Let R be the region in the first quadrant enclosed by the hyperbola #x^2 -y^2= 9#, the x-axis , the line x=5, how do you find the volume of the solid generated by revolving R about the x-axis?

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Answer 1 · 2016-06-24T14:08:57

#44/3 pi#

Desmos

A small vertical strip width #Delta x# within R will have revolved volume as follows

#Delta V = pi y^2 Delta x#

So
#V = pi \int_3^5 y^2 \ dx#

#= pi \int_3^5 x^2 - 9 \ dx#

#= pi [ (x^3)/3 - 9x ]_3^5 #

#= pi [{ (5^3)/3 - 9(5)} - {(3^3)/3 - 9(3)} ]#

#= 44/3 pi#