What is the volume of the solid produced by revolving #f(x)=1/x, x in [1,4] #around the x-axis?

1 Answer
Apr 1, 2018

#(3pi)/4# units#""^3#

Explanation:

Volume of revolution about the #x# axis is given by, #V=piinty^2dx#, where #y# is a function of #x#

In this case #y=f[x]#=#1/x#, so #V=piint[1/x]^2dx#.

#1/x^2# can be written as #x^-2#, this can be integrated using the general power rule, #intx^ndx=x^[n+1]/[n+1 #, giving #int1/x^2dx=-1/x#

Therefore the volume will #=[-pi/4]-[-pi/1]=(3pi)/4# units#""^3#