Set up the integral for the volume of revolution

Question

I essentially have a circle which I am asked to revolve about the y-axis, namely:

$(x-2)^2+y^2=1$

about

$i)$ $The$ $y$$-$$axis$

$ii)$ $y=2$

I'm having a little trouble setting up the integral however. I suspect I am supposed to solve for y since that makes the integral nicer. I'm not really sure where to proceed from there though...

Both of these look like a torus but I'm not sure is I can make that assumption.

I'm asked to set up the integral for the volume of revolution in both cases.

Answer 1

Hints:

$1$: We know that the volume of solid of revolution about the y-axis is given by: $$V = \int_{c}^{d} \pi x^2 \mathrm {d}y$$

In our question, we have, $$V = \int_{-1}^{1} \pi (\sqrt {1-y^2}+2)^2 \mathrm {d}y $$

$2$: We also know that the volume of solid of revolution about the line $y=c $ is given by: $$V = \int_{a}^{b} \pi (y-c)^2 \mathrm {d}x $$

In our question, we have, $$V = \int_{1}^{3} (\sqrt {1-(x-2)^2}-2)^2 \mathrm {d}x $$

Hope you can take it from here.