I want to revolve the region bounded by $y=1$, $x^2=y^2+y$ about the $x$ axis. For this, I need to isolate $y$ in terms of $x$. Then, $V = \pi\int^{\sqrt2}_0 (1-y^2)dx$.
Is it possible to set $y$ and $x$ as functions of $t$? Then, how would I proceed?
I also thought if there was something like implicit integration which would let me approach this problem, because you can implicitly find the derivative.
Lastly, is there any relation of the volume with the inverse function, because I know how to find the integral of an inverse function?
$y^2+y=x^2$
Add $1/4$ on both sides
$(y+\frac12)^2=x^2+ \frac14$
And finally,
$$y=\sqrt{x^2+\frac14} -\frac12$$