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In order to evaluate the given 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 of the given ellipsoid.

Let \phi (x,y,z)=ax^{2}+by^{2}+cz^{2}=1\, --(1)

Then \nabla \phi =[i{\frac  {\partial }{\partial x}}+j{\frac  {\partial }{\partial y}}+k{\frac  {\partial }{\partial z}}](ax^{2}+by^{2}+cz^{2})=2(axi+byj+czk)\,

Therefore,n={\frac  {\nabla \phi }{|\nabla \phi |}}={\frac  {2(axi+byj+czk)}{{\sqrt  {(4a^{2}x^{2}+4b^{2}y^{2}+4c^{2}z^{2})}}}}={\frac  {axi+byj+czk}{{\sqrt  {a^{2}x^{2}+b^{2}y^{2}+c^{2}z^{2}}}}}\, --(2)

Now,F\cdot n=(a^{2}x^{2}+b^{2}y^{2}+c^{2}z^{2})^{{\frac  {-1}{2}}}={\frac  {1}{{\sqrt  {a^{2}x^{2}+b^{2}y^{2}+c^{2}z^{2}}}}}\, --(3)

For the purpose,we take F=xi+yj+zk\, --(4)

Then,using (2) and (4),we obtain

F\cdot n=(xi+yj+zk)\cdot {\frac  {(axi+byj+czk)}{{\sqrt  {(a^{2}x^{2}+b^{2}y^{2}+c^{2}z^{2})}}}}={\frac  {a^{2}x^{2}+b^{2}y^{2}+c^{2}z^{2}}{{\sqrt  {(a^{2}x^{2}+b^{2}y^{2}+c^{2}z^{2})}}}}={\frac  {1}{{\sqrt  {(a^{2}x^{2}+b^{2}y^{2}+c^{2}z^{2})}}}}\,

Now,\iint _{S}(a^{2}x^{2}+b^{2}y^{2}+c^{2}z^{2})^{{\frac  {-1}{2}}}dS=\iint _{S}F\cdot ndS=\iiint _{V}{\mathrm  {div}}FdV\, by the divergence theorm,V being the volume enclosed by S.

=\iiint _{V}[{\frac  {\partial }{\partial x}}(x)+{\frac  {\partial }{\partial y}}(y)+{\frac  {\partial }{\partial z}}(z)]dV\, using (4) and (1)

=3\iiint _{V}dV=3V=3\times Volumeofellipsoid\,

=3{\frac  {4}{3}}\pi [{\frac  {1}{{\sqrt  {a}}}}{\frac  {1}{{\sqrt  {b}}}}{\frac  {1}{{\sqrt  {c}}}}]={\frac  {4\pi }{{\sqrt  {abc}}}}\,

[Remember that the volume of the ellipsoid is taken in the above step]

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