# How do you use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by y =2/x, y=0, x(1)=1 , x(2)=6 revolved about the y-axis?

Jan 31, 2017

$V = 20 \pi$

#### Explanation:

If you imagine an almost infinitesimally thin vertical line of thickness $\delta x$ between the $x$-axis and the curve at some particular $x$-coordinate it would have an area:

$\delta A \approx \text{width" xx "height} = y \delta x = f \left(x\right) \delta x$ If we then rotated this infinitesimally thin vertical line about $O y$ then we would get an infinitesimally thin cylinder (imagine a cross section through a tin can), then its volume $\delta V$ would be given by:

$\delta V \approx 2 \pi \times \text{radius" xx "thickness} = 2 \pi x \delta A = 2 \pi x f \left(x\right) \delta x$

If we add up all these infinitesimally thin cylinders then we would get the precise total volume $V$ given by:

$V = {\int}_{x = a}^{x = b} 2 \pi x f \left(x\right) \mathrm{dx}$

So for this problem we have:

$V = {\int}_{1}^{6} 2 \pi x \left(\frac{2}{x}\right) \mathrm{dx}$
$\setminus \setminus \setminus = {\int}_{1}^{6} 4 \pi \mathrm{dx}$
$\setminus \setminus \setminus = 4 \pi \setminus {\int}_{1}^{6} \setminus \mathrm{dx}$
$\setminus \setminus \setminus = 4 \pi {\left[x\right]}_{1}^{6}$
$\setminus \setminus \setminus = 4 \pi \left(6 - 1\right)$
$\setminus \setminus \setminus = 20 \pi$