91Ó°ÊÓ

Consider the given question,

$\mathbf{F}(x,y,z)=⟨yz,xz,xyâŸ\copyright$

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,

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

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

$$\begin{gathered} \begin{array}{r}\frac{\mathrm{∂}z}{\mathrm{∂}x}=\frac{\mathrm{∂}}{\mathrm{∂}x}\left(x^2−y^2\right)\\ =2x\end{array} \\ \begin{array}{r}\frac{\mathrm{∂}z}{\mathrm{∂}y}=\frac{\mathrm{∂}}{\mathrm{∂}y}\left(x^2−y^2\right)\\ =−2y\end{array} \end{gathered}$$

On calculating the value,

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

If $z=z\left(x,y\right)$, then the following vector is normal to the surface.

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

Then the desired normal vector will be,

role="math" localid="1650313791536" $\begin{array}{r}\mathbf{n}=\frac{1}{âˆ\yen \mathbf{v}âˆ\yen }\mathbf{v}\\ =\frac{1}{âˆ\yen ⟨2x,2y,1âŸ\copyright âˆ\yen }⟨2x,2y,1âŸ\copyright \\ =\frac{1}{\sqrt{4x^2+4y^2+1}}⟨2x,2y,1âŸ\copyright \end{array}$

The value of $\mathbf{F}(x,y,z)â‹\dots \mathbf{n}$will be,

role="math" localid="1650314091675" $\begin{array}{r}\mathbf{F}(x,y,z)â‹\dots \mathbf{n}=⟨yz,xz,xyâŸ\copyright â‹\dots \frac{1}{\sqrt{4x^2+4y^2+1}}⟨2x,2y,1âŸ\copyright \\ =\frac{1}{\sqrt{4x^2+4y^2+1}}⟨yz,xz,xyâŸ\copyright â‹\dots ⟨2x,2y,1âŸ\copyright \\ =\frac{1}{\sqrt{4x^2+4y^2+1}}(yzâ‹\dots 2x+xzâ‹\dots 2y+xyâ‹\dots 1)\\ \begin{array}{r}=\frac{1}{\sqrt{4x^2+4y^2+1}}(4xyz+xy)\\ =\frac{xy(4z+1)}{\sqrt{4x^2+4y^2+1}}......\left(i\right)\end{array}\end{array}$

Substitute the values of zin equation (i),

On substituting the value of z in equation (ii),

$\begin{array}{r}{∫}_S \mathbf{F}(x,y,z)â‹\dots \mathbf{n}dS={âˆ\neg }_D \left(\mathbf{F}\right(x,y,z)â‹\dots \mathbf{n})\left(\sqrt{{\left(\frac{\mathrm{∂}z}{\mathrm{∂}x}\right)}^2+{\left(\frac{\mathrm{∂}z}{\mathrm{∂}y}\right)}^2+1}\right)dA\\ ={âˆ\neg }_D \left(\frac{xy\left(4x^2−4y^2+1\right)}{\sqrt{4x^2+4y^2+1}}\right)\left(\sqrt{4x^2+4y^2+1}\right)dA\\ ={âˆ\neg }_D xy\left(4x^2−4y^2+1\right)dA\\ ={âˆ\neg }_D \left(4x^{3}y−4x{y}^3\right.+xy)dA…….(iii)\end{array}$

The region of integration D is the region in the xy-plane bounded by the x-axis and the parabola with equation y as shown in the figure below,

Viewed it as a y-simple region, then the region of integration will be,

$D=\left\{(x,y)âˆ\pounds −1≤x≤1,0≤y≤1−x^2\right\}$

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

$\begin{array}{r}{∫}_s \mathbf{F}(x,y,z)â‹\dots \mathbf{n}dS\\ ={âˆ\neg }_D \left(4x^{3}y−4x{y}^3+xy\right)dA\\ ={∫}_{−1}^1 {∫}_0^{1−x^2} \left(4x^{3}y−4x{y}^3+xy\right)dydx\\ ={∫}_{−1}^1 {\left[2x^3{y}^2−x{y}^4+\frac{x{y}^2}{2}\right]}_0^{1−x^2}dx\\ ={∫}_{−1}^1 \left[\left(2x^3{\left(1−x^2\right)}^2−x{\left(1−x^2\right)}^4+\frac{x{\left(1−x^2\right)}^2}{2}\right)−\left(2x^3(0{)}^2−x(0{)}^4+\frac{x(0{)}^2}{2}\right)\right]dx\\ ={∫}^1\left(2x^3−4x^5+2x^7−x+4x^3−6x^5+4x^7−x^9+\frac{1}{2}x−x^3+\frac{1}{2}{x}^5\right)dx\end{array}$

On solving the above equation,

$\begin{array}{r}={∫}_{−1}^1 \left(−x^9+6x^7−\frac{19}{2}{x}^5+5x^3−\frac{1}{2}x\right)dx\\ ={\left[−\frac{x^{10}}{10}+6â‹\dots \frac{x^8}{8}−\frac{19}{2}â‹\dots \frac{x^6}{6}+5â‹\dots \frac{x^4}{4}−\frac{1}{2}â‹\dots \frac{x^2}{2}\right]}_{−1}^1\\ ={\left[−\frac{1}{10}{x}^{10}+\frac{3}{4}{x}^8−\frac{19}{12}{x}^6+\frac{5}{4}{x}^4−\frac{1}{4}{x}^2\right]}_{−1}^1\\ =\left(−\frac{1}{10}â‹\dots 1^{10}+\frac{3}{4}â‹\dots 1^8−\frac{19}{12}â‹\dots 1^6+\frac{5}{4}â‹\dots 1^4−\frac{1}{4}â‹\dots 1^2\right)−\left(−\frac{1}{10}(−1{)}^{10}+\frac{3}{4}(−1{)}^8−\frac{19}{12}(−1{)}^6+\frac{5}{4}(−1{)}^4−\frac{1}{4}(−1{)}^2\right)\\ \begin{array}{r}=\left(−\frac{1}{10}+\frac{3}{4}−\frac{19}{12}+\frac{5}{4}−\frac{1}{4}\right)−\left(−\frac{1}{10}+\frac{3}{4}−\frac{19}{12}+\frac{5}{4}−\frac{1}{4}\right)\\ =0\end{array}\end{array}$