calculus - Evaluate the double integral by changing to polar coordinates for $x^2+y^2leq4$

Change the double integral $\iint_D \sqrt{4-x^2-y^2} \, dx \, dy$ where $D = \{(x,y):x^2+y^2\leq4,y\geq0\}$ by changing to polar coordinates $r, \phi$

So am I right in thinking the limits would be $0$ and $4$ for $x$ and $y$?

Converting the integral would be

\begin{align}
& \int_0^4 \int_0^4 \sqrt{4-x^2-y^2} \, dx \, dy = \iint_D \sqrt{4-r^2\cos^2\phi-r^2\sin^2\phi} \ |r| \, dx \, dy \\[10pt]
= {} & \iint_D \sqrt{4-r^2} \, |r| \, dx \, dy
\end{align}

I am unsure how to change the coordinates?