1. ## Conservative field

I have one problem to prove the field is conservative
$\displaystyle F(x,y,z) = yzi + xzj + yzk$

But for me this field not is conservative.
I am mistaken?

2. It seems like we're doing similar things! Do you have exams in May as well?

For this I would use the Component test for Conservative fields.

let $\displaystyle M(x,y,z)=yz$,$\displaystyle N(x,y,z)=xz$ and $\displaystyle P(x,y,z)=yz$.

For a field to be conservative $\displaystyle \frac{\partial P}{\partial y}=\frac{\partial N}{\partial z}$, $\displaystyle \frac{\partial M}{\partial z}=\frac{\partial P}{ \partial x}$ and $\displaystyle \frac{\partial N}{ \partial x}=\frac{\partial M}{\partial y}$.

But $\displaystyle \frac{\partial P}{\partial y}=z$ and $\displaystyle \frac{\partial N}{\partial z}=x$.

But we also know that $\displaystyle x=z$ since $\displaystyle yz=yz$.

Hence $\displaystyle \boxed{\frac{\partial P}{\partial y}=\frac{\partial N}{\partial z}}$

Try it for the others and see what you get. For me, it doesn't appear that $\displaystyle \frac{\partial M}{\partial z}=\frac{\partial P}{ \partial x}$ holds.....

3. Originally Posted by Showcase_22
It seems like we're doing similar things! Do you have exams in May as well?

For this I would use the Component test for Conservative fields.

let $\displaystyle M(x,y,z)=yz$,$\displaystyle N(x,y,z)=xz$ and $\displaystyle P(x,y,z)=yz$.

For a field to be conservative $\displaystyle \frac{\partial P}{\partial y}=\frac{\partial N}{\partial z}$, $\displaystyle \frac{\partial M}{\partial z}=\frac{\partial P}{ \partial x}$ and $\displaystyle \frac{\partial N}{ \partial x}=\frac{\partial M}{\partial y}$.

But $\displaystyle \frac{\partial P}{\partial y}=z$ and $\displaystyle \frac{\partial N}{\partial z}=x$.

But we also know that $\displaystyle x=z$ since $\displaystyle yz=yz$.

Hence $\displaystyle \boxed{\frac{\partial P}{\partial y}=\frac{\partial N}{\partial z}}$

Try it for the others and see what you get. For me, it doesn't appear that $\displaystyle \frac{\partial M}{\partial z}=\frac{\partial P}{ \partial x}$ holds.....
I have exams in end april.

I solve the rotational and value is different 0 hence i think that not is conservative

4. I don't think it is either.

I've had the opportunity to check $\displaystyle \frac{\partial M}{\partial z}=\frac{\partial P}{ \partial x} \Rightarrow \ y=0$ which isn't always true.

5. Originally Posted by Showcase_22
I don't think it is either.

I've had the opportunity to check $\displaystyle \frac{\partial M}{\partial z}=\frac{\partial P}{ \partial x} \Rightarrow \ y=0$ which isn't always true.
$\displaystyle rot = (\frac{ \partial (yz)}{ \partial y} - \frac{ \partial (xz)}{ \partial z})i + (\frac{ \partial (yz)}{ \partial z} - \frac{ \partial (yz)}{ \partial x})j + (\frac{ \partial (xz)}{ \partial x} - \frac{ \partial (yz)}{ \partial y})k$

$\displaystyle rot = (z-x)i + (y-0)j + (0)k$ is different 0

6. exactly!

How did you prove it was not conservative? I'm new to this stuff myself and would love to see how other people do these types of questions.