Question

Search results for If v=(y²+z²-x²) î+ (z²+x²-y²) j^+(x²+y²-z²) k^, then find to it's divergence v and curl v? 

Answer 1

div $\overrightarrow{V}=\frac{\partial}{\partial\>x}(y^2+z^2-x^2)+\frac{\partial}{\partial\>y}(z^2+x^2-y^2)+\frac{\partial}{\partial\>z}(x^2+y^2-z^2)$

$=-2-2y-2z$

$=-2(x+y+z)$

$Curl{\overrightarrow{V}}$ .$\begin{vmatrix} i & j&k\\ \frac{\partial}{\partial\>x}&\frac{\partial}{\partial\>y}&\frac{\partial}{\partial\>z} \\ y^2+z^2-x^2&z^2+x^2-y^2&x^2+y^2-z^2 \end{vmatrix}$

$=(2y-2z)\utilde{i}+(2z-2x)\utilde{j}+(2x-2y)\utilde{k}$

$=2(y-z)\utilde{i}-2(x-z)\utilde{j}+2(x-y)\utilde{k}$