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The Helmholtz Theorem says that if

[tex]\nabla \circ \vec{F} = 0[/tex] and

[tex]\nabla \times \vec{F} = \vec{C}(\vec{r})[/tex] then

[tex]\vec{F} = \nabla \times \vec{A}[/tex] where

[tex]\vec{A} = \frac{1}{4\pi}\int\int\int\frac{\vec{C}(\vec{r}')}{\left|\vec{r} - \vec{r}'\right|}dx'dy'dz'[/tex]

The problem gave [tex]\vec{F_1}=x^2\hat{z}[/tex].

The divergence of [tex]\vec{F}_1[/tex] is zero, and the curl of [tex]\vec{F}_1[/tex] is [tex]-2x\hat{y}[/tex].

My problem is finding the vector potential, [tex]\vec{A}[/tex]. This looks like a monster to do since none of the integral tricks I know seem to work.

So any help on solving this would be very appreciated. Here is the integral:

[tex]A_y = \frac{1}{4\pi}\int\int\int \frac{2x'}{\left|\vec{r} - \vec{r}'\right|}dx'dy'dz'[/tex]

thanks in advance!

-LD

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# Homework Help: Int x' dV/|r-r'| ?

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