we get OD /sVXB.da=0/(Jf+)·da Using Stokes Theorem, this is written as OD 允B·d=A0(Jf+) da ot where C is the curve bounding the surface et c to be a circle of radius r. Note that Je+ oD is homogeneous. So aD 27B=10(J+)4) nr at B=②(y+ aD at