${\mathbf{\iiint }}^{\mathbf{\prime }}\mathbf{curlV}\mathbf{)}\mathbf{\times }\mathbf{\text{nd}}\mathbf{\sigma }$over any surface whose bounding curve is in the $\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{}\mathbf{y}\mathbf{)}$plane, where $\mathbf{V}\mathbf{=}\left(x-x^{2}z\right)\mathbf{i}\mathbf{+}\left({\mathrm{yz}}^3-y^2\right)\mathbf{j}\mathbf{+}\left(x^{2}y-\mathrm{xz}\right)\mathbf{k}$.

The given equation is$\mathrm{V}=\left(\mathrm{x}-{\mathrm{x}}^2\mathrm{z}\right)\mathrm{i}+\left({\mathrm{yz}}^3-{\mathrm{y}}^2\right)\mathrm{j}+\left({\mathrm{x}}^2\mathrm{y}-\mathrm{xz}\right)\mathrm{k}$

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.

In this problem calculate $\iint \left(\text{curlV}\right)\times \text{nd}\mathrm{\sigma }$where $\mathrm{V}=\left(\mathrm{x}-{\mathrm{x}}^2\mathrm{z}\right)\mathrm{i}+\left({\mathrm{yz}}^3-{\mathrm{y}}^2\right)\mathrm{j}+\left({\mathrm{x}}^2\mathrm{y}-\mathrm{xz}\right)\mathrm{k}$for any surface whose boundary lies on the $\mathrm{x}-\mathrm{y}$plane. First calculate the curl.

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

$\left({\mathrm{x}}^2-2{\mathrm{yz}}^2\right)\mathrm{i}+\left(-{\mathrm{x}}^2-2\mathrm{xy}+\mathrm{z}\right)\mathrm{j}+(0-0)\mathrm{z}$

Since any surface can be chosen whose boundary lays on the $\mathrm{x}-\mathrm{y}$plane, choose a flat surface that lays on the $\mathrm{x}-\mathrm{y}$plane entirely. Its normal vector is then always Since the curl of has no component, the dot product $(\mathrm{Curl}\mathrm{V})$is zero, which gives the equation $\iint \left(\text{curl V)}\right)\times \text{nd}\mathrm{\sigma }=0$.

Hence, the solution derived is $\iint \left(\text{curl V)}\right)\times \text{nd}\mathrm{\sigma }=0$.