find volume created by revolving the region between $x=y^2$ and $y=x^2$ about the line $x=-1$.
My attempt is since to solve for $x$ first correct? so we have $x=y^2, x = \sqrt{y}$ so we. integrate from $0$ to $1$? I know its top function minus bottom but I forget do we do this
$$\pi \int_0^1 (y+1)^2 - \sqrt{y+1} dy$$
or is it this one? forget where to subtract the line really.
$$\pi \int_0^1 y^2 +1 - \sqrt{y} +1dy$$
thanks in advance I am studying for GRE subject, learned calc II in 2008 forgot most of it sadly. Hopefully the top integral is the correct set up but not sure.
It would help to draw the region first. I am not sure how to do it here. But anyway, the formula the OP is looking for is the washer method formula (which I had incorrectly called the cylindrical shell method, thank you @MathLover for pointing it out), which gives in this case
$$ \pi \int_0^1 \left( (\sqrt{y} + 1)^2 - (y^2 + 1)^2 \right) dy. $$