Potential Function for a gradient field.

[Ted123]

Homework Statement

[PLAIN]http://img576.imageshack.us/img576/4968/vec0.jpg

The Attempt at a Solution

(i) is not irrotational and (ii) is - I wish it was the other way round!

Can anyone help my construct a potential function $\phi (x,y,z)$ for (ii)?

 

[Dick]

It's a lot easier than it looks. Integrate F1 dx to get started. You'll find the part of the potential function you get explains the three hardest parts in that expression. Then try F2 dy. The answer is built to look intimidating.

 

[HallsofIvy]

To integrate
$$\int \frac{yz dx}{\sqrt{x^2y^2z^2+ 1}}$$
let $xyz= tan(\theta)$. Then $yzdx= sec^2(\theta)$, $\sqrt{x^2y^2z^2+ 1}= \sqrt{tan^2(\theta)+ 1}= sec(\theta)$ and the integral becomes
$$\int sec(\theta)d\theta)= \int \frac{d\theta}{cos(\theta)}$$
$$= \int \frac{cos(\theta)d\theta)}{cos^2(\theta)}= \int \frac{cos(\theta)d\theta}{1- sin^2(\theta)}$$
and now you can let $u= sin(\theta)$. Don't forget that the "constant of integration" may be a function of y and z.

 

[Ted123]

To integrate
$$\int \frac{yz dx}{\sqrt{x^2y^2z^2+ 1}}$$
let $xyz= tan(\theta)$. Then $yzdx= sec^2(\theta)$, $\sqrt{x^2y^2z^2+ 1}= \sqrt{tan^2(\theta)+ 1}= sec(\theta)$ and the integral becomes
$$\int sec(\theta)d\theta)= \int \frac{d\theta}{cos(\theta)}$$
$$= \int \frac{cos(\theta)d\theta)}{cos^2(\theta)}= \int \frac{cos(\theta)d\theta}{1- sin^2(\theta)}$$
and now you can let $u= sin(\theta)$. Don't forget that the "constant of integration" may be a function of y and z.

So I get $\displaystyle \phi (x,y,z) = \int \frac{1}{1-u^2}\;du = \text{sinh}^{-1} (xyz) + f(y,z)$

Hence $\displaystyle \frac{\partial}{\partial y} \bigg\{ \text{sinh}^{-1} (xyz)\bigg\} + \frac{\partial f(y,z)}{\partial y} = \frac{xz}{\sqrt{x^2y^2z^2 + 1}} + \frac{2yz^2}{y^2z^2 + 1}$

$\displaystyle \frac{xz}{\sqrt{x^2y^2z^2 + 1}} + \frac{\partial f(y,z)}{\partial y} = \frac{xz}{\sqrt{x^2y^2z^2 + 1}} + \frac{2yz^2}{y^2z^2 + 1}$

$\displaystyle \frac{\partial f(y,z)}{\partial y} = \frac{2yz^2}{y^2z^2 + 1}$

$\displaystyle f(y,z) = \int \frac{2yz^2}{y^2z^2 + 1}\;dy = \ln (y^2z^2 + 1) + g(z)$

So $\phi (x,y,z) = \text{sinh}^{-1} (xyz) + \ln (y^2z^2 + 1) + g(z)$

$\displaystyle \frac{\partial}{\partial z} \bigg\{ \text{sinh}^{-1} (xyz) + \ln (y^2z^2 + 1) + g(z) \bigg \} = \frac{xy}{\sqrt{x^2y^2z^2 + 1}} + \frac{2y^2z}{y^2z^2 + 1} + \frac{1}{z^2 +1}$

$\displaystyle \frac{xy}{\sqrt{x^2y^2z^2 + 1}} + \frac{2y^2z}{y^2z^2 + 1} + g'(z) = \frac{xy}{\sqrt{x^2y^2z^2 + 1}} + \frac{2y^2z}{y^2z^2 + 1} + \frac{1}{z^2 +1}$

$\displaystyle g'(z) = \frac{1}{z^2 +1}$

$\displaystyle g(z) = \int \frac{1}{z^2 +1}\;dz = \tan^{-1} (z) + c$

So $\phi (x,y,z) = \text{sinh}^{-1} (xyz) + \ln (y^2 z^2 + 1) + \tan ^{-1} (z) + c\;\;\;\;\;\text{(}c\;\text{constant)}$

 

[Dick]

Right. That wasn't so hard, was it?