1. ## Conservative Fields

Hello again, I have no idea about how to solve this, could you help me? Thank you so much!

Let $x\in{\mathbb{R}^3}$ and $r= \left\|{x}\right\|$. Consider the field $f(x)=r^px$ where p is a real constant.

(a) Specify on which regions of $\mathbb{R}^3$ this field is conservative.
(b) On those regions where the field is conservative, find the potential of f. (Consider separately the case when p = 2).

Thanks!!

2. What kind of norm on R 3 are we using the standard norm? Also, is that just x time r^p or is that an x vector?

3. The original post said " $x\in R^3$ so x must be a vector.

Writing $\vec{x}= x\vec{i}+ y\vec{j}+ z\vec{k}$, with the standard norm, $f(\vec{x})= r^p\vec{x}= (x^2+ y^2+ z^2)^{p/2}x\vec{i}+ y\vec{j}+ z\vec{k}$.

To show that this is a "conservative field" you must show that
$\frac{\partial }{\partial x}((x^2+ y^2+ z^2)^{p/2}y)=$ $\frac{\partial}{\partial y}((x^2+ y^2+ z^2)^{p/2}x)$

$\frac{\partial }{\partial x}((x^2+ y^2+ z^2)^{p/2}z)= \frac{\partial}{\partial z}((x^2+ y^2+ z^2)^{p/2}x)$

$\frac{\partial }{\partial y}((x^2+ y^2+ z^2)^{p/2}z)= \frac{\partial}{\partial z}((x^2+ y^2+ z^2)^{p/2}y)$

4. Hello, I don't understand why I have to calcule those partials derivatives. I mean, I thought I had to calculte this derivatives:

$\dfrac{{\partial f(\overrightarrow{x})}}{{\partial x}}$
$\dfrac{{\partial f(\overrightarrow{x})}}{{\partial y}}$
$\dfrac{{\partial f(\overrightarrow{x})}}{{\partial z}}$

And show that they are the same ??

I don't know, thank you so much for your help

5. If you are familiar with Green's Theorem, it should make some sense on why you would calculate those partial derivatives.

With a conservative field, all the partials are equal because when you run the integral, you will be evaluating 0.

For example, you would have dm/dy=dn/dx in the integral you would then do dm/dy-dn/dx which if they are equal you obtain zero and have a conservative field.

6. I don't know Green's Theorem.

I think I have dificulties with the condition for a field to be conservative.

If $\dfrac{{\partial F_k}}{{\partial x_i}}=\dfrac{{\partial F_i}}{{\partial x_k}}$ the the field F is conservative.

So in this case I can say that $F:\mathbb{R}^3\rightarrow{R}^3$ with

$F_1=(x^2+y^2+z^2)^{p/2}x$
$F_2=(x^2+y^2+z^2)^{p/2}y$
$F_3=(x^2+y^2+z^2)^{p/2}z$

and $F(x,y,z)=(F_1,F_2,F_3)$

And that way what I have to prove is what you told me, is way of thinking correct?

But this partial derivatives are always the same, right? So the field is conservative for all (x,y,z) in R^3?

PS: I'm sorry if my english is not very good, I'm from Venezuela