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

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Answer 1 · 2016-11-17T02:10:24

# V = pi[x+1/3x^3]_a^b #

The volume of revolution about #Ox# is given by;

# V = int_a^b piy^2dx #

so with #f(x)=sqrt(1+x^2)#, then:

# V = int_a^b pi(sqrt(1+x^2))^2dx #
# :. V = piint_a^b 1+x^2 dx #
# :. V = pi[x+1/3x^3]_a^b #

As you have not specified the #x#-coordinate bounds this is as far as we can proceed

Answer 2 · 2016-11-17T02:25:05

Nov 17, 2016

See answer below:
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