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

Question

3406 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-02T13:25:01

#= pi^2#

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]}