What is the surface area of the solid created by revolving #f(x)=sqrt(x^3)# for #x in [1,2]# around the x-axis?

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Answer 1 · 2016-03-17T18:02:57

24.93

Surface area of solid of revolution about x axis is given by the formula

#S= int_a^b 2pi y ds = int_a^b 2piy sqrt (1+(dy/dx)^2) dx #. In the present case ,
#dy/dx = 3/2 x^(1/2)#, hence #sqrt (1+(dy/dx)^2)=sqrt ( 1+9x/4)#

#S= 2pi int_1^2 x^(3/2)sqrt (1+9x/4) dx #

=#pi int_1^2 x^(3/2) sqrt (9x+4) dx#

This integration is too long to write it here. hence
use integral calculator to solve the integral

= 24.93