Answer
The region bounded by
$y= x^4 ,\ \ y=0 ,\ \ x= 0,\ \ x=3 $
rotates about $y-$axis
Work Step by Step
Given
$$
\int_0^3 2 \pi x^5 d x
$$
Rewrite the integral as the following
$$
\int_0^3 2 \pi x(x^4) d x
$$
Compare with
$$V= \int_a^b 2\pi r(x) h(x)dx$$
Here $r(x)= x,\ \ h(x)= x^4$. This represents the volume of the generated solid when the region bounded by
$$y= x^4 ,\ \ y=0 ,\ \ x= 0,\ \ x=3 $$
rotates about $y-$axis
