Circulation and the flux in a field Problem:

Find the circulation and flux of the field $F=x^2i+y^2j$ around and across the closed semicircular path that consists of the semicircular arch $r_1(t)=(a\cos(t))i+(a\sin(t))j$, $0\le t\le\pi$, followed by the line segment $r_2(t)=ti$, $-a\le t\le a$.

Then I compute the followings:
$r_1(t)=(a\cos(t))i+(a\sin(t))j\implies r_1'(t)=(-a\sin(t))i+(a\cos(t))j$
$F(r_1(t))=(a^2\cos^2t)i+(a^2\sin^2t)j$
$F(r_1(t))\cdot r_1'(t)=-a^3\sin t\cos^2 t+a^3\sin^2t\cos t$
Circulation along $r_1=\int_{-a}^aa^3(\sin^2 t\cos t-\sin t\cos^2t)dt=\frac{2}{3}a^3\sin^3a$
Flux along $r_1=\int_0^{\pi}(a\cos t)(a\cos t)-(a\sin t)(-a\sin t)dt=a^2\pi$
$r_2(t)=ti\implies r_2'(t)=i$
$F(r_2(t))=t^2i$
$F(r_2(t))\cdot r_2'(t)=t^2$
Circulation along $r_2=\int_{-a}^at^2dt=\frac{2}{3}a^3$
Flux along $r_2=\int_{-a}^a(t(0)-(0)(1))dt=0$
Hence the total circulation$=\frac{2}{3}a^3\sin^3a+\frac{2}{3}a^3=\frac{2}{3}a^3(1+\sin^3a)$, and the total flux$=a^2\pi+0=a^2\pi$.
My questions are:

*

*Are my computations valid? (I mean the steps here, not the arithmetics)


*It seems that the calculation of the flux is independent of the field. Is that normal?

The total circulation should be zero because $\mathbf{F}$ is a conservative field,
$\nabla U=\mathbf{F}$ with $U(x,y)=\frac{x^3}{3}+\frac{y^3}{3}$, and the path is closed.
In your work, the limits of the integral for $r_1$ should be $0$ and $\pi$ (not $-a$ and $a$),
$$\int_{0}^\pi a^3(\sin^2 t\cos t-\sin t\cos^2t)dt=-\frac{2a^3}{3}.$$
By the divergence theorem, the total flux is
$$\iint_{D} (2x+2y)dxdy=0+2\iint_{D} ydxdy=2\int_{r=0}^a\int_{\theta=0}^{\pi}r^2\sin(\theta)dr d\theta=\frac{4a^3}{3}$$
where $D$ is the semidisc $\{(x,y): x^2+y^2\leq a^2, y\geq 0\}.$
In your work, the flux accross $r_1$ should be (you missed the squares),
$$\int_0^{\pi}(a\cos t)^2(a\cos t)+(a\sin t)^2(a\sin t)dt=\frac{4a^3}{3}.$$
