Find the volume of the solid obtained by rotating the region underneath the graph of f(x) = (x)/sqrt(x^3+1) about the y-axis over the interval (1, 10)

Answer 1

using shells, we have

v = ∫[1,10] 2πrh dx
where r = x and x=y=x/√(x^3+1)

v = 2π∫[1,10] x(x/√(x^3+1)) dx
= 2π (2/3 √(x^3+1)) [1,10]
= 4π/3 (√1001 - √2)