Evaluate the line integrals:

${\int }_C\mathbf{F}(x,y,z)·d\mathbf{r}$where

$\mathbf{F}(x,y,z)=\left(yz{e}^{xyz}+2\right)\mathbf{i}+\left(xz{e}^{xyz}-1\right)\mathbf{j}+xy{e}^{xyz}\mathbf{k}$

$C$ is curve of intersection$z=\sqrt{x^2+y^2}$and the plane$z-x+y=10$

The given vector field is $\mathbf{F}(x,y,z)=\left(yz{e}^{xyz}+2\right)\mathbf{i}+\left(xz{e}^{xyz}-1\right)\mathbf{j}+xy{e}^{xyz}\mathbf{k}$

Equation of surface is$z=\sqrt{x^2+y^2}$and that of plane is$z-x+y=10$.

If curve $C$is graph of vector function $r\left(t\right)$with $P=\mathbf{r}\left(a\right)\text{and}Q=\mathbf{r}\left(b\right)$

and $F$is a conservative field with $\mathbf{F}=\nabla f$on an open, simply connected and connected domain having curve, then

${\int }_C\mathbf{F}·d\mathbf{r}=f\left(Q\right)-f\left(P\right)$

Vector field is $\mathbf{F}(x,y,z)=\left(yz{e}^{xyz}+2\right)\mathbf{i}+\left(xz{e}^{xyz}-1\right)\mathbf{j}+xy{e}^{xyz}\mathbf{k}$

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 conservative if

$\frac{\partial F_1}{\partial y}=\frac{\partial F_2}{\partial x},\frac{\partial F_1}{\partial z}=\frac{\partial F_3}{\partial x},\frac{\partial F_3}{\partial y}=\frac{\partial F_2}{\partial z}$

For given vector field

$F_1(x,y,z)=yz{e}^{xyz}+2$

$\frac{\partial F_1}{\partial y}=\frac{\partial }{\partial y}\left(yz{e}^{xyz}+2\right)=z{e}^{xyz}(1+xyz)$

$\frac{\partial F_1}{\partial z}=\frac{\partial }{\partial z}\left(yz{e}^{xyz}+2\right)=y{e}^{xyz}(1+xyz)$

$F_2(x,y,z)=xz{e}^{xyz}-1$

$\frac{\partial F_2}{\partial x}=\frac{\partial }{\partial x}\left(xz{e}^{xzz}-1\right)=z{e}^{xyz}(1+xyz)$

$\frac{\partial F_2}{\partial z}=\frac{\partial }{\partial z}\left(xz{e}^{xyz}-1\right)=x{e}^{xyz}(1+xyz)$

$F_3(x,y,z)=xy{e}^{xyz}$

$\frac{\partial F_3}{\partial x}=\frac{\partial }{\partial x}\left(xy{e}^{xz}\right)=y{e}^{xyz}(1+xyz)$

role="math" localid="1653737386103" $\frac{\partial F_3}{\partial y}=\frac{\partial }{\partial y}\left(xy{e}^{xyz}\right)=x{e}^{xyz}(1+xyz)$

Hence,

$\frac{\partial F_1}{\partial y}=\frac{\partial F_2}{\partial x},\frac{\partial F_1}{\partial z}=\frac{\partial F_3}{\partial x},\frac{\partial F_3}{\partial y}=\frac{\partial F_2}{\partial z}$

Vector Field is conservative.

As $F$is conservative, there exists a potential function $f$such that $\Delta f=\mathbf{F}$

$\Rightarrow \frac{\partial f}{\partial x}=yz{e}^{xyz}+2$

$\frac{\partial f}{\partial y}=xz{e}^{xz}-1$

$\frac{\partial f}{\partial z}=xy{e}^{xyz}$

Integrating wrt $x,y,z$

$f(x,y,z)=\int \frac{\partial f}{\partial x}dx+A(y,z)$

$=\int \left(yz{e}^{xyz}+2\right)dx+A(y,z)$

$=e^{xyz}+2x+\alpha +A(y,z)$

$f(x,y,z)=\int \frac{\partial f}{\partial y}dy+B(x,z)$

$=\int \left(xz{e}^{xyz}-1\right)dy+B(x,z)$

$=e^{xyz}-y+\beta +B(x,z)$

and

$f(x,y,z)=\int \frac{\partial f}{\partial z}dz+C(x,y)$

$=\int \left(xy{e}^{xyz}\right)dz+C(x,y)$

$=e^{xyz}+y+C(x,y)$

It is concluded that

$f(x,y,z)=e^{xyz}+2x-y$

Hence, the function is potential function for$F$.

The intersection of surface and curve is an ellipse, so it is a closed curve.

Here

$P=Q$

$\Rightarrow f\left(P\right)=f\left(Q\right)$

or $f\left(P\right)-f\left(Q\right)=0$

Using Fundamental Theorem of Line Integral to evaluate integral, ${\int }_C\mathbf{F}·d\mathbf{r}$as follows,

${\int }_C\mathbf{F}(x,y,z)·d\mathbf{r}=f\left(Q\right)-f\left(P\right)=0$

Hence, required integral is zero.