Answer
$${\text{The integral diverges}}$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \frac{{\sqrt x }}{{\root 3 \of {{x^2} + 1} }},\,\,\,{\text{Interval }}\left[ {0,\infty } \right) \cr
& {\text{Calculate the volume using the disk method }}V = \int_a^b {\pi f{{\left( x \right)}^2}dx} \cr
& V = \int_0^\infty {\pi {{\left( {\frac{{\sqrt x }}{{\root 3 \of {{x^2} + 1} }}} \right)}^2}dx} \cr
& {\text{Then}}{\text{,}} \cr
& V = \int_0^\infty {\pi \frac{x}{{{{\left( {{x^2} + 1} \right)}^{2/3}}}}dx} \cr
& V = \frac{\pi }{2}\mathop {\lim }\limits_{b \to \infty } \int_0^b {{{\left( {{x^2} + 1} \right)}^{ - 2/3}}\left( {2x} \right)dx} \cr
& {\text{Evaluate the integral}} \cr
& V = \frac{\pi }{2}\mathop {\lim }\limits_{b \to \infty } \left[ {3{{\left( {{x^2} + 1} \right)}^{1/3}}} \right]_0^b \cr
& V = \frac{{3\pi }}{2}\mathop {\lim }\limits_{b \to \infty } \left[ {\root 3 \of {{b^2} + 1} - \root 3 \of 1 } \right] \cr
& V = \frac{{3\pi }}{2}\mathop {\lim }\limits_{b \to \infty } \left[ {\root 3 \of {{b^2} + 1} - 1} \right] \cr
& {\text{Evaluate the limit when }}b \to \infty \cr
& V = \frac{{3\pi }}{2}\left[ {\root 3 \of {{\infty ^2} + 1} - 1} \right] \cr
& V = \infty \cr
& {\text{The integral diverges}} \cr} $$
