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