What is the volume of the solid produced by revolving f(x)=1/sqrt(1+x^2) around the x-axis?

Aug 2, 2016

$= {\pi}^{2}$

Explanation:

i'll put the graph in at the end as its already crashed me once

for a small element of $y = f \left(x\right)$ of width dx revolved around the x axis, the volume dV is

$\mathrm{dV} = \pi {y}^{2} \times \mathrm{dx}$

So
$V = \pi {\int}_{- \infty}^{\infty} \setminus {y}^{2} \setminus \mathrm{dx}$

$= 2 \pi {\int}_{0}^{\infty} \setminus {y}^{2} \setminus \mathrm{dx}$ .....due to symmetry

The question is whether the integral will converge so we plough on

$= 2 \pi {\lim}_{t \to \infty} {\int}_{0}^{t} \setminus \frac{1}{{x}^{2} + 1} \setminus \mathrm{dx}$

$= 2 \pi {\lim}_{t \to \infty} {\left[\setminus {\tan}^{- 1} x \setminus\right]}_{0}^{t}$

$= 2 \pi \left(\frac{\pi}{2} - 0\right)$

$= {\pi}^{2}$

the graph ....showing the symmetry etc follows

graph{1/sqrt(1+x^2) [-5, 5, -2.5, 2.5]}

And a plot of Arctan x
graph{arctan (x) [-5, 5, -2.5, 2.5]}