91影视

Consider the following vector field:

$\mathbf{F}(x,y,z)=x{2}^z\ln 2\mathbf{i}+(x-y+z)\mathbf{j}+(2y+4z)\mathbf{k}.$

The objective is to evaluate the integral${鈭�}_s\mathbf{F}(x,y,z)路\mathbf{n}dS$, where $S$is the surface of the first octant region $W$bounded by the coordinate planes and the plane $x+y+2z=2$, and $\mathbf{n}$is the outwards-pointing normal vector.

Use Divergence Theorem to evaluate this integral.

Divergence Theorem states that,

"Let $W$be a bounded region in ${\mathrm{鈩�}}^3$whose boundary $S$is a smooth or piecewise-smooth closed oriented surface. If a vector field $\mathbf{F}(x,y,z)$is defined on an open region containing $W$, then,${鈭�}_S\mathbf{F}(x,y,z)路\mathbf{n}dS={鈭�}_{W}div\mathbf{F}(x,y,z)dV"$

where $\mathbf{n}$is the outwards unit normal vector."

First find the divergence of the vector field $\mathbf{F}=x{2}^z\ln 2\mathbf{i}+(x-y+z)\mathbf{j}+(2y+4z)\mathbf{k}$.

The divergence of a vector field $\mathbf{F}(x,y,z)=F_1(x,y,z)\mathbf{i}+F_2(x,y,z)\mathbf{j}+F_3(x,y,z)\mathbf{k}$is defined as follows: $div\mathbf{F}(x,y,z)=\left(\frac{鈭�}{鈭�x}\mathbf{i}+\frac{鈭�}{鈭�y}\mathbf{j}+\frac{鈭�}{鈭�z}\mathbf{k}\right)路\left(F_1\mathbf{i}+F_2\mathbf{j}+F_3\mathbf{k}\right)$$=\frac{鈭�F_1}{鈭�x}+\frac{鈭�F_2}{鈭�y}+\frac{鈭�F_3}{鈭�z}$.

$\text{Then, the divergence of the vector field}$$\mathbf{F}(x,y,z)=x{2}^z\ln 2\mathbf{i}+(x-y+z)\mathbf{j}+(2y+4z)\mathbf{k}\text{will be,}$

$div\mathbf{F}(x,y,z)=\left(\frac{鈭�}{鈭�x}\mathbf{i}+\frac{鈭�}{鈭�y}\mathbf{j}+\frac{鈭�}{鈭�z}\mathbf{k}\right)路\left(x{2}^z\ln 2\mathbf{i}+(x-y+z)\mathbf{j}+(2y+4z)\mathbf{k}\right)$

$=\frac{鈭�}{鈭�x}\left(x{2}^z\ln 2\right)+\frac{鈭�}{鈭�y}(x-y+z)+\frac{鈭�}{鈭�z}(2y+4z)$

$=2^2\ln 2+(-1)+4$

$=2^2\ln 2+3$

The region is bounded by the surface $S$, where$S$is the surface of the first octant region $W$bounded by the coordinate planes and the plane $x+y+2z=2$.

Here the region of integration will be,

$R=\left\{(x,y,z)鈭�0鈮�x鈮�2,0鈮�y鈮�2-x,0鈮�z鈮�\frac{2-x-y}{2}\right\}.$

$\text{Now, use Divergence Theorem (1) to evaluate the integral}$

role="math" localid="1650791737749" ${鈭�}_S\mathbf{F}(x,y,z)路\mathbf{n}dS\text{asfollows:}$

${鈭�}_S\mathbf{F}路\mathbf{n}dS={鈭�}_{R}div\mathbf{F}dV$

$={鈭�}_0^2{鈭�}_0^{2-{鈭�}_0^{2-x-1}}{鈭�}^2\left(2^2\ln 2+3\right)dzdydx$

$={鈭�}_0^2{鈭�}_0^{2-x}\left[{鈭�}_0^{2-x-y}\left(2^z\ln 2+3\right)dz\right]dydx$

$={鈭�}_0^2{鈭�}_0^{2-x}{\left[2^2+3z\right]}_0^{2-x-y}dydx$

$={鈭�}_0^2{鈭�}_0^{2-y}\left[\left(2^{2-x}+3\left(\frac{2-x-y}{2}\right)\right)-\left(2^a+3路0\right)\right]dydx$

$={鈭�}_0^2{鈭�}_0^{2-x}\left(2^{\frac{2-y-y}{2}}+\frac{3}{2}(2-x-y)-1\right)dydx.$

Simplify the last integral Sas follows:

${鈭�}_S\mathbf{F}-\mathbf{n}dS$

$={鈭�}_0^2{鈭�}_0^{2-x}\left(2^{\frac{2-x-x}{2}}+\frac{3}{2}(2-x-y)-1\right)dydx$

$={鈭�}_0^2\left[{鈭�}_0^{2-x}\left(2^{\frac{2-x-y}{2}}+\frac{3}{2}(2-x-y)-1\right)dy\right]dx$

$={鈭�}_0^2{\left[-\frac{2^{\frac{2-x-y}{2}}}{\ln 2}+2y-\frac{3xy}{2}-\frac{3y^2}{4}\right]}_0^{2-x}dx$

$={鈭�}_0^2\left[-\frac{2^{\frac{2-x-(2-x)}{2}}}{\ln 2}+2(2-x)-\frac{3x(2-x)}{2}-\frac{3(2-x{)}^2}{4}\right)$

$\left.-\left(-\frac{2^{\frac{2-x-0}{2}}}{\ln 2}+2路0-\frac{3x路0}{2}-\frac{3路0^2}{4}\right)\right]dx$

$={鈭�}_0^2\left[\left(-\frac{2^{\frac{2-x-(2-x)}{2}}}{\ln 2}+2(2-x)-\frac{3x(2-x)}{2}-\frac{3(2-x{)}^2}{4}\right)\right.$

$\left.-\left(-\frac{2^{\frac{2-x-0}{2}}}{\ln 2}+2路0-\frac{3x-0}{2}-\frac{3路0^2}{4}\right)\right]dx$

$={鈭�}_0^2\left(\frac{2^{\frac{2-x}{2}}-1}{\ln 2}+1-2x+\frac{3x^2}{4}\right)dx$

$={\left[x-x^2+\frac{x^3}{4}-\frac{2^{3-x}}{(\ln 2{)}^2}-\frac{2x}{\ln 2}\right]}_0^2$

$=\left(2-2^2+\frac{2^3}{4}-\frac{2^{3-\frac{2}{2}}}{(\ln 2{)}^2}-\frac{2路2}{\ln 2}\right)-\left(0-0^2+\frac{0^3}{4}-\frac{2^{3路\frac{0}{2}}}{(\ln 2{)}^2}-\frac{2路0}{\ln 2}\right)$

$=\frac{4(1-\ln 2)}{(\ln 2{)}^2}$

$\text{Therefore, the required integral is}$

${鈭�}_s\mathbf{F}(x,y,z)路\mathbf{n}dS=\frac{4(1-\ln 2)}{(\ln 2{)}^2}$