The solid lies above the dik $D$ whose boundary circle has equation $x^2{y}^2=2x$ ec$,$ after completing the square,

$(x-{)}^2{y}^2=$

In polar coordinates we have $x^2{y}^2=r^2$ and $x=rcos(),$ so the boundary circle becomes $r^2=2rcos(),$ or $r=2cos().$ Thus the dik $D$ is given by

$D=\{(r\frac{,}{b}ar())|-\frac{x}{2}\frac{x}{2},0r2cos()\}$

and, by this formula, we have the following.