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Given F=yi+2xj-zk\, --(1) Given surface is plane \phi (x,y,z)=2x=6-y=0\, --(2)

Note that the plane (2) is perpendicular to xy-plane.So,we shall not project the given surface on xy-plane. Hence,in the present problem, we project surface S on xz-plane.Then,clearly,the projection of surface S is the rectangle on xz-plane bounded by z=0 to z=4 and x=0 to x=3.

Therefore,Required surface integral=\iint _{S}F\cdot ndS=\iint _{S}F\cdot n{\frac  {dxdz}{|n\cdot j|}}\, --(3) where n =unit normal vector to surface S={\frac  {\nabla \phi }{|\nabla \phi |}}\, --(4)

From(2),\nabla \phi =[i{\frac  {\partial \phi }{\partial x}}+j{\frac  {\partial \phi }{\partial y}}+k{\frac  {\partial \phi }{\partial z}}]=2i+j\,

Therefore,from (4)n={\frac  {2i+j}{{\sqrt  {(4+1}}}}={\frac  {2i+j}{{\sqrt  {5}}}}\,,So n\cdot j={\frac  {1}{{\sqrt  {5}}}}\, --(5)

Also,F\cdot n=(yi+2xj-zk)\cdot {\frac  {2i+j}{{\sqrt  {5}}}}={\frac  {2y+2x}{{\sqrt  {5}}}}\, --(6)

Using (5) and (6),the required surface integral =\int _{{z=0}}^{{4}}\int _{{x=0}}^{{3}}(2y+2x)dxdz=\int _{0}^{4}\int _{0}^{3}[2(6-2x)+2x]dxdz\,

=[\int _{0}^{3}(12-2x)dx]\times [\int _{0}^{4}dz]=[12x-x^{2}]_{0}^{3}\times [z]_{0}^{4}=108\, on simplification.

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