Question

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

Answer 1

`= 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(x)` of width dx revolved around the x axis, the volume dV is

`dV = pi y^2 times dx`

So
`V = pi int_(-oo)^(oo) \ y^2 \ dx`

`= 2pi int_(0)^(oo) \ y^2 \ dx` .....due to symmetry

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

`= 2pi lim_(t to oo) int_(0)^(t) \ 1/(x^2 + 1) \ dx`

`= 2pi lim_(t to oo) [ \ tan^(-1) x \ ]_(0)^(t)`

`= 2pi ( pi /2 - 0)`

`= 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]}