What is the surface area of the solid created by revolving #f(x)=sqrt(4x)# for #x in [0,1]# around the x-axis?

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Answer 1 · 2016-06-26T11:44:00

#= 2pi#

the thin vertical strip width #Delta x# and between #y = 0# and #y = sqrt{4x}# will, when revolved round the x-axis, have volume

#Delta V = pi y^2 Delta x#

So

#V = pi int_0^1 \ y^2 dx#

#= pi int_0^1 \ 4x \ dx# ........ as #y = sqrt{4x}#

#= 4pi int_0^1 \ x \ dx#

#= 4pi [ x^2/2]_0^1 #

#= 2pi#