$\mathbf{F}(x,y,z)=z\cos yz\mathbf{j}+z\sin yz\mathbf{k}$, where S is the portion of the plane with equation $2x-8y-10z=42$ that lies on the positive side of the rectangle with corners$\left(0,-\mathrm{\pi },0\right)(0,\pi ,0),(0,\pi ,\pi ),(0,-\pi ,\pi )$in theyz-plane.

Consider the given question,

$\mathbf{F}(x,y,z)=z\cos yz\mathbf{j}+z\sin yz\mathbf{k}$

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

${\int }_S \mathbf{F}(x,y,z)\cdot \mathbf{n}dS={\iint }_D \left(\mathbf{F}\right(x,y,z)\cdot \mathbf{n})\left(\sqrt{{\left(\frac{\mathrm{\partial }x}{\mathrm{\partial }y}\right)}^2+{\left(\frac{\mathrm{\partial }x}{\mathrm{\partial }z}\right)}^2+1}\right)dA......\left(i\right)$

Here, the surface S is the portion of the following plane,

$$\begin{gathered} 2x=42+8y+10z \\ x=\frac{1}{2}(42+8y+10z) \\ x=21+4y+5z \end{gathered}$$

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

$$\begin{gathered} \frac{\mathrm{\partial }\mathrm{x}}{\mathrm{\partial }\mathrm{y}}=\frac{\mathrm{\partial }}{\mathrm{\partial }\mathrm{y}}\left(21+4y+5z\right) \\ =4 \end{gathered}$$

Find $\frac{\mathrm{\partial }\mathrm{x}}{\mathrm{\partial }\mathrm{y}}$ when the first partial derivates of z are given below,

$\begin{array}{r}\frac{\mathrm{\partial }x}{\mathrm{\partial }z}=\frac{\mathrm{\partial }}{\mathrm{\partial }z}(21+4y+5z)\\ =5\end{array}$

Then,

$$\begin{gathered} \sqrt{{\left(\frac{\mathrm{\partial }\mathrm{x}}{\mathrm{\partial }\mathrm{y}}\right)}^2+{\left(\frac{\mathrm{\partial }\mathrm{x}}{\mathrm{\partial }\mathrm{z}}\right)}^2+1}=\sqrt{4^2+5^2+1} \\ =\sqrt{16+25+1} \\ =\sqrt{42} \end{gathered}$$

The normal vector of a plane $2x-8y-10z=42$is given below,

$\mathbf{v}=⟨2,-8,-10⟩$.

For the value of z, it gives the following vector perpendicular to this surface,

$\begin{array}{r}n=\frac{1}{\parallel ⟨2,-8,-10⟩\parallel }⟨2,-8,-10⟩\\ =\frac{1}{\sqrt{2^2+(-8{)}^2+(-10)}}⟨2,-8,-10⟩\\ =\frac{1}{2\sqrt{42}}⟨2,-8,-10⟩\\ =\frac{1}{\sqrt{42}}⟨1,-4,-5⟩\end{array}$

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

role="math" localid="1650341559809" $\begin{array}{r}\mathbf{F}(x,y,z)\cdot \mathbf{n}=⟨0,z\cos yz,z\sin yz⟩\cdot \frac{1}{\sqrt{42}}⟨1,-4,-5⟩\\ \begin{array}{r}=\frac{1}{\sqrt{42}}(0\cdot 1+z\cos yz\cdot (-4)+z\sin yz\cdot (-5\left)\right)\\ =\frac{1}{\sqrt{42}}(-4z\cos yz-5z\sin yz)......\left(ii\right)\end{array}\end{array}$

On substituting the values in equation (ii),

$\begin{array}{r}{\int }_S \mathbf{F}(x,y,z)\cdot \mathbf{n}dS={\iint }_D \left(\mathbf{F}\right(x,y,z)\cdot \mathbf{n})\left(\sqrt{{\left(\frac{\mathrm{\partial }x}{\mathrm{\partial }y}\right)}^2+{\left(\frac{\mathrm{\partial }x}{\mathrm{\partial }z}\right)}^2+1}\right)dA\\ ={\iint }_D \left(\frac{1}{\sqrt{42}}(-4z\cos yz-5z\sin yz)\right)\left(\sqrt{42}\right)dA\\ ={\iint }_D (-4z\cos yz-5z\sin yz)dA\\ ={\iint }_D z(-4\cos yz-5\sin yz)dA\dots \dots (iii)\end{array}$

The region of integration D is the region on the positive side of the rectangle with the given corners in theyz-plane as shown in the following figure,

Here, the region of integration will be,

$D=\left\{\right(y,z)\mid -\pi \le y\le \pi ,0\le z\le \pi \}$

Using equation (iii), the flux of the vector field through the surface S,

$\begin{array}{r}{\int }_S \mathbf{F}(x,y,z)\cdot \mathbf{n}dS={\int }_D z(-4\cos yz-5\sin yz)dA\\ \begin{array}{r}={\int }_0^{\pi } {\int }_{-\pi }^{\pi } z(-4\cos yz-5\sin yz)dydz\\ \begin{array}{r}={\int }_0^{\pi } [-4\sin yz+5\cos yz{]}_{-\pi }^{\pi }dz\\ ={\int }_0^{\pi } \left[\right(-4\sin \pi y+5\cos \pi y)-(-4\sin (-\pi y)+5\cos (-\pi y)\left)\right]dz\\ ={\int }_0^{\pi } \left[\right(-4\sin \pi y+5\cos \pi y)-(4\sin \pi y+5\cos \pi y\left)\right]dz\end{array}\end{array}\end{array}$

On continuing the above equation,

$\begin{array}{r}{\int }_S \mathbf{F}(x,y,z)\cdot \mathbf{n}dS={\int }_0^{\pi } -8\sin \pi ydz\\ ={\left[\frac{-8\cos \pi y}{\pi }\right]}_0^{\pi }\\ =\left(\frac{-8\cos \pi y}{\pi }\right)-\left(\frac{-8\cos \pi y}{\pi }\right)\\ \begin{array}{r}=\frac{-8\cos {\pi }^2}{\pi }+\frac{8}{\pi }\\ =\frac{8}{\pi }\left(1-\cos {\pi }^2\right)\end{array}\end{array}$