Because the volume of this rotated solid converges while its surface area diverges. Just because Volume is "one-dimension more" than Area doesn't mean it's "larger." For example, a 0.1m by 0.1m square has an area of 0.01 m^2, but a 0.1 x 0.1 x 0.1 cube has a volume of 0.001 m^3. In your problem, the V_n simple converges quicker than A_n.
Your topic heading said "conceptual question," but here's the math explanation as well if you're curious:
[tex]A = \int_1^{\infty} \frac{dx}{x} \ = \ {lim}_{b \rightarrow \ \infty} ( \ ln|x|_1^{b} \ ) = b - 0 = b = \infty.[/tex]
On the other hand,
[tex]V = \int_1^{\infty} \pi(\frac{1}{x})^2 * dx = \pi \int_1^{\infty}\frac{dx}{x^2} = \pi * {lim}_{b \rightarrow \ \infty} [ \ -x^{-1}|_1^{b} \ ] = 0 - (-\pi) = \pi.[/tex]