Shell method to compute volume

Question

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?

Answer 1

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.

Answer 2

Yes, your integral is correct as the shell integration: $$ V=2\pi\int_a^b (x-h)f(x)dx,\quad \textrm{for } h\le a<b $$ where $h=-2$, $a=0$ and $b=1$ and $f(x)=x-x^2$ in your example.

It is useful to draw a picture.