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

Feb 21, 2017

Volume $= \pi \ln 2$

#### 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:

$V = {\int}_{0}^{1} 2 \pi x \frac{1}{{x}^{2} + 1} \mathrm{dx}$
$\setminus \setminus \setminus = \pi {\int}_{0}^{1} \frac{2 x}{{x}^{2} + 1} \mathrm{dx}$
 \ \ \ = pi ln|x^2+1)|_0^1
$\setminus \setminus \setminus = \pi \left(\ln 2 - \ln 1\right)$
$\setminus \setminus \setminus = \pi \ln 2$