Find volume of solid obtained by rotating the regin formed by $\displaystyle y=\frac{1}{x}$ and $x$ axis and line $x=3$ to $x=4$ about $x$ axis using Shell method is
What i try:: 
I have tried using Disk Method:
Required volume $\displaystyle V =\pi\int^{4}_{3}y^2dx=\pi\int^{4}_{3}x^{-2}dx$
$\displaystyle V =-\pi\bigg(\frac{1}{x}\bigg)\bigg|^{4}_{3}=\frac{\pi}{12}$ cubic unit
But i did not understand How do i solve it Using Cylindrical method.
Please help me , Thanks
Using the shell method in that problem is perverse and unnatural, but it can be done; you just have to do it in two parts. For $0\le y\le\frac14$ the height of the shell at $y$ is $4-3=1$, but for $\frac14\le y\le\frac13$ the height of the shell is $\frac1y-3$, because the horizontal line at height $y$ cuts the curve $y=\frac1x$ at $x=\frac1y$.
Thus, the volume is
$$2\pi\int_0^{1/4}1\cdot y\,dy+2\pi\int_{1/4}^{1/3}\left(\frac1y-3\right)y\,dy\,.$$
Of course the first integral is just the volume of the cylinder of radius $\frac14$ and height $1$, which we could have calculated directly as $\pi\left(\frac14\right)^2\cdot1=\frac{\pi}{16}$, but evaluating the integral and getting the same result offers some assurance that we’re on the right track.
I’ll leave it to you to finish off that last integral and verify that you get the same answer as with the disk method.