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

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.
• Are you sure you got the right curve? Looks more like the surface gotten by revolving $y=1/x$ instead of $y=1/x^2$. Anyway, the judgement day is here. – Jyrki Lahtonen Oct 6 '18 at 17:11
For positive $$x$$, $$\sqrt{1+x^{-4}}$$ is always greater than $$1$$. Then $$A>2\pi\int_1^\infty \frac{1}{x}dx=2\pi(\ln\infty-\ln 1)=\infty$$
• Good concept :.... It depends on the problem. The idea is that if you can solve a related but simpler integral that converges/diverges and compare your desired function with that, without calculating the integral. Suppose my simplified function $f_s$ is always positive in a certain interval $a$ to $b$. Then if $f(x)>f_s(x)$ you get $$\int_a^b f(x)dx>\int_a^b f_s(x)dx$$ – Andrei Oct 6 '18 at 17:39
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.