$\displaystyle \int_{0}^{1}\int_{x}^{\sqrt{x}}\int_{0}^{\sqrt{x-y^{2}}}dzdydx$

i tried to traw the projection of the top interval in the z-y plane

$\displaystyle z=\sqrt{x-y^{2}}$ -> $\displaystyle z^2+y^2=x$

so we have a circle

but the radius changes from x to $\displaystyle \sqrt{x}$.

which is not possible because when we scan the ring it goes from the biiger radius to the smaller one.

then i thought to try in anyway ,and i wrote the integral in polar

$\displaystyle \int_{0}^{1}\int_{0}^{2\pi}\int_{x}^{\sqrt{x}}rdrd \theta dz$

but i dont know how mathmatickly to change the intervals?

how to solve it in a polar way?