Use the Fundamental Theorem of Line Integrals, if applicable, to evaluate the integrals in Exercises 37–44. Otherwise, show that the vector field is not conservative.

$\mathbf{F}(x,y,z)=y{z}^{xy}\ln z\mathbf{i}+x{z}^{xy}\ln z\mathbf{j}+\frac{z^{xy}}{z}\mathbf{k}$, with C any curve from $(0,0,1)\mathrm{to}(2,16,3)$.

Use the Fundamental Theorem of Line Integrals, if applicable, to evaluate the integrals in the given exercises. Otherwise, show that the vector field is not conservative.

$\mathbf{F}(x,y,z)=y{z}^{xy}\ln z\mathbf{i}+x{z}^{xy}\ln z\mathbf{j}+\frac{z^{xy}}{z}\mathbf{k}$, with C any curve from $(0,0,1)\mathrm{to}(2,16,3)$.

$$\begin{gathered} \frac{d\mathbf{F}(x,y,z)}{dy}=\frac{d}{dy}y{z}^{xy}\ln z\frac{d\mathbf{F}(x,y,z)}{dx}=\frac{d}{dx}x{z}^{xy}\mathrm{In}z \\ \frac{d\mathbf{F}(x,y,z)}{dy}=\ln z\left(y\frac{d}{dy}{z}^{xy}+z^{xy}\frac{d}{dy}y\right)\frac{d\mathbf{F}(x,y,z)}{dx}=\ln z\left(x\frac{d}{dx}{z}^{xy}+z^{xy}\frac{d}{dx}x\right) \\ \frac{d\mathbf{F}(x,y,z)}{dy}=\ln z\left(yx{z}^{xy}+z^{xy}·1\right)\frac{d\mathbf{F}(x,y,z)}{dx}=\ln z\left(xy{z}^{xy}+z^{xy}·1\right) \\ \frac{d\mathbf{F}(x,y,z)}{dy}=\ln z\left(yx{z}^{xy}+z^{xy}\right)\frac{d\mathbf{F}(x,y,z)}{dx}=\ln z\left(xy{z}^{xy}+z^{xy}\right) \end{gathered}$$

Since, $\frac{d\mathbf{F}(x,y,z)}{dy}=\frac{d\mathbf{F}(x,y,z)}{dx}$, so the given function is conservative.

The give points are: $(0,0,1)\mathrm{to}(2,16,3)$

We find a potential function for F:

$$\begin{gathered} f(x,y,z)=\int y{z}^{xy}\ln zdx \\ f(x,y,z)=y\ln z\int z^{xy}dx \\ \mathrm{Let}xy=t \\ ydx=dt \\ f(x,y,z)=\ln z\int z^{t}dt \\ f(x,y,z)=\ln z·z^{xy} \\ f(x,y,z)=z^{xy} \end{gathered}$$

$$\begin{gathered} {\int }_{\mathrm{C}}\mathbf{F}(x,y,z)·\mathrm{d}\mathbf{r}=3^{2·16}-1^{0·0} \\ {\int }_{\mathrm{C}}\mathbf{F}(x,y,z)·\mathrm{d}\mathbf{r}=3^{32}-1^0 \\ {\int }_{\mathrm{C}}\mathbf{F}(x,y,z)·\mathrm{d}\mathbf{r}=3^{32}-1 \end{gathered}$$