${\iint }_{S}curl\mathbf{F}(x,y,z)·\mathbf{n}d\mathbf{S}$, where S is the portion of the hyperbolic paraboloid $z=x^2-y^2$ that lies inside the elliptical cylinder $4x^2+9y^2=36$ with upwards-pointing normal vector and$\mathbf{F}(x,y,z)=(1-yz\sin (xyz\left)\right)\mathbf{i}-$ $(1+xz\sin (xyz\left)\right)\mathbf{j}+(1-xy\sin (xyz\left)\right)\mathbf{k}$.

Take a look at the vector field below.:

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

The goal is to assess the integral.${\iint }_{S}curl\mathbf{F}(x,y,z)·\mathbf{n}dS$, where the surface Siis defined as:

The surface S is the hyperbolic paraboloid's portion.$z=x^2-y^2$that lies inside the cylinder $4x^2+9y^2=36$with upwards-pointing normal vector.

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

A vector field's curl$\mathbf{F}(x,y,z)=F_1(x,y,z)\mathbf{i}+F_2(x,y,z)\mathbf{j}+F_3(x,y,z)\mathbf{k}$is defined as follows:

$\mathrm{curl}\mathbf{F}(x,y,z)=\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ \frac{\mathrm{\partial }}{\mathrm{\partial }x}& \frac{\mathrm{\partial }}{\mathrm{\partial }y}& \frac{\mathrm{\partial }}{\mathrm{\partial }z}\\ F_1(x,y,z)& F_2(x,y,z)& F_3(x,y,z)\end{array}\right|$

$=\left[\frac{\mathrm{\partial }{F}_3}{\mathrm{\partial }y}-\frac{\mathrm{\partial }{F}_2}{\mathrm{\partial }z}\right]\mathbf{i}-\left[\frac{\mathrm{\partial }{F}_3}{\mathrm{\partial }x}-\frac{\mathrm{\partial }{F}_1}{\mathrm{\partial }z}\right]\mathbf{j}+\left[\frac{\mathrm{\partial }{F}_2}{\mathrm{\partial }x}-\frac{\mathrm{\partial }{F}_1}{\mathrm{\partial }y}\right]\mathbf{k}$

]]The vector field's curl is then applied. $\mathbf{F}(x,y,z)$will be,

$\mathrm{curl}\mathbf{F}(x,y,z)=\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ \frac{\mathrm{\partial }}{\mathrm{\partial }x}& \frac{\mathrm{\partial }}{\mathrm{\partial }y}& \frac{\mathrm{\partial }}{\mathrm{\partial }z}\\ (1-yz\sin (xyz\left)\right)& -(1+xz\sin (xyz\left)\right)& (1-xy\sin (xyz\left)\right)\end{array}\right|$

$=\left[\frac{\mathrm{\partial }}{\mathrm{\partial }y}(1-xy\sin (xyz\left)\right)-\frac{\mathrm{\partial }}{\mathrm{\partial }z}(-(1+xz\sin \left(xyz\right)\left)\right)\right]\mathbf{i}$

$-\left[\frac{\mathrm{\partial }}{\mathrm{\partial }x}(1-xy\sin (xyz\left)\right)-\frac{\mathrm{\partial }}{\mathrm{\partial }z}(1-yz\sin (xyz\left)\right)\right]\mathbf{j}$

$+\left[\frac{\mathrm{\partial }}{\mathrm{\partial }x}(-(1+xz\sin \left(xyz\right)\left)\right)-\frac{\mathrm{\partial }}{\mathrm{\partial }y}(1-yz\sin (xyz\left)\right)\right]\mathbf{k}$

$$\begin{gathered} 0i+oj+ok \\ =0 \end{gathered}$$

Then, the value of curl$\mathbf{F}(x,y,z)·\mathbf{n}$will be,

$$\begin{gathered} \mathbf{F}(x,y,z)\cdot \mathbf{n}=\mathbf{0}\cdot \mathbf{n} \\ \mathbf{=}\mathbf{0} \end{gathered}$$

As a result, the required integral is

$$\begin{gathered} {\iint }_S \mathrm{curl}\mathbf{F}(x,y,z)\cdot \mathbf{n}dS={\iint }_S 0dS \\ =0 \end{gathered}$$

As a result, the required integral is${\iint }_{S}curl\mathbf{F}(x,y,z)·\mathbf{n}dS=0$.