91Ó°ÊÓ

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 $âˆ\neg \left(\text{curlV}\right)×\text{nd}\mathrm{σ}$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.

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

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

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 $âˆ\neg \left(\text{curl V)}\right)×\text{nd}\mathrm{σ}=0$.

Hence, the solution derived is $âˆ\neg \left(\text{curl V)}\right)×\text{nd}\mathrm{σ}=0$.