91Ó°ÊÓ

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,

${∫}_S \mathbf{F}(x,y,z)â‹\dots \mathbf{n}dS={âˆ\neg }_D \left(\mathbf{F}\right(x,y,z)â‹\dots \mathbf{n})\left(\sqrt{{\left(\frac{\mathrm{∂}x}{\mathrm{∂}y}\right)}^2+{\left(\frac{\mathrm{∂}x}{\mathrm{∂}z}\right)}^2+1}\right)dA......\left(i\right)$

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

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

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

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

Then,

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

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

$\mathbf{v}=⟨2,−8,−10âŸ\copyright$.

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

$\begin{array}{r}n=\frac{1}{âˆ\yen ⟨2,−8,−10âŸ\copyright âˆ\yen }⟨2,−8,−10âŸ\copyright \\ =\frac{1}{\sqrt{2^2+(−8{)}^2+(−10)}}⟨2,−8,−10âŸ\copyright \\ =\frac{1}{2\sqrt{42}}⟨2,−8,−10âŸ\copyright \\ =\frac{1}{\sqrt{42}}⟨1,−4,−5âŸ\copyright \end{array}$

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

role="math" localid="1650341559809" $\begin{array}{r}\mathbf{F}(x,y,z)â‹\dots \mathbf{n}=⟨0,z\cos yz,z\sin yzâŸ\copyright â‹\dots \frac{1}{\sqrt{42}}⟨1,−4,−5âŸ\copyright \\ \begin{array}{r}=\frac{1}{\sqrt{42}}(0â‹\dots 1+z\cos yzâ‹\dots (−4)+z\sin yzâ‹\dots (−5\left)\right)\\ =\frac{1}{\sqrt{42}}(−4z\cos yz−5z\sin yz)......\left(ii\right)\end{array}\end{array}$

On substituting the values in equation (ii),

$\begin{array}{r}{∫}_S \mathbf{F}(x,y,z)â‹\dots \mathbf{n}dS={âˆ\neg }_D \left(\mathbf{F}\right(x,y,z)â‹\dots \mathbf{n})\left(\sqrt{{\left(\frac{\mathrm{∂}x}{\mathrm{∂}y}\right)}^2+{\left(\frac{\mathrm{∂}x}{\mathrm{∂}z}\right)}^2+1}\right)dA\\ ={âˆ\neg }_D \left(\frac{1}{\sqrt{42}}(−4z\cos yz−5z\sin yz)\right)\left(\sqrt{42}\right)dA\\ ={âˆ\neg }_D (−4z\cos yz−5z\sin yz)dA\\ ={âˆ\neg }_D z(−4\cos yz−5\sin yz)dA……(iii)\end{array}$

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,

$D=\left\{\right(y,z)âˆ\pounds −\mathrm{Ï€}≤y≤\mathrm{Ï€},0≤z≤\mathrm{Ï€}\}$

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

$\begin{array}{r}{∫}_S \mathbf{F}(x,y,z)â‹\dots \mathbf{n}dS={∫}_D z(−4\cos yz−5\sin yz)dA\\ \begin{array}{r}={∫}_0^{\mathrm{Ï€}} {∫}_{−\mathrm{Ï€}}^{\mathrm{Ï€}} z(−4\cos yz−5\sin yz)dydz\\ \begin{array}{r}={∫}_0^{\mathrm{Ï€}} [−4\sin yz+5\cos yz{]}_{−\mathrm{Ï€}}^{\mathrm{Ï€}}dz\\ ={∫}_0^{\mathrm{Ï€}} \left[\right(−4\sin \mathrm{Ï€}y+5\cos \mathrm{Ï€}y)−(−4\sin (−\mathrm{Ï€}y)+5\cos (−\mathrm{Ï€}y)\left)\right]dz\\ ={∫}_0^{\mathrm{Ï€}} \left[\right(−4\sin \mathrm{Ï€}y+5\cos \mathrm{Ï€}y)−(4\sin \mathrm{Ï€}y+5\cos \mathrm{Ï€}y\left)\right]dz\end{array}\end{array}\end{array}$

On continuing the above equation,

$\begin{array}{r}{∫}_S \mathbf{F}(x,y,z)â‹\dots \mathbf{n}dS={∫}_0^{\mathrm{Ï€}} −8\sin \mathrm{Ï€}ydz\\ ={\left[\frac{−8\cos \mathrm{Ï€}y}{\mathrm{Ï€}}\right]}_0^{\mathrm{Ï€}}\\ =\left(\frac{−8\cos \mathrm{Ï€}y}{\mathrm{Ï€}}\right)−\left(\frac{−8\cos \mathrm{Ï€}y}{\mathrm{Ï€}}\right)\\ \begin{array}{r}=\frac{−8\cos {\mathrm{Ï€}}^2}{\mathrm{Ï€}}+\frac{8}{\mathrm{Ï€}}\\ =\frac{8}{\mathrm{Ï€}}\left(1−\cos {\mathrm{Ï€}}^2\right)\end{array}\end{array}$