Question

Will award medal!!!!
find a) the curl and b) the divergence of the vector field.

Answer 1

\[F(x,y,z,)=\frac{ 1 }{ \sqrt(x^2+y^2+z^2 )} (x i+yj+zk)\]

Answer 2

@iambatman

Answer 3

we can solve this using a determinant

Answer 4

can u guys help me plz http://openstudy.com/study#/updates/55366a0ee4b0634ee6726953

Answer 5

$$
\Large {\\ {\bf F}(x,y,z,)=\frac{(x {\bf i}+y{\bf j}+z{\bf k})}{ \sqrt{x^2+y^2+z^2 }}
\\~\\

\\
curl {~\bf F} =\nabla \times {~\bf F}
\\= \begin{vmatrix} {\bf i} & {\bf j}& {\bf k}\\ \frac{\partial }{\partial x } &\frac{\partial }{\partial y } & \frac{\partial }{\partial z }
\\\frac{x }{ \sqrt{x^2+y^2+z^2 }} & \frac{y }{ \sqrt{x^2+y^2+z^2 }} &\frac{z }{ \sqrt{x^2+y^2+z^2 }}



\end{vmatrix}

}$$

Answer 6

\[\vec F (x,y,z) = \frac{1}{\sqrt{\rho}}<x,y,z>; \ \ \rho = x^2 + y^2 + z^2\]

CURL
F is conservative if there exists scalar function V such that \[\vec F = \vec{\nabla} V\]
\[\vec{\nabla} (\rho^{\frac{1}{2}}) = <\frac{1}{2} \rho^{-\frac{1}{2}}(2x),\frac{1}{2} \rho^{-\frac{1}{2}}(2y),\frac{1}{2} \rho^{-\frac{1}{2}}(2z)>\]
\[\vec{F} = \vec{\nabla} (\rho^{\frac{1}{2}}) = \frac{1}{\sqrt{\rho}}<x,y,z>\]
F is conservative so curl F = 0.

DIV
\[\nabla \bullet \vec{F} = \frac{∂}{∂x}\frac{x}{\sqrt{\rho}} + \frac{∂}{∂y}\frac{y}{\sqrt{\rho}} + \frac{∂}{∂z}\frac{z}{\sqrt{\rho}}\]
\[\frac{∂}{∂x}\frac{x}{\sqrt{\rho}} = \frac{\sqrt{\rho} - x\frac{x}{\sqrt{\rho}} }{\rho} = \frac{1}{\sqrt{\rho}} - \frac{x^2}{\rho^{\frac{3}{2}}}\]
\[\vec{\nabla} \bullet \vec{F} = \frac{3}{\sqrt{\rho}} - \frac{x^2 + y^2 + z^2}{\rho^\frac{3}{2}} = \frac{2}{\sqrt{x^2 + y^2 + z^2}}\]