Calculating a surface integral over a regular shape

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Oussama GASSAB
Oussama GASSAB 2019-11-22
编辑： Oussama GASSAB 2019-11-24
let's suppose a function given in polar coordinate F(r,phi) and our purpose is to calculate the surface integral, say F(r,phi)dA over the region S defined by
S={ |Z|<b , |z-z0|>a } where Z=r*exp(1j*phi). it means the regions between the circles |Z|=b , |z-z0|=a . However, F(r,phi) has singularities inside the circle |z-z0|=a . therefore, we are not able to use integral(over circle |Z|<b )-integral(over circle |z-z0|<a) .
your help and consideration are much appreciated.
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David Goodmanson
David Goodmanson 2019-11-24
Hi Oussama,
I am assuming that z and Z are basically the same thing, is that correct? Are z0 and 'a' such that the 'a' circle is totally contained in the b circle? Or the other way round? Is z0 real, or can it be complex?
Oussama GASSAB
Oussama GASSAB 2019-11-24
yes z is the same as Z (typing mistake). yes exactly the z0 and 'a' are chosen such that the circle |Z-z0|=a is totally inside the bigger circle |Z|=b.

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Oussama GASSAB
Oussama GASSAB 2019-11-24
编辑：Oussama GASSAB 2019-11-24
i have solved by using the following, where F(x,y)=F(r,phi). i have used x-y coordinate and quad2d.
the result was so accurate.
y1= @(x) -sqrt(b^2-x.^2) ; y2=@(x) sqrt(b^2-x.^2) ;
y3= @(x) -sqrt(a^2-(x-z0).^2) ; y4= @(x) sqrt(a^2-(x-z0).^2) ;
intg1 = quad2d(F,-b,z0-a,y1,y2) ;
intg2 = quad2d(F,z0-a,z0+a,y1,y3) ;
intg3 = quad2d(F,z0-a,z0+a,y4,y2) ;
intg4 = quad2d(F,z0+a,b,y1,y2) ;
integ_total=intg1+intg2+intg3+intg4 ;

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