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.
Best Answer
If you look at the graph of $f(x), g(x)$, and eyeball the line $y = 3$,
you'll see the inner radius will be $r_i = 3-(x^3 + 1) = 2-x^3$ and the outer radius will be $r_o = 3-x^2$.
Then our integrand will be of the form $\pi\big(r_0^2 - r_i^2)$.
That gives us the integral $$\pi \int_0^1 \left((3-x^2)^2 - (2 - x^3)^2\right)\,dx$$