Step 1

Consider the surface in xy-plane as z = g(x,y).

Step 2

Obtain the value of \(\displaystyle{z}_{{x}}{\quad\text{and}\quad}{z}_{{y}}{a}{s}{z}_{{x}}={g}_{{x}}{\quad\text{and}\quad}{z}_{{y}}={g}_{{y}}\) respectively

Thus, the parametric form becomes \(\displaystyle{\left\langle{z}_{{x}},{z}_{{y}},{1}\right\rangle}={\left\langle{g}_{{x}},{g}_{{y}},{1}\right\rangle}\)

If the vector product of two vectors is obtained, the direction of the vector product is perpendicular to both the vectors.

The components \(\displaystyle{z}_{{x}},{\quad\text{and}\quad}{z}_{{y}}\) lie on the xy- plane, hence the vector product \(\displaystyle{z}_{{x}}\times{z}_{{y}}\), lie perpendicular to the xy- plane.

Therefore, the normal to the surface is the same as that of xy- plane.

Here, surface is just a plane.

Hence, there is only one normal vector at every point of the plane.

Thus, the normal vector in Stokes’ theorem becomes the unit vector that results in circulation form of Green’s theorem.

Consider the surface in xy-plane as z = g(x,y).

Step 2

Obtain the value of \(\displaystyle{z}_{{x}}{\quad\text{and}\quad}{z}_{{y}}{a}{s}{z}_{{x}}={g}_{{x}}{\quad\text{and}\quad}{z}_{{y}}={g}_{{y}}\) respectively

Thus, the parametric form becomes \(\displaystyle{\left\langle{z}_{{x}},{z}_{{y}},{1}\right\rangle}={\left\langle{g}_{{x}},{g}_{{y}},{1}\right\rangle}\)

If the vector product of two vectors is obtained, the direction of the vector product is perpendicular to both the vectors.

The components \(\displaystyle{z}_{{x}},{\quad\text{and}\quad}{z}_{{y}}\) lie on the xy- plane, hence the vector product \(\displaystyle{z}_{{x}}\times{z}_{{y}}\), lie perpendicular to the xy- plane.

Therefore, the normal to the surface is the same as that of xy- plane.

Here, surface is just a plane.

Hence, there is only one normal vector at every point of the plane.

Thus, the normal vector in Stokes’ theorem becomes the unit vector that results in circulation form of Green’s theorem.