Question

If the velocity vector in a two dimensional flow field is given by $\overrightarrow{v}=2xy\hat{i}+(2y^2-x^2)\hat{j}$, the vorticity vector, curl $\overrightarrow{v}$ will be

• A. $2y^2\hat{j}$
• B. $6y\hat{k}$
• C. zero
• D. $-4x\hat{k}$

Answer 1

The correct option is D $-4x\hat{k}$

$\overrightarrow{V}=2xy\hat{i}+(2y^2-x^2)\hat{j}$
$Curl\overrightarrow{V}=▽\times V$
$=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ 2xy& 2y^2-x^2& 0\end{array}\right|$
$=\left[\frac{\partial }{\partial y}\right(0)-\frac{\partial }{\partial z}(2y^2-x^2)]\hat{i}$
$+\left[\frac{\partial }{\partial z}\right(2xy)-\frac{\partial }{\partial x}(0\left)\right]\hat{j}$
$+[\frac{\partial }{\partial x}(2y^2-x^2)-\frac{\partial }{\partial y}2xy]\hat{k}$
$=0\hat{i}+\hat{j}+(-2x-2x)\hat{k}$
$curl\overrightarrow{V}=-4x\hat{k}$