Magnetic Force On this disk !!

We go at a distance y from the center of the disc in the y-direction and take an elemental thin rod of width dy.Its length will be 2 $\sqrt{R^2-y^2}$ . Therefore its area is 2 $\sqrt{R^2-y^2}$ dy and the charge on it will be 2 $\sqrt{R^2-y^2}$ $\sigma$ dy. Now the magnetic force acting on it will be dF=QvBsin $\alpha$ . Looking at the figure we can say that $\alpha$ = $\pi$ \2 - $\theta$ - $\beta$ .Thus sin $\alpha$ =cos( $\theta$ + $\beta$ ).Solving this we get sin $\alpha$ = $\frac{Ly}{\sqrt{(x^2+y^2)((L+x)^2+y^2)}}$ . Also, B= $\frac{\mu_{0}I}{2\pi\sqrt{(L+x)^2+y^2}}$ and v= $\omega\sqrt{x^2+y^2}$ .Putting these values in place of dF we finally get dF= $\frac{IL\mu_{0}\sigma\omega\sqrt{R^2-y^2}ydy}{\pi((l+x)^{2}+y^{2})}$ . Integrating this expression from y=0 to y=R we get $\frac{IL\mu_{0}\sigma\omega}{\pi}$ $\int\dfrac{y\sqrt{R^2-y^2}}{y^2+\left(x+L\right)^2}dy$ = $\frac{IL\mu_{0}\sigma\omega}{\pi}$ [ $\dfrac{\sqrt{x^2+2Lx+R^2+l^2}\ln\left(\frac{\left|\sqrt{R^2-y^2}-\sqrt{x^2+2Lx+R^2+L^2}\right|}{\sqrt{R^2-y^2}+\sqrt{x^2+2Lx+R^2+L^2}}\right)}{2}+\sqrt{R^2-y^2}$ ] Note that the expression given above is the indefinite integral.We would have to put the limits from y=0 to y=+R and then, we would get the final answer.