Calculate the div and rot:
Correct?
The divergence is correct
Not sure what you have in mind for rotF
but it is 5z^2 i +3xz^2 j +4 y^4x k
The middle row of the determinant are the operators d/dx d/dy and d/dz
and the last row are the x , y , and z components of F not the parial derivatives
The symbol $\displaystyle \bigtriangledown$ means divergent ?
If I have: $\displaystyle \bigtriangledown . ( \bigtriangledown x F)$ and have F how can I calculate ?
How can I verify that the vector projection $\displaystyle r$ have the property:
1) $\displaystyle \bigtriangledown ||r|| = \frac{r}{||r||}$
is an operator and means d/dx i +d/dy j +d/dz k where d/dx d/dy d/dz ar the partial derivative operators
f is the gradient where f is a scalar valued function
X F is the curl or rotation where F is a vector field
*F is the divergence where F is a vector field * means dot product
From the previous identity * ( x F) = 0
r= xi +yj +zk
||r|| = (x^2 +y^2 +z^2)^(1/2)
Compute ||r|| and the result follows directly
Thanks, you're helping me a lot
check my learning:$\displaystyle r = xi + yj + zk$
1) $\displaystyle \bigtriangledown . (FxG)$
$\displaystyle F(x,y,z) = yzi + xzj + xyk$
$\displaystyle G(x,y,z) = xyj + xyzk$
My solution:
$\displaystyle FxG = (x^2yz^2 - x^2y^2)i + (-xy^2z^2)j + (xy^2z)k$
$\displaystyle div = (2xyz^2 - 2xy^2)i + (-2xyz^2)j + (xy^2)k$
2) $\displaystyle \bigtriangledown . (\bigtriangledown x F)$
$\displaystyle F(x,y,z) = sen(x)i + cos(x-y)j + zk$
My solution:
$\displaystyle rotF = 0i - 0j + (-sen(x-y))k$
$\displaystyle div(rotF) = 0i + 0j + 0k$
3) Check that: $\displaystyle \bigtriangledown ||r|| = \frac{r}{||r||}$
My solution:
$\displaystyle div||r|| = \frac{x}{\sqrt{x^2+y^2+z^2}} + \frac{y}{\sqrt{x^2+y^2+z^2}} + \frac{z}{\sqrt{x^2+y^2+z^2}}$
$\displaystyle \bigtriangledown ||r|| = \frac{r}{||r||}$ is correct