91Ó°ÊÓ

Consider the given question,

$$\begin{gathered} \mathit{F}\left(x,y,z\right)=cos\left(xyz\right)\mathit{i}+\mathit{j}-yz\mathit{k}, \\ z=y^3-y^2 \end{gathered}$$

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 is given below,

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

Now first find localid="1650308407203" $\frac{\mathrm{∂}z}{\mathrm{∂}x},\frac{\mathrm{∂}z}{\mathrm{∂}\mathrm{y}}$. The first partial derivates of z are given below,

$$\begin{gathered} \frac{\mathrm{∂}z}{\mathrm{∂}x}=\frac{\mathrm{∂}}{\mathrm{∂}\mathrm{x}}\left(y^3-y^2\right) \\ =0 \\ \frac{\mathrm{∂}\mathrm{z}}{\mathrm{∂}\mathrm{y}}=\frac{\mathrm{∂}}{\mathrm{∂}\mathrm{y}}\left(y^3-y^2\right) \\ =3y^2-2y \end{gathered}$$

Then,$\begin{array}{r}\sqrt{{\left(\frac{\mathrm{∂}z}{\mathrm{∂}x}\right)}^2+{\left(\frac{\mathrm{∂}z}{\mathrm{∂}y}\right)}^2+1}=\sqrt{0^2+{\left(3y^2−2y\right)}^2+1}\\ =\sqrt{9y^4−12y^3+4y^2+1}\end{array}$

The choice of n should have pointing in the positive z-direction. Then the following vector is given below,

$\left(\mathbf{i}+0\mathbf{j}+\frac{\mathrm{∂}z}{\mathrm{∂}x}\mathbf{k}\right)×\left(0\mathbf{i}+\mathbf{j}+\frac{\mathrm{∂}z}{\mathrm{∂}y}\mathbf{k}\right)=−\frac{\mathrm{∂}z}{\mathrm{∂}x}\mathbf{i}−\frac{\mathrm{∂}z}{\mathrm{∂}y}\mathbf{j}+1\mathbf{k}$is normal to the surface.

Now, for the value of z,it gives the following vector perpendicular to the surface,

$\begin{array}{r}\mathbf{v}=−\frac{\mathrm{∂}z}{\mathrm{∂}x}\mathbf{i}−\frac{\mathrm{∂}z}{\mathrm{∂}y}\mathbf{j}+1\mathbf{k}\\ =0\mathbf{i}−\left(3y^2−2y\right)\mathbf{j}+1\mathbf{k}\end{array}$

Then, the desired normal vector is given below,

localid="1650311860352" $\begin{array}{r}\mathbf{n}=\frac{\mathbf{v}}{âˆ\yen \mathbf{v}âˆ\yen }\\ =\frac{0\mathbf{i}−\left(3y^2−2y\right)\mathbf{j}+1\mathbf{k}}{∥\left(0\mathbf{i}−\left(3y^2−2y\right)\mathbf{j}+1\mathbf{k}\right)∥}\\ =\frac{0\mathbf{i}−\left(3y^2−2y\right)\mathbf{j}+1\mathbf{k}}{\sqrt{0^2+{\left(−3y^2+2y\right)}^2+1^2}}\\ =\frac{0\mathbf{i}−\left(3y^2−2y\right)\mathbf{j}+1\mathbf{k}}{\sqrt{9y^4−12y^3+4y^2+1}}\end{array}$

The value of $\mathbf{F}(x,y,z)â‹\dots \mathbf{n}$is given below,

role="math" localid="1650308842698" $$\begin{gathered} \mathbf{F}(x,y,z)â‹\dots \mathbf{n}=\left(\cos \right(xyz)\mathbf{i}+\mathbf{j}−yz\mathbf{k})â‹\dots \left(\frac{0\mathbf{i}−\left(3y^2−2y\right)\mathbf{j}+1\mathbf{k}}{\sqrt{9y^4−12y^3+4y^2+1}}\right) \\ \begin{array}{r}=\frac{1}{\sqrt{9y^4−12y^3+4y^2+1}}\left(\cos \right(xyz)\mathbf{i}+\mathbf{j}−yz\mathbf{k})â‹\dots \left(0\mathbf{i}−\left(3y^2−2y\right)\mathbf{j}+1\mathbf{k}\right)\\ =\frac{1}{\sqrt{9y^4−12y^3+4y^2+1}}\left(\cos \left(xyz\right)â‹\dots 0+1â‹\dots \left(−3y^2+2y\right)−yzâ‹\dots 1\right)\\ =\frac{−3y^2+2y−yz}{\sqrt{9y^4−12y^3+4y^2+1}}\end{array} \end{gathered}$$

Then,$$\begin{gathered} \mathbf{F}(x,y,z)â‹\dots \mathbf{n}=\frac{−3y^2+2y−y\left(y^3−y^2\right)}{\sqrt{9y^4−12y^3+4y^2+1}} \\ =\frac{−y^4+y^3−3y^2+2y}{\sqrt{9y^4−12y^3+4y^2+1}} \end{gathered}$$

Substitute the values in equation (i),

$D=\left\{\right(x,y)âˆ\pounds −3≤x≤2,−1≤y≤1\}$

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

role="math" localid="1650309281896" $\begin{array}{r}{∫}_S \mathbf{F}(x,y,z)â‹\dots \mathbf{n}dS={âˆ\neg }_D \left(\mathbf{F}\right(x,y,z)â‹\dots \mathbf{n})\left(\sqrt{{\left(\frac{\mathrm{∂}z}{\mathrm{∂}x}\right)}^2+{\left(\frac{\mathrm{∂}z}{\mathrm{∂}y}\right)}^2+1}dA\right.\\ ={∫}_{−1}^1 {∫}_{−3}^2 \left(\frac{−y^4+y^3−3y^2+2y}{\sqrt{9y^4−12y^3+4y^2+1}}\right)\left(\sqrt{9y^4−12y^3+4y^2+1}\right)dxdy\\ \begin{array}{r}={∫}_{−1}^1 \left(−y^4+y^3−3y^2+2y\right)\left[{∫}_{−3}^2 dx\right]dy\\ ={∫}_{−1}^1 \left(−y^4+y^3−3y^2+2y\right)[x{]}_{−3}^{2}dy\end{array}\end{array}$

On solving the above equation,

$\begin{array}{r}{∫}_S \mathbf{F}(x,y,z)â‹\dots \mathbf{n}dS=5{\left[−\frac{y^5}{5}+\frac{y^4}{4}−y^3+y^2\right]}_{−1}^1\\ \left.=5\left[\left(−\frac{1^5}{5}+\frac{1^4}{4}−1^3+1^2\right)−\left(−\frac{(−1{)}^5}{5}+\frac{(−1{)}^4}{4}−(−1{)}^3+(−1{)}^2\right]\right]−\left(\frac{1}{5}+\frac{1}{4}+1+1\right)\right]\\ \begin{array}{r}=5\left[−\frac{12}{5}\right]\\ =−12\end{array}\end{array}$