Show that the solid generated by the revolution of the region has finite volume but infinite surface area

Question

Q:Let R be the region to the right of $x=1$ that is bounded by the $x$-axis and the curve $y=\frac{1}{x}$. Show that the solid generated bt the revolution of the region about the $x$-axis has finite volume but infinite surface area.
My Approach:$$ V = \pi \int_1^\infty \frac{1}{x^2} dx=\pi\lim_{t \to \infty}\int_1^t\frac{1}{x^2} dx=\pi $$Hence i can said that the revolution of the region has finite volume.But i haven't a very good idea about improper integral then how could I prove$$A=2\pi\int_1^\infty \frac{1}{x}\sqrt{1+\frac{1}{x^4}} dx=divergent$$Any hints or solution will be appreciated.
Thanks in advance.

Answer 1

Lower-bound the surface area integral with another divergent integral. $\sqrt{1+1/x^4}>1$ for all $x\ge 1$, so the surface area integral is greater than $$\int_1^\infty\frac 1x\,dx$$ which diverges.

Answer 2

For positive $x$, $\sqrt{1+x^{-4}}$ is always greater than $1$. Then $$A>2\pi\int_1^\infty \frac{1}{x}dx=2\pi(\ln\infty-\ln 1)=\infty$$