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


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 $y$."

  • $\begingroup$ Thank you very much John! Great to know that I'm understanding this. $\endgroup$ – mrybak834 Feb 5 '15 at 12:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.