91影视

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,

$\begin{array}{r}{鈭�}_S鈥�\mathbf{F}(x,y,z)鈰�\mathbf{n}dS={鈭�}_D鈥�\left(\mathbf{F}\right(x,y,z)鈰�\mathbf{n})鈭�z_x脳z_y鈭�dA\\ ={鈭�}_D鈥�\left(\mathbf{F}\right(x,y,z)鈰�\mathbf{n})\left(\sqrt{{\left(\frac{\mathrm{鈭�}z}{\mathrm{鈭�}x}\right)}^2+{\left(\frac{\mathrm{鈭�}z}{\mathrm{鈭�}y}\right)}^2+1}dA\right.鈥�鈥�.\left(1\right)\end{array}$

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

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

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

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

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

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

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

$\begin{array}{r}v=鉄�12,鈭�9,3鉄�\\ =3鉄�4,鈭�3,1鉄�\end{array}$

The desired normal vector will be,

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

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),

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

Substituting these values in equation (i),

$D=\left\{\right(x,y)鈭�2鈮�x鈮�3,5鈮�y鈮�10\}$

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

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

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