Monday, February 16, 2009

Surface Integral explained

The best way to motivate surface integral would be to use the Gauss's Law which says that int(int E. n ds)) = q/epsilon. Where E is electrostatic field and n is a unit normal vector, ds is the infinitesimal area of the surface. Surface integral is a double integral. All surface integral are of the form int(int F.n ds)) where F is a vector field and n is the unit normal vector. To understand this one need to first understand the unit normal vector. The unit normal vector is a vector which is normal to the surface. To derive the unit normal vector for a surface oriented over x-y plane we define two vectors u and v and take their cross product at a particular pt on the surface. Its remarkable that this normal vector comes out to be
(-del(f)/dx-del(f)/dy+1)/ sqrt( (del(f)/dx)^2+(del(f)/dy)^2+1)

and doesn't contain any information about u and v we considered. The above expression is easy to derive by considering two planes one parallel to x-z plane and other parallel to y-z plane. The angle the surface makes is del(f)/dx and del(f)/dy. ux and vy are the components along the x axis and y axis for the vector u and v respectively. The component along z axis can be calculated by ux*del(f)/dx and vy*del(f)/dy.
and the surface integral for the scalar function becomes

int int f(z) sqrt( (del(f)/dx)^2+(del(f)/dy)^2+1) dx dy

The one for the vector field is slightly more complex its just that you are writing the components of the field and multiplying with the unit vector.



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