Question

double integral sqrt(x^2+y^2)dxdy on the domain x^2+y^2=4

Answer 1

\[\int\int_R\sqrt{x^2+y^2}~dx~dy\]
where \(R:=\left\{(x,y)~:~-2\le x\le2,~-2\le y\le2\right\}\)

Convert to polar coordinates:
\[x^2+y^2=r^2\\
dx~dy=r~dr~d\theta\]
The region \(R\) is just a circle with radius 2, which can be defined as follows in polar:
\[R:=\left\{(r,\theta)~:~0\le r\le2,~0\le\theta\le2\pi\right\}\]
So your integral is equivalent to
\[\int_0^{2\pi}\int_0^2\sqrt{r^2}~r~dr~d\theta=\int_0^{2\pi}\int_0^2r^2~dr~d\theta\]

Answer 2

ty. i was really having a hard time w/ understanding the polar coordinate conversion.

Answer 3

You're welcome