, where S is the region of the plane with equation $12x-9y+3z=10$, where $2\le x\le 3$and $5\le y\le 10$, with n pointing upwards.

Consider the given question,

If a surface S is the graph of $z=z\left(x,y\right)$, then the Flux of $\mathit{F}\left(x,y,z\right)$ through S,

$\begin{array}{r}{\int }_S \mathbf{F}(x,y,z)\cdot \mathbf{n}dS={\iint }_D \left(\mathbf{F}\right(x,y,z)\cdot \mathbf{n})∥z_x\times z_y∥dA\\ ={\iint }_D \left(\mathbf{F}\right(x,y,z)\cdot \mathbf{n})\left(\sqrt{{\left(\frac{\mathrm{\partial }z}{\mathrm{\partial }x}\right)}^2+{\left(\frac{\mathrm{\partial }z}{\mathrm{\partial }y}\right)}^2+1}dA\right.\dots \dots .\left(1\right)\end{array}$

Surface S is the portion of the following plane, $\begin{array}{r}3z=10-12x+9y\\ z=\frac{10}{3}-4x+3y\end{array}$.

Now first find $\frac{\mathrm{\partial }z}{\mathrm{\partial }x},\frac{\mathrm{\partial }z}{\mathrm{\partial }\mathrm{y}}$. The first partial derivates of z are given below,

$\begin{array}{r}\frac{\mathrm{\partial }z}{\mathrm{\partial }x}=\frac{\mathrm{\partial }}{\mathrm{\partial }x}\left(\frac{10}{3}-4x+3y\right)\\ =-4\end{array}$

On finding the value of $\frac{\mathrm{\partial }z}{\mathrm{\partial }y}$,

$\begin{array}{r}\frac{\mathrm{\partial }z}{\mathrm{\partial }y}=\frac{\mathrm{\partial }}{\mathrm{\partial }y}\left(\frac{10}{3}-4x+3y\right)\\ =3\end{array}$

Then,$\begin{array}{r}\sqrt{{\left(\frac{\mathrm{\partial }z}{\mathrm{\partial }x}\right)}^2+{\left(\frac{\mathrm{\partial }z}{\mathrm{\partial }y}\right)}^2+1}=\sqrt{(-4{)}^2+3^2+1}\\ =\sqrt{16+9+1}\\ =\sqrt{26}\end{array}$

The normal vector of a plane $12x-9y+3z=10$is given below,

$\begin{array}{r}v=⟨12,-9,3⟩\\ =3⟨4,-3,1⟩\end{array}$

The desired normal vector will be,

role="math" localid="1650310517434" $\begin{array}{r}\mathbf{n}=\frac{\mathbf{v}}{\parallel \mathbf{v}\parallel }\\ =\frac{3⟨4,-3,1⟩}{\parallel 3⟨4,-3,1⟩\parallel }\\ =\frac{⟨4,-3,1⟩}{\sqrt{4^2+(-3{)}^2+1^2}}\\ =\frac{⟨4,-3,1⟩}{\sqrt{26}}\end{array}$

Then, the value of $\mathit{F}\left(x,y,z\right)\mathbf{.}\mathit{n}$will be,

role="math" localid="1650310588904"

Substitute the value of z in equation (ii),

$\begin{array}{r}\mathbf{F}(x,y,z)\cdot \mathbf{n}=\frac{1}{\sqrt{26}}\left(4x\ln \left(xz\right)-15z+\frac{1}{y^2+1}\right)\\ =\frac{1}{\sqrt{26}}\left(4x\ln \left(x\cdot \left(\frac{10}{3}-4x+3y\right)\right)-15\cdot \left(\frac{10}{3}-4x+3y\right)+\frac{1}{y^2+1}\right)\\ \\ \\ =\frac{1}{\sqrt{26}}\left(4x\ln x+4x\ln \left(\frac{10}{3}-4x+3y\right)-50+60x-45y+\frac{1}{y^2+1}\right).\end{array}$

Substituting these values in equation (i),

$D=\left\{\right(x,y)\mid 2\le x\le 3,5\le y\le 10\}$

Then, the flux of the vector field through the surface S is given below,

$\begin{array}{r}{\int }_S \mathbf{F}(x,y,z)\cdot \mathbf{n}dS={\iint }_D \left(\mathbf{F}\right(x,y,z)\cdot \mathbf{n})\left(\sqrt{{\left(\frac{\mathrm{\partial }z}{\mathrm{\partial }x}\right)}^2+{\left(\frac{\mathrm{\partial }z}{\mathrm{\partial }y}\right)}^2+1}\right)dA\\ ={\int }_2^3 {\int }_5^{30} \left(4x\ln x+4x\ln \left(\frac{10}{3}-4x+3y\right)-50+60x-45y+\frac{1}{y^2+1}\right)dydx\end{array}$

Therefore, we can say that the required flux of the vector field through the surface Sis evaluated by this integral.