Cartesian to polar integral conversion

[djoe]

Hi;
I ve to convert this double integral from Cartesian to polar coordinates

\[\iint\]f(x,y)dxdy -a<x<+a ; -a<y<+a
somebody helps me, analytically or numerically (with a matlab code)
Thanks

 

[albbg]

To convert from cartesian to polar coordinates you have change the function as:

f(x,y)dxdy to \[f(\rho\cos\theta,\rho\sin\theta)\rho{d}\rho{d}\theta\]

Now we have calculate the new two intervals. We know

x=\[\rho\cos\theta\]
y=\[\rho\sin\theta\]

then, I suppose [-a,a] is the interval of both iinternal and external integrals then "a" is included in the range, so:

-a<=\[\rho\cos\theta\]<=a
-a<=\[\rho\sin\theta\]<=a

Of course both \[\cos\theta\] and \[\sin\theta\] function goes from -1 to 1 in a 2\[\pi\] period.
The modulus \[\rho\] is a radius in the polar plane, then when \[\rho\]=a we have the maximum extension (+a or -a depending on the quadrant). The minimum extension will be \[\rho\]=0, that is a point, so:

0<=\[\rho\]<=a interval of the external integral
0<=\[\theta\]<=2\[\pi\] interval of the internal integral

 

[djoe]

Thank you albbg
but it is still ambigeous: rho can take values >a exple: when x=a and y=a; rho=sqrt(a^2+b^2)>a
I think 0<=rho<=sqrt(a^2+b^2); doesn't it?
now the problem is with theta range!
please try it again

 

[djoe]

sorry
take b=a
and rho=sqrt(2)*a >a