vector calculus/ conservative field

[mathstresser]

Determine whether or not the vector field is conservative.
If it is conservative, find a function f such that F = del f.

. . .\(\displaystyle \L F(x,y,z)\, =\, 3z^2\hat{i}\, +\, cosy\hat{j}\, +\, 2xz\hat{k}\)

Is that conservative? (I think it is.)

How do I find a function f such that F = del f? I have some work, but I might be completely wrong, and it might confuse you...

With the matrix, I get:

. . .\(\displaystyle \L (2xz\, d/dy\, -\, cosy d/dz)\hat{i}\,\)
. . . . .\(\displaystyle \L -\, (2xz\, d/dx\, -\, 3z^2\, d/dz)\hat{j}\,\)
. . . . . . .\(\displaystyle \L +\, (cosy\, d/dx\, -\, 3z^2\, d/dy)\hat{k}\)

. . .\(\displaystyle \L 0\hat{i}i\, -\, (2z\, -\, 6z)\, +\, 0\, =\, 4z\)

I do the partial derivatives and get:

. . .\(\displaystyle \L fx\, =\, 3z^2\)
. . .\(\displaystyle \L fy\, =\, cosy\)
. . .\(\displaystyle \L fz\, =\, xy\)

But the example I was following then gets 0 for gy(y, z). So, I can't follow the example anymore.
___________________________
Edited by stapel -- Reason for edit: formatting

 

[galactus]

You ought to preview your post before you hit submit.

You can't put [b][/b] inside [tex][/tex]. fractions are made by 

\frac{a}{b}

 

[pka]

\(\displaystyle F(x,y,z) = 3z^2 i + \cos (y)j + 2xzk\) is not conservative.

Because \(\displaystyle \L \frac{{\delta (2xz)}}{{\partial x}} \not= \frac{{\delta (3z^2 )}}{{\partial z}}.\)