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?
Best Answer
İf you think $z=\sqrt{4-x^2 -y^2}$ and $0\leq y$ , it shows a semi-sphere. $$\int_{-2}^2 \int_0^{2\sqrt{4-x^2}} \int_0^\sqrt{4-x^2 - z^2} \, dy \, dz \, dx$$
Converting to polar coordinates in double integral;
Note:Hence there is a symmetry , we can think like $z \geq 0$ instead of $y \geq0$ $$ \int_0^{2\pi} \int_0^2 \sqrt{4-r^2}\,r \, dr \, d\theta $$