Electric field in a semi-circle (non-uniformly charged)

[BitterX]

Homework Statement

A plastic semi-circle/arc with radius R has a non-uniformly distributed charge upon it.
The density of the charge/length is $\lambda = \lambda_0 sin\theta$
θ is 0 in the middle of the arc

find the electric field in the point where the radius come from

Homework Equations

$dE=\frac{kdQ}{R^2}$
$\phi = \int{Edr}$

The Attempt at a Solution

charge differential to angle differential is :
$dQ=\lambda R = \lambda_0 sin\theta d\theta$

for the field I separated for x and y
$E_x =\frac{k\lambda_0}{R} \int_{0}^{\pi}sin^2\theta d\theta=\frac{k\lambda_0}{R}=\frac{k\lambda_0\pi}{2R}$

the same way I get
$E_y= \frac{k\lambda_0}{R}\int_{0}^{\pi}sin\theta cos\theta d\theta=0$

this is counter-intuitive for me, I thought the electric field should be pointing in the y direction.
I guess I need to have my answers inverted, but would be glad to understand why, and maybe shed some light on my mistakes...edit: I think I've found my mistake... I should integrate once from 0 to $\frac{\pi}{2}$ and once from 0 to $-\frac{\pi}{2}$
and add what I get

 

[jbsteele]

from each one, so E_x =\frac{k\lambda_0}{R} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}sin^2\theta d\theta=\frac{k\lambda_0}{R}=\frac{k\lambda_0\pi}{R} and E_y= \frac{k\lambda_0}{R}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}sin\theta cos\theta d\theta=\frac{k\lambda_0\pi}{R} so the electric field is pointing in both directions with the same magnitude.