Rotating Square Loop

For a square loop of side $a$ in a magnetic field $mathbf{B}=B_0[1-(x^2/l^2)]*hat x$ rotating with angular velocity $omega$ please find $E(t)$ induced in loop.

In text it says that the plate is square so I guess l=a in the definition of B

$B=B_0 (1-x^2/l^2 )*x ̂$

Cylindrical coordinates

$x=r*cos(ωt)$  and $y=r*sin(⁡ωt)$

$dS=a*dr$

The flux is

$dϕ=B ⃗*dS ⃗=B_0 (1-(r^2  cos^2⁡ωt)/l^2 )*(a*dr)*cos ⁡ωt$

(last $cos ⁡ωt$ comes from the scalar product.)

$ϕ_{tot} (t)=B_0 a*cos ⁡ωt ∫_0^a (1-(r^2  cos^2⁡ωt)/l^2 )*dr$

$ϕ_{tot} (t)=B_0 a*cos⁡ ωt*(r-r^3/(3l^2 )*cos^2⁡ ωt)|_0^a$ 

$ϕ_{tot} (t)=B_0 a*cos⁡ ωt (a-a^3/(3l^2 )*cos^2⁡ ωt)$

$U=-(dϕ_{tot})/dt=$

$=B_0 a^2*sin ⁡ωt  (1- a^2/(3l^2 )*cos^2 ω⁡t)+B_0 a^2  cos⁡ ωt*((2 ωa^2)/(3l^2 )*sin⁡ ωt*cos⁡ ωt)$

$U(t)=B_0 a^2*sin⁡ ωt*[1-a^2/(3l^2 )*cos^2⁡ ωt+2ω*a^2/(3l^2 )*cos^2⁡ ωt]$

$U(t)= B_0 a^2*sin⁡ ωt[1-a^2/(3l^2 ) (2ω-1)*cos^2⁡ ωt]$

$B(x)$ is constant for $a/l$ (it does depend just on $l$).