Question

Curl of vector $overrightarrowF=x^2{z}^2\hat{i}-2x{y}^{2}z\hat{j}+2y^2{z}^3\hat{k}$

• A. $(4y{z}^3+2x{y}^2)\hat{i}+2x^{2}z\hat{j}-2y^{2}z\hat{k}$
• B. $(4y{z}^3+2x{y}^2)\hat{i}-2x^{2}z\hat{j}-2y^{2}z\hat{k}$
• C. $2x{z}^2\hat{i}-4xyz\hat{j}+6y^2{z}^2\hat{k}$
• D. $2x{z}^2\hat{i}+4xyz\hat{j}+6y^2{z}^2\hat{k}$

Answer 1

The correct option is A $(4y{z}^3+2x{y}^2)\hat{i}+2x^{2}z\hat{j}-2y^{2}z\hat{k}$

$curl\overrightarrow{F}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ x^2{z}^2& -2x{y}^z& 2y^2{z}^3\end{array}\right|$
$=\left[\frac{\partial }{\partial y}\right(2y^2{z}^3)-\frac{\partial }{\partial z}(-2x{y}^{2}z\left)\right]\hat{i}$
$+\left[\frac{\partial }{\partial z}\right(x^2{z}^2)-\frac{\partial }{\partial x}(2y^2{z}^3\left)\right]\hat{j}$
$+\left[\frac{\partial }{\partial x}\right(-2x{y}^{2}z)-\frac{\partial }{\partial y}(x^2{z}^2\left)\right]\hat{k}$
$curl\overrightarrow{F}=(4y{z}^3+2x{y}^2)\hat{i}+\left(2x^{2}z\right)\hat{j}-\left(2y^{2}z\right)\hat{k}$