Let $R$ be the region under the graph of
$$y = 18 − x^2 \text{ for } 0 \leq x \leq 2.$$
Use the Shell Method to compute the volume of rotation of $R$ about the $x$-axis as a sum of two integrals along the y-axis. Hint: The shells generated depend on whether $y$ is in $[0, 14]$ or $y$ is in $[14, 18]$.
I'm new to the shell method and don't quite understand how it works fully. I understand that the formula is set up as
$$2 \pi \int x \cdot f(x) \, dx$$
but I'm still struggling and not sure where to go. Any help is appreciated, sorry for the poor formatting, I'm new and will try to fix it when I figure out how later today.
$$V=2\pi\int\limits_{0}^{14}y.2.dy+2\pi\int\limits_{14}^{18}y\sqrt{18-y}dy$$ You can evaluate the second integral by applying substiution $18-y=t^2$.
On the other hand, via disc method volume can be found by the integral $$V=\pi\int_{0}^{2}(18-x^2)^2dx$$