VC4.33

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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= $\iint_S F\cdot n dS=\iint_S F\cdot n \frac{dx dz}{|n\cdot j|}\,$ --(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 = $\int_{z=0}^{4}\int_{x=0}^{3}(2y+2x)dx dz=\int_0^4\int_0^3[2(6-2x)+2x]dx dz\,$

= $[\int_0^3(12-2x)dx]\times[\int_0^4dz]=[12x-x^2]_0^3\times[z]_0^4=108\,$ on simplification.

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