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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?

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    $\begingroup$ are you sure end points of first integral? it sould be from $0$ to $2$ $\endgroup$ – corcia candy Jan 17 '16 at 17:40
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    $\begingroup$ and if you change coordinates, then you should change $dxdy$ to polar coordinates $drd\theta$ $\endgroup$ – corcia candy Jan 17 '16 at 17:41
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    $\begingroup$ The integral is over the upper semicircular region of radius $2$ centered at the origin. Therefore it is equal to $\int_0^2 dr\ r\sqrt{4-r^2}\int_0^\pi d\phi=8 \pi/3$. The original limits of integration should be $$ \int_{-2}^2 dx\int_0^{\sqrt{4-x^2}}dy\ . $$ $\endgroup$ – Pierpaolo Vivo Jan 17 '16 at 17:58
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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 $y \geq0$ $$ \int_0^{2\pi} \int_0^2 \sqrt{4-r^2}\,r \, dr \, d\theta $$

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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 $dxdy = rdrd\theta$

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