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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.

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    $\begingroup$ 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. $\endgroup$ – Jyrki Lahtonen Oct 6 '18 at 17:11
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    $\begingroup$ See also here $\endgroup$ – Jyrki Lahtonen Oct 6 '18 at 17:15
  • $\begingroup$ thanks @Jyrki Lahtonen. Yeah i did type mistake :) $\endgroup$ – emonHR Oct 6 '18 at 17:30
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For positive $x$, $\sqrt{1+x^{-4}}$ is always greater than $1$. Then $$A>2\pi\int_1^\infty \frac{1}{x}dx=2\pi(\ln\infty-\ln 1)=\infty$$

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  • $\begingroup$ Thanks @Andrei. I really suffer with improper integral to show divergent or convergent ':( can you suggest me how to make a good concept about comparison test $\endgroup$ – emonHR Oct 6 '18 at 17:27
  • $\begingroup$ 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 $$\int_a^b f(x)dx>\int_a^b f_s(x)dx$$ $\endgroup$ – Andrei Oct 6 '18 at 17:39
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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.

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