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

Question

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?

Answer 1

Equation of the circle is $x^2+y^2=r^2$

Since you have $ x^2+y^2=4$

That means radius of your circle is $2$

So the following integral will become

$$\iint_D \sqrt{4-x^2-y^2} \, dx \, dy = \int_0^\pi \, d\theta \int_0^2\:r\sqrt{4-r^2}\ dr$$

$\theta$ is from $0$ to $\pi$ because you have only upper half of a circle. $\sqrt{4-r^2}$ gets multiplied on $r$ because you need to take into account $dxdy = rdrd\theta$

Answer 2

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 $y \geq0$ $$ \int_0^{2\pi} \int_0^2 \sqrt{4-r^2}\,r \, dr \, d\theta $$