Evaluate the integrals. In some cases, Stokes’ Theorem will help; in other cases, it may be preferable to evaluate the integrals directly.

${\int }_C \mathbf{F}(x,y,z)\times d{\mathbf{r}}_{}$, where $C$is the boundary of the triangle in the plane $y=2$and with vertices $(1,0,2),$$(0,2,0),$and $(0,2,1)$, with the normal vector pointing in the positive y direction and$\mathbf{F}(x,y,z)=3yz\mathbf{i}+e^x\mathbf{j}+x^{2}z\mathbf{k}$

Consider the next vector field:

$\mathbf{F}(x,y,z)=3yz\mathbf{i}+e^x\mathbf{j}+x^{2}z\mathbf{k}$

The objective is to gouge the road integral ${\int }_C \mathbf{F}(x,y,z)\times d\mathbf{r}\mathbf{,}$where the curve $C$is defined as follows:

The curve $C$is that the boundary of Triangulum within the plane $y=2$and with vertices $(1,0,2)$,$(0,2,0)$and $(0,2,1)$, with the normal vector pointing within the positive $y$direction.

The plane $y=2$is parallel to the $xz$-plane, so a unit normal vector is$j$.
Then, the value of $\mathrm{curl}\mathbf{F}(x,y,z)\times \mathbf{n}$ are,

$\mathrm{curl}\mathbf{F}(x,y,z)\times \mathbf{n}=\left(0\mathbf{i}-(2xz-3y)\mathbf{j}+\left(e^x-3z\right)\mathbf{k}\right)\times \mathbf{j}$

$=-(2xz-3y)$

$=-2xz+3y$

Because therefore the worth of$y=2$, $\mathrm{curl}\mathbf{F}(x,y,z)\cdot \mathbf{n}$are visiting be,

$\begin{array}{r}\mathrm{curl}\mathbf{F}(x,y,z)\times \mathbf{n}=-2xz+3y\end{array}$

$\begin{array}{r}=-2xz+3\times 2\\ \end{array}$

$=-2xz+6$

The surface $S$is bounded by $C$, where $C$ is that constellation within the plane $y=2$and with vertices $(1,2,0),$$(0,2,0),$and $(0,2,1).$

Hence, the region of integration $D$ is the trianglewith vertices $(1,0),$$(0,0),$and $(0,1)$ within the $xz$plane as shown in following figure:

$\begin{array}{r}{\int }_C \mathbf{F}(x,y,z)\times d\mathbf{r}={\iint }_S \mathrm{curl}\mathbf{F}(x,y,z)\times \mathbf{n}dS\end{array}$

$\begin{array}{r}={\iint }_D (-2xz+6)dA\end{array}$

$\begin{array}{r}={\int }_0^1 {\int }_0^{1-x} (-2xz+6)dzdx\\ \end{array}$

$={\int }_0^1 \left[{\int }_0^{1-x} (-2xz+6)dz\right]dx$

$\begin{array}{r}={\int }_0^1 {\left[-x{z}^2+6z\right]}_0^{1-x}dx\end{array}$

$\begin{array}{r}={\int }_0^1 \left[\left(-x(1-x{)}^2+6(1-x)\right)-\left(-x\cdot 0^2+6\cdot 0\right)\right]dx\\ \end{array}$

$\begin{array}{r}={\int }_0^1 \left[\left(-x\left(1-2x+x^2\right)+6-6x\right)-0\right]dx\\ \end{array}$

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

$\begin{array}{r}={\int }_0^1 \left(-x^3+2x^2-7x+6\right)dx\end{array}$

$\begin{array}{r}={\left[-\frac{x^4}{4}+\frac{2x^3}{3}-\frac{7x^2}{2}+6x\right]}_0^1\end{array}$

$\begin{array}{r}=\left(-\frac{1^4}{4}+\frac{2\times 1^3}{3}-\frac{7\times 1^2}{2}+6\times 1\right)-\left(-\frac{0^4}{4}+\frac{2\times 0^3}{3}-\frac{7\times 0^2}{2}+6\times 0\right)\\ \end{array}$

$=\left(-\frac{1}{4}+\frac{2}{3}-\frac{7}{2}+6\right)-0$

$=\frac{35}{12}.$

Therefore,the desired integral is ${\int }_C \mathrm{F}(x,y,z)\times d\mathbf{r}=\frac{35}{12}$