The 19/3 is just a nuisance constant. Don't know why they stuffed it in there. I'll just do the integral with a constant k, and it will make things easier to read.
y = kx(x-1)
crosses th x-axis at x=0 and 1. So, using discs of area πr^2 and thickness dx, the volume v is just
v = ∫[0,1] πr^2 dx
where r=y=k(x^2-x)
v = k^2 π∫[0,1] (x^2-x)^2 dx
= k^2 π/30
Now just plug in your value for k.
Find the volume of the solid obtained by revolving the region bounded by y=(19/3)x-(19/3)x^2 and the x-axis around the x-axis.
Keep answer in terms of pi
PLease explain each step
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PS Did you see my typo, and that it makes no difference in the answer?
1 answer