From Example Problems
Jump to: navigation, search

To evaluate the given surface integral,we first re-write it in the form \iint _{S}F\cdot ndS\,,where n is a unit normal vector to the closed surface S whose equation is x^{2}+y^{2}+z^{2}=a^{2}\,

Given surface S is \phi (x,y,z)=x^{2}+y^{2}+z^{2}=1\, --(1)

Therefore \nabla \phi =[i{\frac  {\partial }{\partial x}}+j{\frac  {\partial }{\partial y}}+{\frac  {\partial }{\partial z}}](x^{2}+y^{2}+z^{2})=2xi+2yj+2zk\,

Therefore,n={\frac  {\nabla \phi }{|nabla\phi |}}={\frac  {2(xi+yj+zk)}{{\sqrt  {4(x^{2}+y^{2}+z^{2})}}}}=xi+yj+zk\, --(2)

We choose F=axi+byj+czk\, --(3)

so that ax^{2}+by^{2}+cz^{2}=(axi+byj+czk)\cdot (xi+yj+zk)\, by (2) and (3)

Therefore,ax^{2}+by^{2}+cz^{2}=F\cdot n\, --(4)

Now,\iint _{S}(ax^{2}+by^{2}+cz^{2})dS=\iint _{S}F\cdot ndS\, by (4)

=\iiint _{V}{\mathrm  {div}}FdV\,,by divergence theorm

=\iiint _{V}{\mathrm  {div}}(axi+byj+czk)dV\,,by (3)

=\iiint _{V}[{\frac  {\partial }{\partial x}}(ax)+{\frac  {\partial }{\partial y}}(by)+{\frac  {\partial }{\partial z}}(cz)]dV\,,by definition of divergence

=\iiint _{V}(a+b+c)dV=(a+b+c)V={\frac  {4\pi }{3}}(a+b+c)\, [Since V=volume of sphere={\frac  {4}{3}}\pi (a)^{3}={\frac  {4\pi }{3}}\,]

Main Page