# 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?

Sep 28, 2016

$= 4 \pi$

#### Explanation:

It's this

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

$V = \pi {\int}_{0}^{2} {y}^{2} \mathrm{dx}$

$= \pi {\int}_{0}^{2} {x}^{3} - 3 {x}^{2} + 4 \mathrm{dx}$

$= \pi {\left[{x}^{4} / 4 - {x}^{3} + 4 x\right]}_{0}^{2}$

$= 4 \pi$