Calculus: volume of revolution about a line other than the $x$-axis. Find the volume of the solid of revolution obtained by rotating the region bounded by 
$f(x) = x^3 + 1$,  $g(x) = x^2$  and $0 ≤ x ≤ 1$ about the line $y = 3$.
I know the gist of the problem, but I'm struggling with the $y = 3$. If it was simply rotating around the $x$ axis I would just plug in the outer and inner radius into $\pi r^2$, but when $y$ is set equal to a number I'm not sure how to handle the problem. An explanation would be helpful. 

When you intersect the described rotational body $B$ with a plane $x={\rm const.}$ you obtain an annulus in this plane of outer radius $3-x^2$ and inner radius $3-(x^3+1)=2-x^3$. The area $A(x)$ of this annulus is given by
$$A(x)=\pi\bigl((3-x^2)^2-(2-x^3)^2\bigr)\qquad(0\leq x\leq 1)\ .$$
Therefore the volume of $B$ is given by
$$V(B)=\int_0^1 A(x)\ dx={142\pi\over35}\ .$$
