$\mathbf{\iint }\mathbf{\left(}\mathbf{curlV}\mathbf{\right)}\mathbf{\times }\mathbf{}\mathbf{ndo}\mathbf{}$over the part of the surface$\mathbf{z}\mathbf{=}\mathbf{9}\mathbf{-}{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{9}{\mathbf{y}}^{\mathbf{2}}$ above the $\mathbf{}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{}\mathbf{y}\mathbf{)}\mathbf{}$plane, if .$\mathbf{V}\mathbf{=}\mathbf{2}\mathbf{xyi}\mathbf{+}\left(x^2-2x\right)\mathbf{j}\mathbf{-}{\mathbf{x}}^{\mathbf{2}}{\mathbf{z}}^{\mathbf{2}}\mathbf{k}$

The given equation is$\mathrm{V}=2\mathrm{xyi}+({\mathrm{x}}^2-2\mathrm{x})\mathrm{j}-{\mathrm{x}}^2{\mathrm{z}}^2\mathrm{k}$

A quantity that has magnitude as well as direction is called a vector. It is typically denoted by an arrow in which the head determines the direction of the vector and the length determines its magnitude.

Calculate the expression $\iint (\text{curlV)×}\mathrm{nd\sigma }$for$\mathrm{V}=2\mathrm{xyi}+({\mathrm{x}}^2-2\mathrm{x})\mathrm{j}-{\mathrm{x}}^2{\mathrm{z}}^2\mathrm{k}$ over the surface $\mathrm{z}=9-{\mathrm{x}}^2-9{\mathrm{y}}^2$above the $\mathrm{x}-\mathrm{y}$plane(i.e., for $\mathrm{z}\ge 0$).This surface is bounded by the ellipse$9-{\mathrm{x}}^2-9{\mathrm{y}}^2=0$ which means that, using Stokes's theorem, the original integral can be calculated over any surface enclosed by this ellipse, which is taken to be the surface on the $\mathrm{x}-\mathrm{y}$plane (i.e., the ellipse itself). The $\mathrm{curl}\mathrm{V}$of is derived as shown below.

$\mathrm{curlV}=\left(\frac{\mathrm{\partial }{\mathrm{V}}_{\mathrm{z}}}{\mathrm{\partial }\mathrm{y}}-\frac{\mathrm{\partial }{\mathrm{V}}_{\mathrm{y}}}{\mathrm{\partial }\mathrm{z}}\right)\mathrm{i}+\left(\frac{\mathrm{\partial }{\mathrm{V}}_{\mathrm{x}}}{\mathrm{\partial }\mathrm{z}}-\frac{\mathrm{\partial }{\mathrm{V}}_{\mathrm{z}}}{\mathrm{\partial }\mathrm{x}}\right)\mathrm{j}+\left(\frac{\mathrm{\partial }{\mathrm{V}}_{\mathrm{y}}}{\mathrm{\partial }\mathrm{x}}-\frac{\mathrm{\partial }{\mathrm{V}}_{\mathrm{x}}}{\mathrm{\partial }\mathrm{y}}\right)\mathrm{k}$

$=(0-0)\mathrm{i}+\left(0-2{\mathrm{xz}}^2\right)\mathrm{j}+(2\mathrm{x}-2-2\mathrm{x})\mathrm{k}$

$=-2\mathrm{K}$

Therefore, the integral of $-2\mathrm{K}$over the area of an ellipse is $\frac{{\mathrm{x}}^2}{9}+{\mathrm{y}}^2=1$.

Its normal vector is and its area is which gives the solution shown below.

$\iint \left(\text{curlV}\right)\times \mathrm{nd\sigma }=\int (-2)\mathrm{k}\times \mathrm{kd\sigma }$

$=-2\left(3\mathrm{\pi }\right)$

$=-6\mathrm{\pi }$

Hence, the solution is $\iint \left(\text{curlV}\right)\times \text{nd}\mathrm{\sigma }=-6\mathrm{\pi }$.