$\displaystyle F(x,y,z) = \frac{x}{x^2+y^2+z^2}i + \frac{y}{x^2+y^2+z^2}j+\frac{z}{x^2+y^2+z^2}k$
(a) Find curl
(b) Find the divergence
Apply the definitions!
(a) $\displaystyle \nabla\times\mathbf{F}=\begin{vmatrix}\mathbf{i} & \mathbf{j}&\mathbf{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z}\\ \frac{x}{x^2+y^2+z^2} & \frac{y}{x^2+y^2+z^2} & \frac{z}{x^2+y^2+z^2}\end{vmatrix}=\dots$
(b) $\displaystyle \nabla\cdot\mathbf{F}=\frac{\partial}{\partial x}\left[\frac{x}{x^2+y^2+z^2}\right]+\frac{\partial}{\partial y}\left[\frac{y}{x^2+y^2+z^2}\right]+\frac{\partial}{\partial z}\left[\frac{z}{x^2+y^2+z^2}\right]=\dots$
Can you finish both of these problems?
To find the derivative we will use the quotient rule correct? So,
$\displaystyle
\frac{\partial}{\partial x}= (x^2+y^2+z^2)-2x^2\\\\
\frac{\partial}{\partial y}= (x^2+y^2+z^2)-2y^2\\\\
\frac{\partial}{\partial x}= (x^2+y^2+z^2)-2z^2\\\\\
$
Then i take the derivative and do the cross product against the original functions.
Ok, i figured out the curl, know i need help with finding the divergence. When i used the formula to calculate the divergence, this is what i did.
$\displaystyle
\frac{x^2+y^2+z^2-2x^2}{(x^2+y^2+z^2)^2} + \frac{x^2+y^2+z^2)-(2y^2)}{(x^2+y^2+z^2)^2} +\frac{(x^2+y^2+z^2-2z^2)}{(x^2+y^2+z^2)^2}
$
Is this correct using the quotient rule? I then simplified the answer but it was still counted as wrong.
The work is correct. Note that
$\displaystyle \frac{x^2+y^2+z^2-2x^2}{(x^2+y^2+z^2)^2} + \frac{x^2+y^2+z^2-2y^2}{(x^2+y^2+z^2)^2} +\frac{x^2+y^2+z^2-2z^2}{(x^2+y^2+z^2)^2}$ $\displaystyle =\frac{3(x^2+y^2+z^2)-2(x^2+y^2+z^2)}{(x^2+y^2+z^2)^2}=\frac{x^2+y^2+z^2 }{(x^2+y^2+z^2)^2}=\frac{1}{x^2+y^2+z^2}$.
Does this make sense?