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

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Answer 1 · 2016-09-28T08:29:39

$= 4 π$ $= 4 pi$

It's this

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but bounded in the first quadrant

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

And so will have volume, when revolved about x-axis, of $δ V = π y^{2} δ x$ $delta V = pi y^2 delta x$

So we can say that

$V = π \int_{0}^{2} y^{2} d x$ $V = pi int_0^2 y^2 dx$

$= π \int_{0}^{2} x^{3} - 3 x^{2} + 4 d x$ $= pi int_0^2 x^3-3x^2+4 dx$

$= π {[\frac{x^{4}}{4} - x^{3} + 4 x]}_{0}^{3}$ $= pi [ x^4/4-x^3+4x ]_0^2$

$= 4 π$ $= 4 pi$

How do you find the volume bounded by y^2=x^3-3x^2+4y2=x3−3x2+4y^2=x^3-3x^2+4 & the lines x=0, y=0 revolved about the x-axis?