Shell method to compute volume I'm currently learning about the shell method to compute the volume of a solid of revolution. I am working on the following problem:
Find the volume of the solid obtained by rotating the region by $y = x - x^2$ and $y = 0$ about the line $x = -2$.
I've plotted the three lines, and I'm still really unsure about how to figure this out. I saw some examples on this website: https://brilliant.org/wiki/shell-method/. However, they do not seem very similar to the problem that I have.
I would appreciate it if someone could please help me with this problem.
I think a cylindrical shell formed by rotating about the line $x = -2$ has radius $2 + x$ and circumference $2\pi(2 - x)$ and height $(x - x^2)$, which gives
$$V = \int_{0}^{1} 2\pi (2 + x)(x - x^2)\mathop{dx} $$
Is this correct?


Your set up is correct. Due to the confusion, I am adding the diagram for your method.
You can use alternate method (washer) to check your working if you already know.
$x-x^2 = y \implies x = \frac{1}{2} \pm \sqrt{\frac{1}{4}-y}$
So, $\frac{5}{2} - \sqrt{\frac{1}{4}-y} \leq r \leq \frac{5}{2} - \sqrt{\frac{1}{4}-y}$
$0 \leq y \leq \frac{1}{4}$
So, $\displaystyle \pi \int_0^{0.25} (2.5+ \sqrt{0.25-y})^2 - (2.5 - \sqrt{0.25-y})^2 dy = \frac{5 \pi}{6}$. This is same that you would get from your integral.
