , which we group into 1 Object & we assign a finite measure to that.
We then use that finite object to measure the size of some other object which will also be finite in terms of that high level object.
(E) In you case , we can think that the Area is infinite but we treat that like a single higher level object of smaller measure. then we use that to measure the volume in terms of that new object. It turns out that we require finite amount of that new object to get the whole volume.
Instead of rotation, if we did translation along the z-axis , we would have got infinite volume.
