VC4.32

From Example Problems
Jump to: navigation, search

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

Then,\nabla \phi =i{\frac  {\partial }{\partial x}}+j{\frac  {\partial }{\partial y}}+k{\frac  {\partial }{\partial z}}=2xi+2yj+2zk\,

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

Now, r=xi+yj+zk,r=|r|={\sqrt  {x^{2}+y^{2}+z^{2}}}\, --(3)

{\frac  {r}{|r|^{3}}}={\frac  {xi+yj+zk}{(x^{2}+y^{2}+z^{2})^{{{\frac  {3}{2}}}}}}\, --(4)

Therefore,\iint _{S}{\frac  {r}{|r|^{3}}}\cdot dS=\iint _{S}{\frac  {r}{|r|^{3}}}\cdot ndS\,

=\iint _{S}{\frac  {xi+yj+zk}{(x^{2}+y^{2}+z^{2})^{{{\frac  {3}{2}}}}}}\cdot {\frac  {xi+yj+zk}{a}}dS\, using (2) and (4)

{\frac  {1}{a}}\iint _{S}{\frac  {x^{2}+y^{2}+z^{2}}{(x^{2}+y^{2}+z^{2})^{{{\frac  {3}{2}}}}}}dS\,

={\frac  {1}{a}}\iint _{S}{\frac  {1}{{\sqrt  {(x^{2}+y^{2}+z^{2})}}}}dS\, ={\frac  {1}{a^{2}}}\iint _{S}dS\, ={\frac  {1}{a^{2}}}\times \,(area of the curved surface of sphere S) ={\frac  {1}{a^{2}}}\times (4\pi a^{2})=4\pi \,

Main Page