Find the outward flux of the given vector field through the specified surface.

$\mathbf{F}(x,y,z)=xyz\mathbf{i}$and S is the portion of the plane with equation$x+y+z=4$in first octant.

The given vector field is $\mathbf{F}(x,y,z)=xyzi$

The portion of plane has equation$x+y+z=4$.

If surface of graph is $z=z(x,y)$, flux is

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

The portion of surface is $x+y+z=4$

$\Rightarrow z=4-x-y$

Now,

$\frac{\partial z}{\partial x}=\frac{\partial }{\partial x}(4-x-y)$

$\frac{\partial z}{\partial x}=-1$

And

$\frac{\partial z}{\partial y}=\frac{\partial }{\partial y}(4-x-y)$

$=-1$

Hence,

$\sqrt{{\left(\frac{\partial z}{\partial x}\right)}^2+{\left(\frac{\partial z}{\partial y}\right)}^2+1}=\sqrt{{\left(-1\right)}^2+{\left(-1\right)}^2+1}$

$\sqrt{3}$

The normal vector for plane $x+y+z=4$is

$v=⟨1,1,1⟩$

Hence, desired normal vector is

$n=\frac{1}{\Vert v\Vert }v$

$=\frac{1}{\sqrt{1^2+1^2+1^2}}⟨1,1,1⟩$

$=\frac{1}{\sqrt{3}}⟨1,1,1⟩$

The value of $\mathbf{F}(x,y,z)·\mathbf{n}$is

$\mathbf{F}(x,y,z)·\mathbf{n}=⟨xyz,0,0⟩·\frac{1}{\sqrt{3}}⟨1,1,1⟩$

$=\frac{1}{\sqrt{3}}⟨xyz,0,0⟩·⟨1,1,1⟩$

$=\frac{xyz-1+0·1+0·1}{\sqrt{3}}$

$=\frac{xyz}{\sqrt{3}}$

Using values of $\mathbf{F}(x,y,z)·\mathbf{n},\sqrt{{\left(\frac{\partial z}{\partial x}\right)}^2+{\left(\frac{\partial z}{\partial y}\right)}^2+1}$, we get

${\int }_s\mathbf{F}(x,y,z)·\mathbf{n}dS={\iint }_0\left(\mathbf{F}\right(x,y,z)·\mathbf{n})\left(\sqrt{{\left(\frac{\partial z}{\partial x}\right)}^2+{\left(\frac{\partial z}{\partial y}\right)}^2+1}\right)dA$

$={\iint }_0\left(\frac{xz}{\sqrt{3}}\right)\left(\sqrt{3}\right)dA$

$={\iint }_{0}xyzdA$

Since $z=4-x-y$, integral becomes

${\int }_S\mathrm{F}(x,y,z)·\mathrm{n}dS={\iint }_{D}xy(4-x-y)dA$

$={\iint }_B\left(4xy-x^{2}y-x{y}^2\right)dA$

The graph of region of integration is:

After applying in region of integration, integral becomes:

$={\int }_0^4{\int }_0^x\left(4xy-x^{2}y-x{y}^2\right)dydx$

$={\int }_0^4{\left[2x{y}^2-\frac{x^2{y}^2}{2}-\frac{x{y}^3}{3}\right]}_0^{4-x}dx$

$={\int }_0^4\left[\left(2x(4-x{)}^2-\frac{x^2(4-x{)}^2}{2}-\frac{x(4-x{)}^3}{3}\right)-\left(2x(0{)}^2-\frac{x^2(0{)}^2}{2}-\frac{x(0{)}^3}{3}\right)\right]dx$

$={\int }_0^4\left[\left(2x\left(16-8x+x^2\right)-\frac{x^2\left(16-8x+x^2\right)}{2}-\frac{x\left(64-48x+12x^2-x^3\right)}{3}\right)-0\right]dx$

$={\int }_0^4\left(-\frac{1}{6}{x}^4+2x^3-8x^2+\frac{32}{3}x\right)dx$

$={\left[-\frac{1}{30}{x}^3+\frac{1}{2}{x}^4-\frac{8}{3}{x}^3+\frac{16}{3}{x}^2\right]}_0^4$

$=\left(-\frac{1}{30}(4{)}^3+\frac{1}{2}(4{)}^4-\frac{8}{3}(4{)}^3+\frac{16}{3}(4{)}^2\right)-\left(-\frac{1}{30}(0{)}^5+\frac{1}{2}(0{)}^4-\frac{8}{3}(0{)}^3+\frac{16}{3}(0{)}^2\right)$

$=\frac{128}{15}$

Hence, required flux is$\frac{128}{15}$