# Why is th $\hat{r}$ component zero in this integral?

I'm trying to evaluate the magnetic field by calculating the Coloumb integral $\overrightarrow{A}$, and later I will take: $$\overrightarrow{B}=\nabla \times \overrightarrow{A}$$

However, in the middle of everything, I get to (cylindrical coordinates):

$$\overrightarrow{A}=\frac{\mu_oI}{4\pi} \oint_{C} \frac{a\hat{\phi'}}{[r²+(z-b)^2-2ar\cos(\phi-\phi')]^{1/2}} d\phi'$$

I should show that the only component of this expression is $\hat{\phi}$, where:

$$\hat{r'}=\hat{x}\cos{\phi'}+\hat{y}\sin{\phi}$$ $$\hat{\phi'}=-\hat{x}\sin{\phi'}+\hat{y}\cos{\phi'}$$ $$\hat{r}=\hat{x}\cos{\phi}+\hat{y}\sin{\phi}$$ $$\hat{\phi}=-\hat{x}\sin{\phi}+\hat{y}\cos{\phi}$$

and that it's leading to only this integral:

$$\overrightarrow{A}=\frac{\mu_oI}{4\pi} \oint_{C} \frac{a\cos{\left(\phi-\phi'\right)}\hat{\phi}}{[r²+(z-b)^2-2ar\cos(\phi-\phi')]^{1/2}} d\phi'$$

But when I rewrite how $\hat{\phi'}$ relates to $\hat{\phi}$ and $\hat{r}$, it's not so obvious that the $\hat{r}$ component disappears. The writer refers to symmetry, but I can't still figure this out. So I basically have this integral instead because I can't show how the $\hat{r}$ component is zero:

$$\overrightarrow{A}=\frac{\mu_oI}{4\pi} \oint_{C} \frac{a\sin{\left(\phi-\phi'\right)}\hat{r}+a\cos{\left(\phi-\phi'\right)}\hat{\phi}}{[r²+(z-b)^2-2ar\cos(\phi-\phi')]^{1/2}} d\phi'$$ Any ideas? I've been stuck at this for a while, and I've tried to show it in Cartesian coordinates without no luck.

Hint: substitute $u=\cos{(\phi - \phi')}$ and integrate.
• Before starting to calculate; does that help to show the $\hat{r}$ component is zero? Nov 10 '13 at 16:54