Question

Find the value of curl F.nds where F=(y^2+z^2-x^2)i+(z^2+x^2-y^2)j+(x^2+y^2-z^2)k

Answer 1

Solution:

The vector F= (y^2+z^2-x^2)i+(z^2+x^2-y^2)j+(x^2+y^2-z^2)k

If , F= A i + B j + Z k

Here, $A=y^2+z^2-x^2, , B=z^2+x^2-y^2, , C=x^2+y^2-z^2,$

Curl (F)=$\bigtriangledown \times \vec{F}$

$\bigtriangledown=\frac{\partial}{\partial x} \times\vec{i}+ \frac{\partial}{\partial y} \times\vec{j} + \frac{\partial}{\partial z}  \times\vec{k}$

$\begin{vmatrix} i  &j  &k \\  \frac{\partial }{\partial x}&\frac{\partial }{\partial y}  &\frac{\partial }{\partial z} \\ y^2+z^2-x^2 &z^2+x^2-y^2  &x^2+y^2-z^2 \end{vmatrix}\\\\= 2 y (\vec{i})-2 z(\vec{i})-2 z(\vec{j})+2 x(\vec{j})+2 x(\vec{k})-2 y(\vec{k})\\\\= \vec{i}(2y-2z)+\vec{j}(2x-2z)+\vec{k}(2x-2y)$