Gabriel's Horn Volume is $\pi$ but the Circle Area at $x = 1$ is $\pi$ We know that Gabriel's Horn has the volume $\pi $ when rotating $f(x) = \frac{1}{x}$ around the $x$-axis for $x \in [1,\infty)$. From disk method, we calculate the volume of this object by summing an infinite sum of a disk with the radius $f(x)$ and a really small height of the disk $dx$.  
But we know that the circle(disk with infinitesimal $dx$) area at the entrance of the Gabriel's Horn is $\pi*\frac{1}{1}^{2} = \pi$; thus the volume of Gabriel's Horn should be bigger than $\pi$ but that, of course, contradicts the integration calculation.  
Where could I possibly go wrong? Is there any conceptual understanding that I miss or misunderstood?
 A: What you're missing here is that the disc in question has no volume at all: instead, you must take all the infinite number of infinitely thin sheets in order to create the whole volume.  To see this in a more immediately graspable situation, let's consider the square pyramid of base length 1 and height 1; the true volume is 1/3 (in cube units) and the area of the base is 1 (in square units).  The cross section as you go up the pyramid decreases quickly enough that it doesn't fill the unit cube at all.  The same is true of Gabriel's Horn.
A: $π$ represents the volume of Gabriel's Horn, and the volume of its cross section at $x = 1$ contributes very little to the total volume of the horn itself, and is far smaller in value than the area of the circle you speak of.
I think this problem lies in a misunderstanding of the relationship between volume and area. The volume of an entire object can be less than the area of one of its sides. Take a right rectangular prism (a box) that has measurements of  $4l \space * 8w \space *0.25h$. Its largest face has an area of $32$ units$^2$, whereas its entire volume is $8$ units$^3$.
This is exactly the case in Gabriel's Horn. While the area of its left-most circular face is indeed $π$, we do not calculate the total volume by summing up the areas of each cross section's circular face, instead we sum up the each cross section's infinitesimally small volume, which we calculate as $π$ using the disk method.
