How do you find the volume bounded by #y^2=x^3-3x^2+4# & the lines x=0, y=0 revolved about the x-axis?

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Answer 1 · 2016-09-28T08:29:39

#= 4 pi#

It's this

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but bounded in the first quadrant

A small element of width #delta x# and reaching from x-axis to the curve will have height #y#

And so will have volume, when revolved about x-axis, of #delta V = pi y^2 delta x#

So we can say that

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

#= pi int_0^2 x^3-3x^2+4 dx#

#= pi [ x^4/4-x^3+4x ]_0^2#

#= 4 pi#