Integrating Half-Loop Radius R: Confused?

[Thomassino]

I can't seem to integrate this correctly. I only need the half loop integration, as i have the correct integration for the infinite lines.

Homework Statement

We start with the top half of a half-loop of radius R, centered at the origin, with infinite line segments traveling in the +/- x directions.

B=\frac{a}{r^3} (-\vec{k} )
and
r=\sqrt{x^2+y^2}

Homework Equations

dF=I (dL x B)

The Attempt at a Solution

My attempt at a solution ends up with me getting confused over how to determine the resultant vector by cross product.

I(dLxB)
<br /> =I Rd\theta (-cos\theta(-\vec{\i})-sin\theta(-\vec{\j})) \times \frac {a}{r^3} (-\vec{k} )<br />
now, d\theta = \pi, and
I integrate \int_{-\pi /2}^{\pi/2}(-cos\theta(-\vec{i})-sin\theta(-\vec{j})

which results in

\int_{-\pi /2}^{\pi/2}(-cos\theta(-\vec{i})-\int_{-\pi /2}^{\pi/2}(sin\theta(-\vec{j})

\left[(-sin\theta(-\vec{i}) \right]_{-\pi /2}^{\pi/2}+\left[(cos\theta(-\vec{j})\right]_{-\pi /2}^{\pi/2}
yielding 2+0=2

I then integrate 2\pi IRa \int_{-R}^{R} \frac{dx}{x^3}

but my final result ends up being

=2IR\pi a [\frac{-1}{2R^2}-\frac{-1}{2R^2}] which is zero and I need it to be one.

Also, I believe the correct answer is
\frac{2Ia}{R^2}(-\vec{j}) which does not have the RΠ factor which I end up with...
Can someone please help me comprehend this better, as it has been many years since I have had calculus and haven't been able to find any examples of this type of integration.

Thank you very much!

 

[Thomassino]

here are some pics

here are the pictures of my problem and, seemingly, my result.