# Solid of revolution about y axis

I am worried that my bounds are incorrect, so I am just making sure that my solution is valid

Question

Let R be the region in the first quadrant bounded by the curve $y = (x − 2)^2$. Compute the volumes V1, V2, and V3 of the solids of revolution obtained by revolving R about the lines x = 0, x = 3, and y = −1, respectively.

For x = 0, aka y-axis, I decided to use the disk method and integrate wrt y.

I first found the bounds, from 0 to $y = (0-2)^2= 4$

and the final integral I came up with was

$$V = \int_0^4 \pi (\sqrt y + 2)^2 dy$$

and then I imagine that for x = 3 you simply subtract 3 from the original radius making it $\sqrt y + 5$.

My concern is with how I am setting up my boundaries and whether I should be changing the variable or not. Just making sure, thank you for any assistance!

What you've done looks fine to me. If you want to use disks for a rotation about the $y$-axis, you pretty much HAVE to change variables. You could, of course, perform a substitution in your displayed integral: $y = (x-2)^2; dy = 2(x-2) dx$ where $x$ goes from 0 to 2, which would turn it back into an integral in $x$, so "have to" is too strong. Perhaps I should say "just about everyone always DOES change the variable to $y$."