Simple Electric Field Problem Solve the Electric Field distance z above a circular loop of radius r. The charge/length = $\lambda$
The arc-length is 2$\pi$r. So the smallest portion of the circle is 2$\pi r \delta \theta$ and charge is therefore 
\begin{align}
q&=2\pi r \delta \theta*\lambda
\\
R&=\sqrt{r^2+z^2}=  \text{constant}
\\
E&= \frac {1}{4 \pi \epsilon _o}\frac {q}{R^2}=\frac {1}{4 \pi \epsilon _o}\frac {2\pi r \delta \theta*\lambda}{\sqrt{r^2+z^2}}\end{align}
And we only need the z component.
$$E =\frac {1}{4 \pi \epsilon _o}\frac {2\pi r \delta \theta*\lambda}{\sqrt{r^2+z^2}}\sin(\theta)=\frac {1}{4 \pi \epsilon _o}\frac {2\pi r \delta \theta*\lambda}{\sqrt{r^2+z^2}} \frac {z}{\sqrt{z^2+r^2}}$$
and everything is constant except for $\delta \theta$
$$E=\frac {1}{4 \pi \epsilon _o}\frac {2\pi r \lambda}{\sqrt{r^2+z^2}} \frac {z}{\sqrt{z^2+r^2}}\int_0^{2\pi}{\delta \theta}$$
So I thought the correct answer must be:
$$E=\frac {1}{4 \pi \epsilon _o}\frac {2\pi r \lambda z}{[r^2+z^2]^{3/2}} \cdot 2\pi$$
But the correct answer does not multiply by 2$\pi$
Correct: $$E=\frac {1}{4 \pi \epsilon _o}\frac {2\pi r \lambda z}{[r^2+z^2]^{3/2}}$$
Why was I wrong? where did I slip up?
Thanks!

The length element should be $r d\theta$ not $2\pi r d\theta$. So the charge element is
$$dq=\lambda r d\theta$$
but not
$$dq=\lambda 2\pi r d\theta.$$
