I can do volume fine about x and y=0, but I get messed up when it's about somewhere other than those.
x+y=3, x=4(y-1)^2...Find the area rotating about the x axis.
Can someone point me in the right direction?
I'm guessing you really mean the volume ...
$\displaystyle 3-y = 4(y-1)^2$
$\displaystyle 3-y = 4y^2 - 8y + 4$
$\displaystyle 0 = 4y^2 - 7y + 1$
$\displaystyle y = \frac{7 \pm \sqrt{33}}{8}$
let $\displaystyle c = \frac{7 - \sqrt{33}}{8}$
$\displaystyle d = \frac{7 + \sqrt{33}}{8}$
$\displaystyle V = 2\pi \int_c^d y[(3-y) - 4(y-1)^2] dy$