Answer
The integral describes the volume of solid obtained by rotating the region
bounded by $f(y)= y^2,\ \ g(y)= y^4,\ \ y=0, \ y=1\ $ about the $y$-axis.
Work Step by Step
Given
$$ \pi \int_{0}^1 (y^4-y^8) dy$$
Compare the integral with
$$V= \int_a^b (f^2(y)-g^2(y))dy $$
We get
$$f (y)= y^2 ,\ \ g(y)= y^4,\ \ \ a= 0,\ \ b=1 $$
The integral describes the volume of solid obtained by rotating the region
bounded by $f(y)= y^2,\ \ g(y)= y^4,\ \ y=0, \ y=1\ $ about the $y$-axis.
